© 2014 Pearson Education, Inc. Sherril Soman Grand Valley State University Lecture Presentation Chapter 15 Acids and Bases.

© 2014 Pearson Education, Inc. Stomach Acid and Heartburn The cells that line your stomach produce hydrochloric acid. –To kill unwanted bacteria –To help break down food –To activate enzymes that break down food If the stomach acid backs up into your esophagus, it irritates those tissues, resulting in heartburn. –Acid reflux

© 2014 Pearson Education, Inc. Curing Heartburn Mild cases of heartburn can be cured by neutralizing the acid in the esophagus. –Swallowing saliva, which contains bicarbonate ion –Taking antacids that contain hydroxide ions and/or carbonate ions

© 2014 Pearson Education, Inc. GERD Chronic heartburn is a problem for some people. GERD (gastroesophageal reflux disease) is chronic leaking of stomach acid into the esophagus. In people with GERD, the muscles separating the stomach from the esophagus do not close tightly, allowing stomach acid to leak into the esophagus. Physicians diagnose GERD by attaching a pH sensor to the esophagus to measure the acidity levels of the fluids over time.

© 2014 Pearson Education, Inc. Properties of Bases Taste bitter alkaloids = plant product that is alkaline  often poisonous Feel slippery Ability to turn red litmus paper blue Ability to neutralize acids

© 2014 Pearson Education, Inc. Indicators Indicators are chemicals that change color depending on the solution’s acidity or basicity. Many vegetable dyes are indicators. –Anthocyanins Litmus –From Spanish moss –Red in acid, blue in base Phenolphthalein –Found in laxatives –Red in base, colorless in acid

© 2014 Pearson Education, Inc. Hydronium Ion The H + ions produced by the acid are so reactive they cannot exist in water. –H + ions are protons! Instead, they react with water molecules to produce complex ions, mainly hydronium ion, H 3 O +. H + + H 2 O  H 3 O + –There are also minor amounts of H + with multiple water molecules, H(H 2 O) n +.

© 2014 Pearson Education, Inc. Arrhenius Acid–Base Reactions The H + from the acid combines with the OH − from the base to make a molecule of H 2 O. –It is often helpful to think of H 2 O as H—OH. The cation from the base combines with the anion from the acid to make a salt. acid + base → salt + water HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l)

© 2014 Pearson Education, Inc. Problems with Arrhenius Theory It does not explain why molecular substances, such as NH 3, dissolve in water to form basic solutions, even though they do not contain OH – ions. It does not explain how some ionic compounds, such as Na 2 CO 3 or Na 2 O, dissolve in water to form basic solutions, even though they do not contain OH – ions. It does not explain why molecular substances, such as CO 2, dissolve in water to form acidic solutions, even though they do not contain H + ions. It does not explain acid–base reactions that take place outside aqueous solution.

© 2014 Pearson Education, Inc. Brønsted–Lowry Acid–Base Theory It defines acids and bases based on what happens in a reaction. Any reaction involving H + (proton) that transfers from one molecule to another is an acid–base reaction, regardless of whether it occurs in aqueous solution or if there is OH − present. All reactions that fit the Arrhenius definition also fit the Brønsted–Lowry definition.

© 2014 Pearson Education, Inc. Brønsted–Lowry Theory The acid is an H + donor. The base is an H + acceptor. –Base structure must contain an atom with an unshared pair of electrons. In a Brønsted–Lowry acid–base reaction, the acid molecule donates an H + to the base molecule. H–A + :B  :A – + H–B +

© 2014 Pearson Education, Inc. Brønsted–Lowry Acids Brønsted–Lowry acids are H + donors. –Any material that has H can potentially be a Brønsted–Lowry acid. –Because of the molecular structure, often one H in the molecule is easier to transfer than others. When HCl dissolves in water, the HCl is the acid because HCl transfers an H + to H 2 O, forming H 3 O + ions. –Water acts as base, accepting H +. HCl(aq) + H 2 O(l) → Cl – (aq) + H 3 O + (aq) acidbase

© 2014 Pearson Education, Inc. Brønsted–Lowry Bases Brønsted–Lowry bases are H + acceptors. –Any material that has atoms with lone pairs can potentially be a Brønsted–Lowry base. –Because of the molecular structure, often one atom in the molecule is more willing to accept H + transfer than others. When NH 3 dissolves in water, the NH 3 (aq) is the base because NH 3 accepts an H + from H 2 O, forming OH – (aq). –Water acts as acid, donating H +. NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH – (aq) base acid

© 2014 Pearson Education, Inc. Amphoteric Substances Amphoteric substances can act as either an acid or a base because they have both a transferable H and an atom with lone pair electrons. Water acts as base, accepting H + from HCl. HCl(aq) + H 2 O(l) → Cl – (aq) + H 3 O + (aq) Water acts as acid, donating H + to NH 3. NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH – (aq)

© 2014 Pearson Education, Inc. Brønsted–Lowry Acid–Base Reactions One of the advantages of Brønsted–Lowry theory is that it illustrates reversible reactions to be as follows: H–A + :B  :A – + H–B + The original base has an extra H + after the reaction, so it will act as an acid in the reverse process. And the original acid has a lone pair of electrons after the reaction, so it will act as a base in the reverse process: :A – + H–B +  H–A + :B

© 2014 Pearson Education, Inc. Conjugate Acid–Base Pairs In a Brønsted-–Lowry acid–base reaction, –the original base becomes an acid in the reverse reaction. –the original acid becomes a base in the reverse process. Each reactant and the product it becomes is called a conjugate pair.

© 2014 Pearson Education, Inc. Arrow Conventions Chemists commonly use two kinds of arrows in reactions to indicate the degree of completion of the reactions. A single arrow indicates that all the reactant molecules are converted to product molecules at the end. A double arrow indicates the reaction stops when only some of the reactant molecules have been converted into products. –  in these notes

© 2014 Pearson Education, Inc. Strong or Weak A strong acid is a strong electrolyte. –Practically all the acid molecules ionize. A strong base is a strong electrolyte. –Practically all the base molecules form OH – ions, either through dissociation or reaction with water. A weak acid is a weak electrolyte. –Only a small percentage of the molecules ionize, . A weak base is a weak electrolyte –only a small percentage of the base molecules form OH – ions, either through dissociation or reaction with water, .

© 2014 Pearson Education, Inc. Weak Acids Weak acids donate a small fraction of their H’s. Most of the weak acid molecules do not donate H to water. Much less than 1% ionized in water [H 3 O + ] << [weak acid]

© 2014 Pearson Education, Inc. Strengths of Acids and Bases Commonly, acid or base strength is measured by determining the equilibrium constant of a substance’s reaction with water. HAcid + H 2 O  Acid − + H 3 O + Base: + H 2 O  HBase + + OH − The farther the equilibrium position lies toward the products, the stronger the acid or base. The position of equilibrium depends on the strength of attraction between the base form and the H +. –Stronger attraction means stronger base or weaker acid.

© 2014 Pearson Education, Inc. General Trends in Acidity The stronger an acid is at donating H, the weaker the conjugate base is at accepting H. Higher oxidation number = stronger oxyacid –H 2 SO 4 > H 2 SO 3 ; HNO 3 > HNO 2 Cation stronger acid than neutral molecule; neutral stronger acid than anion –H 3 O + > H 2 O > OH − ; NH 4 + > NH 3 > NH 2 − –Trend in base strength opposite

© 2014 Pearson Education, Inc. Autoionization of Water Water is amphoteric; it can act either as an acid or a base. –Therefore, there must be a few ions present. About 2 out of every 1 billion water molecules form ions through a process called autoionization. H 2 O  H + + OH – H 2 O + H 2 O  H 3 O + + OH – All aqueous solutions contain both H 3 O + and OH –. –The concentration of H 3 O + and OH – are equal in water. –[H 3 O + ] = [OH – ] = 10 −7 M at 25 °C

© 2014 Pearson Education, Inc. Ion Product of Water The product of the H 3 O + and OH – concentrations is always the same number. The number is called the ion product of water and has the symbol K w. –Also know as the dissociation constant of water [H 3 O + ] × [OH – ] = K w = 1.00 × 10 −14 at 25 °C –If you measure one of the concentrations, you can calculate the other. As [H 3 O + ] increases the [OH – ] must decrease so the product stays constant. –Inversely proportional

© 2014 Pearson Education, Inc. Acid Ionization Constant, K a Acid strength is measured by the size of the equilibrium constant when it reacts with H 2 O. The equilibrium constant for this reaction is called the acid ionization constant, K a. –larger K a = stronger acid

© 2014 Pearson Education, Inc. Acidic and Basic Solutions All aqueous solutions contain both H 3 O + and OH – ions. Neutral solutions have equal [H 3 O + ] and [OH – ]. –[H 3 O + ] = [OH – ] = 1.00 × 10 −7 Acidic solutions have a larger [H 3 O + ] than [OH – ]. –[H 3 O + ] > 1.00 × 10 −7 ; [OH – ] < 1.00 × 10 −7 Basic solutions have a larger [OH – ] than [H 3 O + ]. –[H 3 O + ] 1.00 × 10 −7

© 2014 Pearson Education, Inc. Measuring Acidity: pH The acidity or basicity of a solution is often expressed as pH. pH = −log[H 3 O + ] Exponent on 10 with a positive sign pH water = −log[10 −7 ] = 7 Need to know the [H 3 O + ] concentration to find pH pH 7 is basic. pH = 7 is neutral. [H 3 O + ] = 10 −pH

© 2014 Pearson Education, Inc. Sig., Figs., and Logs When you take the log of a number written in scientific notation, the digits before the decimal point come from the exponent on 10, and the digits after the decimal point come from the decimal part of the number. log(2.0 x 10 6 ) = log(10 6 ) + log(2.0) = 6 + 0.30303… = 6.30303... Because the part of the scientific notation number that determines the significant figures is the decimal part, the sig. figs. are the digits after the decimal point in the log. log(2.0 × 10 6 ) = 6.30

© 2014 Pearson Education, Inc. What Does the pH Number Imply? The lower the pH, the more acidic the solution; the higher the pH, the more basic the solution. 1 pH unit corresponds to a factor of 10 difference in acidity. Normal range of pH is 0 to 14. pH 0 is [H 3 O + ] = 1 M; pH 14 is [OH – ] = 1 M. pH can be negative (very acidic) or larger than 14 (very alkaline).

© 2014 Pearson Education, Inc. pOH Another way of expressing the acidity/basicity of a solution is pOH. pOH = −log[OH  ], [OH  ] = 10 −pOH pOH water = −log[10 −7 ] = 7 Need to know the [OH  ] concentration to find pOH pOH 7 is acidic; pOH = 7 is neutral. pH + pOH = 14.0

© 2014 Pearson Education, Inc. pKpK A way of expressing the strength of an acid or base is pK. pK a = −log(K a ), K a = 10 −pKa pK b = −log(K b ), K b = 10 −pKb The stronger the acid, the smaller the pK a. –Larger K a = smaller pK a –Because it is the –log The stronger the base, the smaller the pK b. –Larger K b = smaller pK b

© 2014 Pearson Education, Inc. [H 3 O + ] and [OH − ] in a Strong Acid or Strong Base Solution There are two sources of H 3 O + in an aqueous solution of a strong acid—the acid and the water. There are two sources of OH − in an aqueous solution of a strong acid—the base and the water. For a strong acid or base, the contribution of the water to the total [H 3 O + ] or [OH − ] is negligible. –The [H 3 O + ] acid shifts the K w equilibrium so far that [H 3 O + ] water is too small to be significant. Except in very dilute solutions, generally < 1 × 10 −4 M

© 2014 Pearson Education, Inc. Finding pH of a Strong Acid or Strong Base Solution For a monoprotic strong acid [H 3 O + ] = [HAcid]. –For polyprotic acids, the other ionizations can generally be ignored. For H 2 SO 4, the second ionization cannot be ignored. –0.10 M HCl has [H 3 O + ] = 0.10 M and pH = 1.00 For a strong ionic base, [OH − ] = (number OH − ) × [Base]. –For molecular bases with multiple lone pairs available, only one lone pair accepts an H; the other reactions can generally be ignored. –0.10 M Ca(OH) 2 has [OH − ] = 0.20 M and pH = 13.30.

© 2014 Pearson Education, Inc. Finding the pH of a Weak Acid There are also two sources of H 3 O + in an aqueous solution of a weak acid—the acid and the water. However, finding the [H 3 O + ] is complicated by the fact that the acid only undergoes partial ionization. Calculating the [H 3 O + ] requires solving an equilibrium problem for the reaction that defines the acidity of the acid. HAcid + H 2 O  Acid  + H 3 O +

© 2014 Pearson Education, Inc. Percent Ionization Another way to measure the strength of an acid is to determine the percentage of acid molecules that ionize when dissolved in water; this is called the percent ionization. –The higher the percent ionization, the stronger the acid. Because [ionized acid] equil = [H 3 O + ] equil

© 2014 Pearson Education, Inc. Relationship between [H 3 O + ] equilibrium and [HA] initial Increasing the initial concentration of acid results in increased [H 3 O + ] at equilibrium. Increasing the initial concentration of acid results in decreased percent ionization. This means that the increase in [H 3 O + ] concentration is slower than the increase in acid concentration.

© 2014 Pearson Education, Inc. Why Doesn’t the Increase in H 3 O + Keep Up with the Increase in HA? The reaction for ionization of a weak acid is as follows: HA (aq) + H 2 O (l)  A − (aq) + H 3 O + (aq) According to Le Châtelier’s principle, if we reduce the concentrations of all the (aq) components, the equilibrium should shift to the right to increase the total number of dissolved particles. –We can reduce the (aq) concentrations by using a more dilute initial acid concentration. The result will be a larger [H 3 O + ] in the dilute solution compared to the initial acid concentration. This will result in a larger percent ionization.

© 2014 Pearson Education, Inc. Finding the pH of Mixtures of Acids Generally, you can ignore the contribution of the weaker acid to the [H 3 O + ] equil. For a mixture of a strong acid with a weak acid, the complete ionization of the strong acid provides more than enough [H 3 O + ] to shift the weak acid equilibrium to the left so far that the weak acid’s added [H 3 O + ] is negligible. For mixtures of weak acids, you generally only need to consider the stronger for the same reasons, as long as one is significantly stronger than the other and their concentrations are similar.

© 2014 Pearson Education, Inc. Strong Bases The stronger the base, the more willing it is to accept H. Use water as the standard acid. For ionic bases, practically all units are dissociated into OH – or accept H’s. Strong electrolyte Multi-OH strong bases completely dissociated

© 2014 Pearson Education, Inc. Weak Bases In weak bases, only a small fraction of molecules accept H’s. Weak electrolyte Most of the weak base molecules do not take H from water. Much less than 1% ionization in water [HO – ] << [weak base] Finding the pH of a weak base solution is similar to finding the pH of a weak acid.

© 2014 Pearson Education, Inc. Base Ionization Constant, K b Base strength is measured by the size of the equilibrium constant when it reacts with H 2 O :Base + H 2 O  OH − + H:Base + The equilibrium constant is called the base ionization constant, K b. –Larger K b = stronger base

© 2014 Pearson Education, Inc. Acid–Base Properties of Ions and Salts Salts are water-soluble ionic compounds. Salts that contain the cation of a strong base and an anion that is the conjugate base of a weak acid are basic. –NaHCO 3 solutions are basic. Na + is the cation of the strong base NaOH. HCO 3 − is the conjugate base of the weak acid H 2 CO 3. Salts that contain cations that are the conjugate acid of a weak base and an anion of a strong acid are acidic. –NH 4 Cl solutions are acidic. NH 4 + is the conjugate acid of the weak base NH 3. Cl − is the anion of the strong acid HCl.

© 2014 Pearson Education, Inc. Anions as Weak Bases Every anion can be thought of as the conjugate base of an acid. Therefore, every anion can potentially be a base. –A − (aq) + H 2 O(l)  HA(aq) + OH − (aq) The stronger the acid, the weaker the conjugate base. An anion that is the conjugate base of a strong acid is pH neutral. Cl − (aq) + H 2 O(l)  HCl(aq) + OH − (aq) An anion that is the conjugate base of a weak acid is basic. F − (aq) + H 2 O(l)  HF(aq) + OH − (aq)

© 2014 Pearson Education, Inc. Relationship between K a of an Acid and K b of Its Conjugate Base Many reference books only give tables of K a values because K b values can be found from them. When you add equations, you multiply the K’s.

© 2014 Pearson Education, Inc. Cations as Weak Acids Some cations can be thought of as the conjugate acid of a weak base. –Others are the counterions of a strong base. Therefore, some cations can potentially be acidic. –MH + (aq) + H 2 O(l)  MOH(aq) + H 3 O + (aq) The stronger the base, the weaker the conjugate acid. –A cation that is the counterion of a strong base is pH neutral. –A cation that is the conjugate acid of a weak base is acidic. NH 4 + (aq) + H 2 O(l)  NH 3 (aq) + H 3 O + (aq) Because NH 3 is a weak base, the position of this equilibrium favors the right.

© 2014 Pearson Education, Inc. Metal Cations as Weak Acids Cations of small, highly charged metals are weakly acidic. –Alkali metal cations and alkali earth metal cations are pH neutral. –Cations are hydrated. Al(H 2 O) 6 3+ (aq) + H 2 O(l)  Al(H 2 O) 5 (OH) 2+ (aq) + H 3 O + (aq)

© 2014 Pearson Education, Inc. Classifying Salt Solutions as Acidic, Basic, or Neutral If the salt cation is the counterion of a strong base and the anion is the conjugate base of a strong acid, it will form a neutral solution. –NaCl Ca(NO 3 ) 2 KBr If the salt cation is the counterion of a strong base and the anion is the conjugate base of a weak acid, it will form a basic solution. –NaF Ca(C 2 H 3 O 2 ) 2 KNO 2

© 2014 Pearson Education, Inc. Classifying Salt Solutions as Acidic, Basic, or Neutral If the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a strong acid, it will form an acidic solution. –NH 4 Cl If the salt cation is a highly charged metal ion and the anion is the conjugate base of a strong acid, it will form an acidic solution. –Al(NO 3 ) 3

© 2014 Pearson Education, Inc. Classifying Salt Solutions as Acidic, Basic, or Neutral If the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a weak acid, the pH of the solution depends on the relative strengths of the acid and base. –NH 4 F because HF is a stronger acid than NH 4 +, K a of NH 4 + is larger than K b of the F − ; therefore, the solution will be acidic.

© 2014 Pearson Education, Inc. Ionization in Polyprotic Acids Because polyprotic acids ionize in steps, each H has a separate K a. K a1 > K a2 > K a3 Generally, the difference in K a values is great enough so that the second ionization does not happen to a large enough extent to affect the pH. –Most pH problems just do first ionization. –Except H 2 SO 4  uses [H 2 SO 4 ] as the [H 3 O + ] for the second ionization. [A 2− ] = K a2 as long as the second ionization is negligible.

Ionization in H 2 SO 4 The ionization constants for H 2 SO 4 are as follows: H 2 SO 4 + H 2 O  HSO 4  + H 3 O + strong HSO 4  + H 2 O  SO 4 2  + H 3 O + K a2 = 1.2 × 10 −2 For most sulfuric acid solutions, the second ionization is significant and must be accounted for. Because the first ionization is complete, use the given [H 2 SO 4 ] = [HSO 4 − ] initial = [H 3 O + ] initial.

© 2014 Pearson Education, Inc. Strengths of Binary Acids The more  + H─X  − polarized the bond, the more acidic the bond. The stronger the H─X bond, the weaker the acid. Binary acid strength increases to the right across a period. –Acidity: H─C < H─N < H─O < H─F Binary acid strength increases down the column. –Acidity: H─F < H─Cl < H─Br < H─I

© 2014 Pearson Education, Inc. Strengths of Oxyacids, H–O–Y The more electronegative the Y atom, the stronger the oxyacid. –HClO > HIO –Acidity of oxyacids decreases down a group. Same trend as binary acids –Helps weaken the H–O bond. The larger the oxidation number of the central atom, the stronger the oxyacid. –H 2 CO 3 > H 3 BO 3 –Acidity of oxyacids increases to the right across a period. Opposite trend of binary acids The more oxygens attached to Y, the stronger the oxyacid. –Further weakens and polarizes the H–O bond –HClO 3 > HClO 2

© 2014 Pearson Education, Inc. Lewis Acid–Base Theory Lewis acid–base theory focuses on transferring an electron pair. –Lone pair  bond –Bond  lone pair Does NOT require H atoms The electron donor is called the Lewis base. –Electron rich; therefore nucleophile The electron acceptor is called the Lewis acid. –Electron deficient; therefore electrophile

© 2014 Pearson Education, Inc. Lewis Bases The Lewis base has electrons it is willing to give away to or share with another atom. The Lewis base must have a lone pair of electrons on it that it can donate. Anions are better Lewis bases than neutral atoms or molecules. –N: < N: − Generally, the more electronegative an atom, the less willing it is to be a Lewis base. –O: < S:

© 2014 Pearson Education, Inc. Lewis Acids They are electron deficient, either from being attached to electronegative atom(s) or not having an octet. They must have an empty orbital willing to accept the electron pair. H + has empty 1s orbital. B in BF 3 has empty 2p orbital and an incomplete octet. Many small, highly charged metal cations have empty orbitals they can use to accept electrons. Atoms that are attached to highly electronegative atoms and have multiple bonds can be Lewis acids.

© 2014 Pearson Education, Inc. Lewis Acid–Base Reactions The base donates a pair of electrons to the acid. It generally results in a covalent bond forming H 3 N: + BF 3  H 3 N─BF 3 The product that forms is called an adduct. Arrhenius and Brønsted–Lowry acid–base reactions are also Lewis.

© 2014 Pearson Education, Inc. Solution a. Because H 2 SO 4 donates a proton to H 2 O in this reaction, it is the acid (proton donor). After H 2 SO 4 donates the proton, it becomes HSO 4 −, the conjugate base. Because H 2 O accepts a proton, it is the base (proton acceptor). After H 2 O accepts the proton, it becomes H 3 O +, the conjugate acid. H 2 SO 4 (aq) + H 2 O(l) HSO 4 − (aq) + H 3 O + (aq) In each reaction, identify the Brønsted–Lowry acid, the Brønsted–Lowry base, the conjugate acid, and the conjugate base. a. b. Example 15.1Identifying Brønsted–Lowry Acids and Bases and Their Conjugates

© 2014 Pearson Education, Inc. Solution b. Because H 2 O donates a proton to HCO 3 − in this reaction, it is the acid (proton donor). After H 2 O donates the proton, it becomes OH −, the conjugate base. Because HCO 3 − accepts a proton, it is the base (proton acceptor). After HCO 3 − accepts the proton, it becomes H 2 CO 3, the conjugate acid. HCO 3 − (aq) + H 2 O(l) H 2 CO 3 (aq) + OH − (aq) For Practice 15.1 In each reaction, identify the Brønsted–Lowry acid, the Brønsted–Lowry base, the conjugate acid, and the conjugate base. a. b. Continued Example 15.1Identifying Brønsted–Lowry Acids and Bases and Their Conjugates

© 2014 Pearson Education, Inc. Solution a. To find [OH − ] use the ion product constant. Substitute the given value for [H 3 O + ] and solve the equation for [OH − ]. Since [H 3 O + ] > [OH − ], the solution is acidic. b. Substitute the given value for [H 3 O + ] and solve the acid ionization equation for [OH − ]. Since [H 3 O + ] < [OH − ], the solution is basic. Calculate [OH − ] at 25 °C for each solution and determine if the solution is acidic, basic, or neutral. a. [H 3 O + ] = 7.5 × 10 −5 M b. [H 3 O + ] = 1.5 × 10 −9 M c. [H 3 O + ] = 1.0 × 10 −7 M Example 15.2Using K w in Calculations

© 2014 Pearson Education, Inc. Solution c. Substitute the given value for [H 3 O + ] and solve the acid ionization equation for [OH − ]. Since [H 3 O + ] = 1.0 × 10 −7 and [OH − ] = 1.0 × 10 −7, the solution is neutral. For Practice 15.2 Calculate [H 3 O + ] at 25 °C for each solution and determine if the solution is acidic, basic, or neutral. a. [OH − ] = 1.5 × 10 −2 M b. [OH − ] = 1.0 × 10 −7 M c. [OH − ] = 8.2 × 10 −10 M Continued Example 15.2Using K w in Calculations

© 2014 Pearson Education, Inc. Solution a. To calculate pH, substitute the given [H 3 O + ] into the pH equation. Since pH < 7, this solution is acidic. Calculate the pH of each solution at 25 °C and indicate whether the solution is acidic or basic. a. [H 3 O + ] = 1.8 × 10 −4 M b. [OH − ] = 1.3 × 10 −2 M Example 15.3Calculating pH from [H 3 O + ] or [OH − ]

© 2014 Pearson Education, Inc. For Practice 15.3 Calculate the pH of each solution and indicate whether the solution is acidic or basic. a. [H 3 O + ] = 9.5 × 10 −9 M b. [OH − ] = 7.1 × 10 −3 M Continued Example 15.3Calculating pH from [H 3 O + ] or [OH − ] b. First use K w to find [H 3 O + ] from [OH − ]. Then substitute [H 3 O + ] into the pH expression to find pH. Since pH > 7, this solution is basic.

© 2014 Pearson Education, Inc. Solution To find the [H 3 O + ] from pH, start with the equation that defines pH. Substitute the given value of pH and then solve for [H 3 O + ]. Since the given pH value was reported to two decimal places, the [H 3 O + ] is written to two significant figures. (Remember that 10 log x = x (see Appendix I ). Some calculators use an inv log key to represent this function.) For Practice 15.4 Calculate the [H 3 O + ] for a solution with a pH of 8.37. Calculate the [H 3 O + ] for a solution with a pH of 4.80. Example 15.4Calculating [H 3 O + ] from pH

© 2014 Pearson Education, Inc. Procedure For… Finding the pH (or [H 3 O + ]) of a Weak Acid Solution To solve these types of problems, follow the outlined procedure. Solution Step 1Write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE table showing the given concentration of the weak acid as its initial concentration. Leave room in the table for the changes in concentrations and for the equilibrium concentrations. (Note that the [H 3 O + ] is listed as approximately zero because the autoionization of water produces a negligibly small amount of H 3 O + ). Find the [H 3 O + ] of a 0.100 M HCN solution. Example 15.5Finding the [H 3 O + ] of a Weak Acid Solution

© 2014 Pearson Education, Inc. Step 2Represent the change in the concentration of H 3 O + with the variable x. Define the changes in the concentrations of the other reactants and products in terms of x. Always keep in mind the stoichiometry of the reaction. Step 3Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x. Continued Example 15.5Finding the [H 3 O + ] of a Weak Acid Solution

© 2014 Pearson Education, Inc. Step 4Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the acid ionization constant (K a ). In many cases, you can make the approximation that x is small (as discussed in Section 14.8). Substitute the value of the acid ionization constant (from Table 15.5) into the K a expression and solve for x. Confirm that the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in the approximation. The ratio should be less than 0.05 (or 5%). Therefore the approximation is valid. Continued Example 15.5Finding the [H 3 O + ] of a Weak Acid Solution

© 2014 Pearson Education, Inc. Step 5Determine the [H 3 O + ] from the calculated value of x and calculate the pH if necessary. [H 3 O + ] = 7.0 × 10 −6 M (pH was not asked for in this problem.) Step 6Check your answer by substituting the computed equilibrium values into the acid ionization expression. The calculated value of K a should match the given value of K a. Note that rounding errors and the x is small approximation could result in a difference in the least significant digit when comparing values of K a. Since the calculated value of K a matches the given value, the answer is valid. For Practice 15.5 Find the H 3 O + concentration of a 0.250 M hydrofluoric acid solution. Continued Example 15.5Finding the [H 3 O + ] of a Weak Acid Solution

© 2014 Pearson Education, Inc. Procedure For… Finding the pH (or [H 3 O + ]) of a Weak Acid Solution To solve these types of problems, follow the outlined procedure. Solution Step 1Write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE table showing the given concentration of the weak acid as its initial concentration. Leave room in the table for the changes in concentrations and for the equilibrium concentrations. (Note that the [H 3 O + ] is listed as approximately zero because the autoionization of water produces a negligibly small amount of H 3 O + ). Find the pH of a 0.200 M HNO 2 solution. Example 15.6Finding the pH of a Weak Acid Solution

© 2014 Pearson Education, Inc. Step 2Represent the change in the concentration of H 3 O + with the variable x. Define the changes in the concentrations of the other reactants and products in terms of x. Always keep in mind the stoichiometry of the reaction. Step 3Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x. Continued Example 15.6Finding the pH of a Weak Acid Solution

© 2014 Pearson Education, Inc. Step 4Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the acid ionization constant (K a ). In many cases, you can make the approximation that x is small (as discussed in Section 14.8). Substitute the value of the acid ionization constant (from Table 15.5) into the K a expression and solve for x. Confirm that the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in the approximation. The ratio should be less than 0.05 (or 5%). Therefore the approximation is valid (but barely so). Continued Example 15.6Finding the pH of a Weak Acid Solution

© 2014 Pearson Education, Inc. Step 5Determine the [H 3 O + ] from the calculated value of x and calculate the pH if necessary. Step 6Check your answer by substituting the computed equilibrium values into the acid ionization expression. The calculated value of K a should match the given value of K a. Note that rounding errors and the x is small approximation could result in a difference in the least significant digit when comparing values of K a. Since the calculated value of K a matches the given value, the answer is valid. For Practice 15.6 Find the pH of a 0.0150 M acetic acid solution. Continued Example 15.6Finding the pH of a Weak Acid Solution

© 2014 Pearson Education, Inc. Solution Step 1Write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE table showing the given concentration of the weak acid as its initial concentration. (Note that the H 3 O + concentration is listed as approximately zero. Although a little H 3 O + is present from the autoionization of water, this amount is negligibly small compared to the amount of HClO 2 or H 3 O + formed by the acid.) Find the pH of a 0.100 M HClO 2 solution. Example 15.7Finding the pH of a Weak Acid Solution in Cases Where the x is small Approximation Does Not Work

© 2014 Pearson Education, Inc. Step 2Represent the change in [H 3 O + ] with the variable x. Define the changes in the concentrations of the other reactants and products in terms of x. Step 3Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x. Continued Example 15.7Finding the pH of a Weak Acid Solution in Cases Where the x is small Approximation Does Not Work

© 2014 Pearson Education, Inc. Step 4Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the acid ionization constant (K a ). Make the x is small approximation and substitute the value of the acid ionization constant (from Table 15.5) into the K a expression. Solve for x. Continued Example 15.7Finding the pH of a Weak Acid Solution in Cases Where the x is small Approximation Does Not Work

© 2014 Pearson Education, Inc. Check to see if the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in the approximation. The ratio should be less than 0.05 (or 5%). Therefore, the x is small approximation is not valid. Continued Example 15.7Finding the pH of a Weak Acid Solution in Cases Where the x is small Approximation Does Not Work

© 2014 Pearson Education, Inc. Step 4aIf the x is small approximation is not valid, solve the quadratic equation explicitly or use the method of successive approximations to find x. In this case, we solve the quadratic equation. Since x represents the concentration of H 3 O +, and since concentrations cannot be negative, we reject the negative root. x = 0.028 Continued Example 15.7Finding the pH of a Weak Acid Solution in Cases Where the x is small Approximation Does Not Work

© 2014 Pearson Education, Inc. Step 5Determine the H 3 O + concentration from the calculated value of x and calculate the pH (if necessary). Step 6Check your answer by substituting the calculated equilibrium values into the acid ionization expression. The calculated value of K a should match the given value of K a. Note that rounding errors could result in a difference in the least significant digit when comparing values of K a. Since the calculated value of K a matches the given value, the answer is valid. For Practice 15.7 Find the pH of a 0.010 M HNO 2 solution. Continued Example 15.7Finding the pH of a Weak Acid Solution in Cases Where the x is small Approximation Does Not Work

© 2014 Pearson Education, Inc. Solution Use the given pH to find the equilibrium concentration of [H 3 O + ]. Then write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE table showing all known concentrations. A 0.100 M weak acid (HA) solution has a pH of 4.25. Find K a for the acid. Example 15.8Finding the Equilibrium Constant from pH

© 2014 Pearson Education, Inc. Use the equilibrium concentration of H 3 O + and the stoichiometry of the reaction to predict the changes and equilibrium concentration for all species. For most weak acids, the initial and equilibrium concentrations of the weak acid (HA) are effectively equal because the amount that ionizes is usually very small compared to the initial concentration. Substitute the equilibrium concentrations into the expression for K a and calculate its value. Continued Example 15.8Finding the Equilibrium Constant from pH

© 2014 Pearson Education, Inc. Solution To find the percent ionization, you must find the equilibrium concentration of H 3 O +. Follow the procedure in Example 15.5, shown in condensed form here. Therefore, [H 3 O + ] = 0.034 M. Find the percent ionization of a 2.5 M HNO 2 solution. Example 15.9Finding the Percent Ionization of a Weak Acid

© 2014 Pearson Education, Inc. Use the definition of percent ionization to calculate it. (Since the percent ionization is less than 5%, the x is small approximation is valid.) For Practice 15.9 Find the percent ionization of a 0.250 M HC 2 H 3 O 2 solution at 25 °C. Continued Example 15.9Finding the Percent Ionization of a Weak Acid

© 2014 Pearson Education, Inc. Solution The three possible sources of H 3 O + ions are HF, HClO, and H 2 O. Write the ionization equations for the three sources and their corresponding equilibrium constants. Since the equilibrium constant for the ionization of HF is about 12,000 times larger than that for the ionization of HClO, the contribution of HF to [H 3 O + ] is by far the greatest. You can therefore just calculate the [H 3 O + ] formed by HF and neglect the other two potential sources of H 3 O +. Write the balanced equation for the ionization of HF and use it as a guide to prepare an ICE table. Find the pH of a mixture that is 0.150 M in HF and 0.100 M in HClO. Example 15.10Mixtures of Weak Acids

© 2014 Pearson Education, Inc. Substitute the expressions for the equilibrium concentrations into the expression for the acid ionization constant (K a ). Since the equilibrium constant is small relative to the initial concentration of HF, you can make the x is small approximation. Substitute the value of the acid ionization constant (from Table 15.5) into the K a expression and solve for x. Confirm that the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in the approximation. The ratio should be less than 0.05 (or 5%). Therefore, the approximation is valid (though barely so). Continued Example 15.10Mixtures of Weak Acids

© 2014 Pearson Education, Inc. Determine the H 3 O + concentration from the calculated value of x and find the pH. For Practice 15.10 Find the ClO − concentration of the above mixture of HF and HClO. Continued Example 15.10Mixtures of Weak Acids

© 2014 Pearson Education, Inc. Solution a. Since KOH is a strong base, it completely dissociates into K + and OH − in solution. The concentration of OH − will therefore be the same as the given concentration of KOH. Use this concentration and K w to find [H 3 O + ]. Then substitute [H 3 O + ] into the pH expression to find the pH. What is the OH − concentration and pH in each solution? a. 0.225 M KOH b. 0.0015 M Sr(OH) 2 Example 15.11Finding the [OH − ] and pH of a Strong Base Solution

© 2014 Pearson Education, Inc. b. Since Sr(OH) 2 is a strong base, it completely dissociates into 1 mol of Sr 2+ and 2 mol of OH − in solution. The concentration of OH − will therefore be twice the given concentration of Sr(OH) 2. Use this concentration and K w to find [H 3 O + ]. Substitute [H 3 O + ] into the pH expression to find the pH. For Practice 15.11 Find the [OH − ] and pH of a 0.010 M Ba (OH) 2 solution. Continued Example 15.11Finding the [OH − ] and pH of a Strong Base Solution

© 2014 Pearson Education, Inc. Solution Step 1Write the balanced equation for the ionization of water by the base and use it as a guide to prepare an ICE table showing the given concentration of the weak base as its initial concentration. Leave room in the table for the changes in concentrations and for the equilibrium concentrations. (Note that you should list the OH − concentration as approximately zero. Although a little OH is present from the autoionization of water, this amount is negligibly small compared to the amount of OH − formed by the base.) Find the [OH − ] and pH of a 0.100 M NH 3 solution. Example 15.12Finding the [OH − ] and pH of a Weak Base Solution

© 2014 Pearson Education, Inc. Step 2Represent the change in the concentration of OH − with the variable x. Define the changes in the concentrations of the other reactants and products in terms of x. Step 3Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x. Continued Example 15.12Finding the [OH − ] and pH of a Weak Base Solution

© 2014 Pearson Education, Inc. Step 4Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the base ionization constant. In many cases, you can make the approximation that x is small (as discussed in Chapter 14). Substitute the value of the base ionization constant (from Table 15.8) into the K b expression and solve for x. Confirm that the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in the approximation. The ratio should be less than 0.05 (or 5%). Therefore, the approximation is valid. Continued Example 15.12Finding the [OH − ] and pH of a Weak Base Solution

© 2014 Pearson Education, Inc. Step 5Determine the OH − concentration from the calculated value of x. Use the expression for K w to find [H 3 O + ]. Substitute [H 3 O + ] into the pH equation to find pH. For Practice 15.12 Find the [OH − ] and pH of a 0.33 M methylamine solution. Continued Example 15.12Finding the [OH − ] and pH of a Weak Base Solution

© 2014 Pearson Education, Inc. Solution a. From Table 15.3, we can see that NO 3 − is the conjugate base of a strong acid (HNO 3 ) and is therefore pH-neutral. b. From Table 15.5 (or from its absence in Table 15.3), we know that NO 2 − is the conjugate base of a weak acid (HNO 2 ) and is therefore a weak base. Classify each anion as a weak base or pH-neutral: a. NO 3 − b. NO 2 − c. C 2 H 3 O 2 − Example 15.13Determining Whether an Anion Is Basic or pH-Neutral

© 2014 Pearson Education, Inc. c. From Table 15.5 (or from its absence in Table 15.3), we know that C2H 3 O 2 − is the conjugate base of a weak acid (HC 2 H 3 O 2 ) and is therefore a weak base. Continued Example 15.13Determining Whether an Anion Is Basic or pH-Neutral

© 2014 Pearson Education, Inc. Solution Step 1Since the Na + ion does not have any acidic or basic properties, you can ignore it. Write the balanced equation for the ionization of water by the basic anion and use it as a guide to prepare an ICE table showing the given concentration of the weak base as its initial concentration. Find the pH of a 0.100 M NaCHO 2 solution. The salt completely dissociates into Na + (aq) and CHO 2 − (aq), and the Na + ion has no acid or base properties. Example 15.14Determining the pH of a Solution Containing an Anion Acting as a Base

© 2014 Pearson Education, Inc. Step 2Represent the change in the concentration of OH − with the variable x. Define the changes in the concentrations of the other reactants and products in terms of x. Step 3Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x. Continued Example 15.14Determining the pH of a Solution Containing an Anion Acting as a Base

© 2014 Pearson Education, Inc. Step 4Find K b from K a (for the conjugate acid). Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for K b. In many cases, you can make the approximation that x is small. Substitute the value of K b into the K b expression and solve for x. Confirm that the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in the approximation. The ratio should be less than 0.05 (or 5%). Continued Example 15.14Determining the pH of a Solution Containing an Anion Acting as a Base

© 2014 Pearson Education, Inc. Step 5Determine the OH − concentration from the calculated value of x. Use the expression for K w to find [H 3 O + ]. Substitute [H 3 O + ] into the pH equation to find pH. For Practice 15.14 Find the pH of a 0.250 M NaC 2 H 3 O 2 solution. Continued Example 15.14Determining the pH of a Solution Containing an Anion Acting as a Base

© 2014 Pearson Education, Inc. Solution a. The C 5 H 5 NH + cation is the conjugate acid of a weak base and is therefore a weak acid. b. The Ca 2+ cation is the counterion of a strong base and is therefore pH-neutral (neither acidic nor basic). c. The Cr 3+ cation is a small, highly charged metal cation and is therefore a weak acid. For Practice 15.15 Classify each cation as a weak acid or pH-neutral. a. Li + b. CH 3 NH 3 + c. Fe 3+ Classify each cation as a weak acid or pH-neutral. a. C 5 H 5 NH + b. Ca 2+ c. Cr 3+ Example 15.15Determining Whether a Cation Is Acidic or pH-Neutral

© 2014 Pearson Education, Inc. Solution a. The Sr 2+ cation is the counterion of a strong base (Sr(OH) 2 ) and is pH-neutral. The Cl − anion is the conjugate base of a strong acid (HCl) and is pH-neutral as well. The SrCl 2 solution is therefore pH-neutral (neither acidic nor basic). b. The Al 3+ cation is a small, highly charged metal ion (that is not an alkali metal or an alkaline earth metal) and is a weak acid. The Br − anion is the conjugate base of a strong acid (HBr) and is pH-neutral. The AlBr 3 solution is therefore acidic. Determine if the solution formed by each salt is acidic, basic, or neutral. a. SrCl 2 b. AlBr 3 c. CH 3 NH 3 NO 3 d. NaCHO 2 e. NH 4 F Example 15.16Determining the Overall Acidity or Basicity of Salt Solutions

© 2014 Pearson Education, Inc. c. The CH 3 NH 3 + ion is the conjugate acid of a weak base (CH 3 NH 2 ) and is acidic. The NO 3 − anion is the conjugate base of a strong acid (HNO 3 ) and is pH-neutral. The CH 3 NH 3 NO 3 solution is therefore acidic. d. The Na + cation is the counterion of a strong base and is pH-neutral. The CHO 2 − anion is the conjugate base of a weak acid and is basic. The NaCHO 2 solution is therefore basic. Continued Example 15.16Determining the Overall Acidity or Basicity of Salt Solutions

© 2014 Pearson Education, Inc. e. The NH 4 + ion is the conjugate acid of a weak base (NH 3 ) and is acidic. The F − ion is the conjugate base of a weak acid and is basic. To determine the overall acidity or basicity of the solution, compare the values of K a for the acidic cation and K b for the basic anion. Obtain each value of K from the conjugate by using K a × K b = K w. Since K a is greater than K b, the solution is acidic. Continued Example 15.16Determining the Overall Acidity or Basicity of Salt Solutions

© 2014 Pearson Education, Inc. For Practice 15.16 Determine if the solution formed by each salt is acidic, basic, or neutral. a. NaHCO 3 b. CH 3 CH 2 NH 3 Cl c. KNO 3 d. Fe(NO 3 ) 3 Continued Example 15.16Determining the Overall Acidity or Basicity of Salt Solutions

© 2014 Pearson Education, Inc. Solution To find the pH, you must find the equilibrium concentration of H 3 O +. Treat the problem as a weak acid pH problem with a single ionizable proton. The second proton contributes a negligible amount to the concentration of H 3 O + and can be ignored. Follow the procedure from Example 15.6, shown in condensed form here. Use K a 1 for ascorbic acid from Table 15.10. Confirm that the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in the approximation. The ratio should be less than 0.05 (or 5%). Calculate the pH from H 3 O + concentration. Find the pH of a 0.100 M ascorbic acid (H 2 C 6 H 6 O 6 ) solution. Example 15.17Finding the pH of a Polyprotic Acid Solution

© 2014 Pearson Education, Inc. Solution Sulfuric acid is strong in its first ionization step and weak in its second. Begin by writing the equations for the two steps. As the concentration of an H 2 SO 4 solution becomes smaller, the second ionization step becomes more significant because the percent ionization increases (as discussed in Section 15.6). Therefore, for a concentration of 0.0100 M, you can’t neglect the H 3 O + contribution from the second step, as you can for other polyprotic acids. You must calculate the H 3 O + contributions from both steps. The [H 3 O + ] that results from the first ionization step is 0.0100 M (because the first step is strong). To determine the [H 3 O + ] formed by the second step, prepare an ICE table for the second step in which the initial concentration of H 3 O + is 0.0100 M. The initial concentration of HSO 4 − must also be 0.0100 M (due to the stoichiometry of the ionization reaction). Find the pH of a 0.0100 M sulfuric acid (H 2 SO 4 ) solution. Example 15.18Dilute H 2 SO 4 Solutions

© 2014 Pearson Education, Inc. Substitute the expressions for the equilibrium concentrations (from the table just shown) into the expression for K a 2. In this case, you cannot make the x is small approximation because the equilibrium constant (0.012) is not small relative to the initial concentration (0.0100). Substitute the value of K a 2 and multiply out the expression to arrive at the standard quadratic form. Continued Example 15.18Dilute H 2 SO 4 Solutions

© 2014 Pearson Education, Inc. Solve the quadratic equation using the quadratic formula. Since x represents a concentration, and since concentrations cannot be negative, we reject the negative root. x = 0.0045 Determine the H 3 O + concentration from the calculated value of x and calculate the pH. Notice that the second step produces almost half as much H 3 O + as the first step—an amount that must not be neglected. This will always be the case with dilute H 2 SO 4 solutions. Continued Example 15.18Dilute H 2 SO 4 Solutions

© 2014 Pearson Education, Inc. Solution To find the [C 6 H 6 O 6 2− ], use the concentrations of [HC 6 H 6 O 6 − ] and H 3 O + produced by the first ionization step (as calculated in Example 15.17) as the initial concentrations for the second step. Because of the 1:1 stoichiometry, [HC 6 H 6 O 6 − ] = [H 3 O + ]. Then solve an equilibrium problem for the second step similar to that of Example 15.6, shown in condensed form here. Use K a 2 for ascorbic acid from Table 15.10. Find the [C 6 H 6 O 6 2− ] of the 0.100 M ascorbic acid (H 2 C 6 H 6 O 6 ) solution in Example 15.17. Example 15.19Finding the Concentration of the Anions for a Weak Diprotic Acid Solution

© 2014 Pearson Education, Inc. Since x is much smaller than 2.8 × 10 −3, the x is small approximation is valid. Therefore, [C 6 H 6 O 6 2− ] = 1.6 × 10 −12 M. For Practice 15.19 Find the [CO 3 2− ] of the 0.050 M carbonic acid (H 2 CO 3 ) solution in For Practice 15.17. Continued Example 15.19Finding the Concentration of the Anions for a Weak Diprotic Acid Solution

© 2014 Pearson Education, Inc. U.S. Fuel Consumption Over 85% of the energy use in the United States comes from the combustion of fossil fuels. –Oil, natural gas, coal Combustion of fossil fuels produces CO 2. CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(g) Natural fossil fuels also contain small amounts of S that burn to produce SO 2 (g). S(s) + O 2 (g) → SO 2 (g) The high temperatures of combustion allow N 2 (g) in the air to combine with O 2 (g) to form oxides of nitrogen. N 2 (g) + 2 O 2 (g) → 2 NO 2 (g)

What Is Acid Rain? Natural rain water has a pH of 5.6. –Naturally slightly acidic due mainly to CO 2 Rain water with a pH lower than 5.6 is called acid rain. Acid rain is linked to damage in ecosystems and structures.

© 2014 Pearson Education, Inc. What Causes Acid Rain? Many natural and pollutant gases dissolved in the air are nonmetal oxides. –CO 2, SO 2, NO 2 Nonmetal oxides are acidic. CO 2 (g) + H 2 O(l)  H 2 CO 3 (aq) 2 SO 2 (g) + O 2 (g) + 2 H 2 O(l)  2 H 2 SO 4 (aq) 4 NO 2 (g) + O 2 (g) + 2 H 2 O(l)  4 HNO 3 (aq) Processes that produce nonmetal oxide gases as waste increase the acidity of the rain. –Natural – volcanoes and some bacterial action –Man made – combustion of fuel

© 2014 Pearson Education, Inc. Weather Patterns The prevailing winds in the United States travel west to east. Weather patterns may cause rain to be acidic in regions other than where the nonmetal oxide is produced. Much of the Northeast United States has rain of very low pH, even though it has very low sulfur emissions, due in part to the general weather patterns.

© 2014 Pearson Education, Inc. Damage from Acid Rain Acids react with metals and materials that contain carbonates. Acid rain damages bridges, cars, and other metallic structures. Acid rain damages buildings and other structures made of limestone or cement. Acidifying lakes affects aquatic life. Soil acidity causes more dissolving of minerals and leaching more minerals from soil, making it difficult for trees.

© 2014 Pearson Education, Inc. Acid Rain Legislation 1990 Clean Air Act attacks acid rain. –Forces utilities to reduce SO 2 The result is acid rain in the Northeast stabilized and beginning to be reduced.

© 2014 Pearson Education, Inc. Sherril Soman Grand Valley State University Lecture Presentation Chapter 15 Acids and Bases.

Similar presentations

Presentation on theme: "© 2014 Pearson Education, Inc. Sherril Soman Grand Valley State University Lecture Presentation Chapter 15 Acids and Bases."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

© 2014 Pearson Education, Inc. Sherril Soman Grand Valley State University Lecture Presentation Chapter 15 Acids and Bases.

Similar presentations

Presentation on theme: "© 2014 Pearson Education, Inc. Sherril Soman Grand Valley State University Lecture Presentation Chapter 15 Acids and Bases."— Presentation transcript:

Similar presentations

About project

Feedback