Chapter 15 Acids and Bases Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro Chapter 15 Acids and Bases Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall

Stomach Acid & 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 GERD = gastroesophageal reflux disease = chronic leaking of stomach acid into the esophagus Tro, Chemistry: A Molecular Approach

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 Tro, Chemistry: A Molecular Approach

Properties of Acids sour taste react with “active” metals i.e., Al, Zn, Fe, but not Cu, Ag, or Au 2 Al + 6 HCl ® 2 AlCl3 + 3 H2 corrosive react with carbonates, producing CO2 marble, baking soda, chalk, limestone CaCO3 + 2 HCl ® CaCl2 + CO2 + H2O change color of vegetable dyes blue litmus turns red react with bases to form ionic salts Tro, Chemistry: A Molecular Approach

Common Acids Tro, Chemistry: A Molecular Approach

Structures of Acids binary acids have acid hydrogens attached to a nonmetal atom HCl, HF Tro, Chemistry: A Molecular Approach

Structure of Acids oxy acids have acid hydrogens attached to an oxygen atom H2SO4, HNO3 Tro, Chemistry: A Molecular Approach

Structure of Acids carboxylic acids have COOH group HC2H3O2, H3C6H5O7 only the first H in the formula is acidic the H is on the COOH Tro, Chemistry: A Molecular Approach

Properties of Bases also known as alkalis taste bitter alkaloids = plant product that is alkaline often poisonous solutions feel slippery change color of vegetable dyes different color than acid red litmus turns blue react with acids to form ionic salts neutralization Tro, Chemistry: A Molecular Approach

Common Bases Tro, Chemistry: A Molecular Approach

Structure of Bases most ionic bases contain OH ions NaOH, Ca(OH)2 some contain CO32- ions CaCO3 NaHCO3 molecular bases contain structures that react with H+ mostly amine groups Tro, Chemistry: A Molecular Approach

Indicators chemicals which change color depending on the acidity/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 Tro, Chemistry: A Molecular Approach

Arrhenius Theory bases dissociate in water to produce OH- ions and cations ionic substances dissociate in water NaOH(aq) → Na+(aq) + OH–(aq) acids ionize in water to produce H+ ions and anions because molecular acids are not made of ions, they cannot dissociate they must be pulled apart, or ionized, by the water HCl(aq) → H+(aq) + Cl–(aq) in formula, ionizable H written in front HC2H3O2(aq) → H+(aq) + C2H3O2–(aq) Tro, Chemistry: A Molecular Approach

Arrhenius Theory HCl ionizes in water, producing H+ and Cl– ions NaOH dissociates in water, producing Na+ and OH– ions Tro, Chemistry: A Molecular Approach

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 a water molecule(s) to produce complex ions, mainly hydronium ion, H3O+ H+ + H2O  H3O+ there are also minor amounts of H+ with multiple water molecules, H(H2O)n+ Tro, Chemistry: A Molecular Approach

Arrhenius Acid-Base Reactions the H+ from the acid combines with the OH- from the base to make a molecule of H2O it is often helpful to think of H2O 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) + H2O(l) Tro, Chemistry: A Molecular Approach

Problems with Arrhenius Theory does not explain why molecular substances, like NH3, dissolve in water to form basic solutions – even though they do not contain OH– ions does not explain how some ionic compounds, like Na2CO3 or Na2O, dissolve in water to form basic solutions – even though they do not contain OH– ions does not explain why molecular substances, like CO2, dissolve in water to form acidic solutions – even though they do not contain H+ ions does not explain acid-base reactions that take place outside aqueous solution Tro, Chemistry: A Molecular Approach

Brønsted-Lowry Theory in a Brønsted-Lowry Acid-Base reaction, an H+ is transferred does not have to take place in aqueous solution broader definition than Arrhenius acid is H donor, base is H acceptor base structure must contain an atom with an unshared pair of electrons in an acid-base reaction, the acid molecule gives an H+ to the base molecule H–A + :B  :A– + H–B+ Tro, Chemistry: A Molecular Approach

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 HCl(aq) is acidic because HCl transfers an H+ to H2O, forming H3O+ ions water acts as base, accepting H+ HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq) acid base Tro, Chemistry: A Molecular Approach

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 NH3(aq) is basic because NH3 accepts an H+ from H2O, forming OH–(aq) water acts as acid, donating H+ NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq) base acid Tro, Chemistry: A Molecular Approach

Amphoteric Substances amphoteric substances can act as either an acid or a base have both transferable H and atom with lone pair water acts as base, accepting H+ from HCl HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq) water acts as acid, donating H+ to NH3 NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq) Tro, Chemistry: A Molecular Approach

Brønsted-Lowry Acid-Base Reactions one of the advantages of Brønsted-Lowry theory is that it allows reactions to be reversible 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 Tro, Chemistry: A Molecular Approach

Conjugate Pairs In a Brønsted-Lowry Acid-Base reaction, the original base becomes an acid in the reverse reaction, and the original acid becomes a base in the reverse process each reactant and the product it becomes is called a conjugate pair the original base becomes the conjugate acid; and the original acid becomes the conjugate base Tro, Chemistry: A Molecular Approach

Brønsted-Lowry Acid-Base Reactions H–A + :B  :A– + H–B+ acid base conjugate conjugate base acid HCHO2 + H2O  CHO2– + H3O+ acid base conjugate conjugate base acid H2O + NH3  HO– + NH4+ acid base conjugate conjugate base acid Tro, Chemistry: A Molecular Approach

Conjugate Pairs In the reaction H2O + NH3  HO– + NH4+ H2O and HO– constitute an Acid/Conjugate Base pair NH3 and NH4+ constitute a Base/Conjugate Acid pair Tro, Chemistry: A Molecular Approach

Ex 15.1a – Identify the Brønsted-Lowry Acids and Bases and Their Conjugates in the Reaction H2SO4 + H2O  HSO4– + H3O+ When the H2SO4 becomes HSO4, it lost an H+  so H2SO4 must be the acid and HSO4 its conjugate base When the H2O becomes H3O+, it accepted an H+  so H2O must be the base and H3O+ its conjugate acid H2SO4 + H2O  HSO4– + H3O+ acid base conjugate conjugate base acid Tro, Chemistry: A Molecular Approach

Ex 15.1b – Identify the Brønsted-Lowry Acids and Bases and Their Conjugates in the Reaction HCO3– + H2O  H2CO3 + HO– When the HCO3 becomes H2CO3, it accepted an H+  so HCO3 must be the base and H2CO3 its conjugate acid When the H2O becomes OH, it donated an H+  so H2O must be the acid and OH its conjugate base HCO3– + H2O  H2CO3 + HO– base acid conjugate conjugate acid base Tro, Chemistry: A Molecular Approach

Practice – Write the formula for the conjugate acid of the following H2O NH3 CO32− H2PO41− Tro, Chemistry: A Molecular Approach

Practice – Write the formula for the conjugate acid of the following H2O H3O+ NH3 NH4+ CO32− HCO3− H2PO41− H3PO4 Tro, Chemistry: A Molecular Approach

Practice – Write the formula for the conjugate base of the following H2O NH3 CO32− H2PO41− Tro, Chemistry: A Molecular Approach

Practice – Write the formula for the conjugate base of the following H2O HO− NH3 NH2− CO32− since CO32− does not have an H, it cannot be an acid H2PO41− HPO42− Tro, Chemistry: A Molecular Approach

Arrow Conventions chemists commonly use two kinds of arrows in reactions to indicate the degree of completion of the reactions a single arrow indicates 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 Tro, Chemistry: A Molecular Approach

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,  Tro, Chemistry: A Molecular Approach

Strong Acids The stronger the acid, the more willing it is to donate H use water as the standard base strong acids donate practically all their H’s 100% ionized in water strong electrolyte [H3O+] = [strong acid] HCl ® H+ + Cl- HCl + H2O® H3O+ + Cl- Tro, Chemistry: A Molecular Approach

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 [H3O+] << [weak acid] HF Û H+ + F- HF + H2O Û H3O+ + F- Tro, Chemistry: A Molecular Approach

Polyprotic Acids often acid molecules have more than one ionizable H – these are called polyprotic acids the ionizable H’s may have different acid strengths or be equal 1 H = monoprotic, 2 H = diprotic, 3 H = triprotic HCl = monoprotic, H2SO4 = diprotic, H3PO4 = triprotic polyprotic acids ionize in steps each ionizable H removed sequentially removing of the first H automatically makes removal of the second H harder H2SO4 is a stronger acid than HSO4 Tro, Chemistry: A Molecular Approach

Increasing Basicity Increasing Acidity Tro, Chemistry: A Molecular Approach

Strengths of Acids & Bases commonly, acid or base strength is measured by determining the equilibrium constant of a substance’s reaction with water HAcid + H2O  Acid-1 + H3O+1 Base: + H2O  HBase+1 + OH-1 the farther the equilibrium position lies to 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 Tro, Chemistry: A Molecular Approach

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 H2SO4 > H2SO3; HNO3 > HNO2 cation stronger acid than neutral molecule; neutral stronger acid than anion H3O+1 > H2O > OH-1; NH4+1 > NH3 > NH2-1 base trend opposite Tro, Chemistry: A Molecular Approach

Acid Ionization Constant, Ka acid strength measured by the size of the equilibrium constant when react with H2O HAcid + H2O  Acid-1 + H3O+1 the equilibrium constant is called the acid ionization constant, Ka larger Ka = stronger acid Tro, Chemistry: A Molecular Approach

Autoionization of Water Water is actually an extremely weak electrolyte therefore there must be a few ions present about 1 out of every 10 million water molecules form ions through a process called autoionization H2O Û H+ + OH– H2O + H2O Û H3O+ + OH– all aqueous solutions contain both H3O+ and OH– the concentration of H3O+ and OH– are equal in water [H3O+] = [OH–] = 10-7M @ 25°C Tro, Chemistry: A Molecular Approach

Ion Product of Water the product of the H3O+ and OH– concentrations is always the same number the number is called the ion product of water and has the symbol Kw [H3O+] x [OH–] = Kw = 1 x 10-14 @ 25°C if you measure one of the concentrations, you can calculate the other as [H3O+] increases the [OH–] must decrease so the product stays constant inversely proportional Tro, Chemistry: A Molecular Approach

Acidic and Basic Solutions all aqueous solutions contain both H3O+ and OH– ions neutral solutions have equal [H3O+] and [OH–] [H3O+] = [OH–] = 1 x 10-7 acidic solutions have a larger [H3O+] than [OH–] [H3O+] > 1 x 10-7; [OH–] < 1 x 10-7 basic solutions have a larger [OH–] than [H3O+] [H3O+] < 1 x 10-7; [OH–] > 1 x 10-7 Tro, Chemistry: A Molecular Approach

Example 15. 2b – Calculate the [OH] at 25°C when the [H3O+] = 1 Example 15.2b – Calculate the [OH] at 25°C when the [H3O+] = 1.5 x 10-9 M, and determine if the solution is acidic, basic, or neutral Given: Find: [H3O+] = 1.5 x 10-9 M [OH] Concept Plan: Relationships: [H3O+] [OH] Solution: Check: The units are correct. The fact that the [H3O+] < [OH] means the solution is basic

Complete the Table [H+] vs. [OH-] Tro, Chemistry: A Molecular Approach

Complete the Table [H+] vs. [OH-] Acid Base [H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14 OH- H+ [OH-]10-14 10-13 10-11 10-9 10-7 10-5 10-3 10-1 100 even though it may look like it, neither H+ nor OH- will ever be 0 the sizes of the H+ and OH- are not to scale because the divisions are powers of 10 rather than units Tro, Chemistry: A Molecular Approach

pH the acidity/basicity of a solution is often expressed as pH pH = -log[H3O+], [H3O+] = 10-pH exponent on 10 with a positive sign pHwater = -log[10-7] = 7 need to know the [H+] concentration to find pH pH < 7 is acidic; pH > 7 is basic, pH = 7 is neutral Tro, Chemistry: A Molecular Approach

Sig. Figs. & Logs when you take the log of a number written in scientific notation, the digit(s) 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 106) = log(106) + log(2.0) = 6 + 0.30303… = 6.30303... since 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 x 106) = 6.30 Tro, Chemistry: A Molecular Approach

pH 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 0 to 14 pH 0 is [H+] = 1 M, pH 14 is [OH–] = 1 M pH can be negative (very acidic) or larger than 14 (very alkaline)

pH of Common Substances 1.0 M HCl 0.0 0.1 M HCl 1.0 stomach acid 1.0 to 3.0 lemons 2.2 to 2.4 soft drinks 2.0 to 4.0 plums 2.8 to 3.0 apples 2.9 to 3.3 cherries 3.2 to 4.0 unpolluted rainwater 5.6 human blood 7.3 to 7.4 egg whites 7.6 to 8.0 milk of magnesia (sat’d Mg(OH)2) 10.5 household ammonia 10.5 to 11.5 1.0 M NaOH 14

Tro, Chemistry: A Molecular Approach

Example 15. 3b – Calculate the pH at 25°C when the [OH] = 1 Example 15.3b – Calculate the pH at 25°C when the [OH] = 1.3 x 10-2 M, and determine if the solution is acidic, basic, or neutral Given: Find: [OH] = 1.3 x 10-2 M pH Concept Plan: Relationships: [H3O+] [OH] pH Solution: Check: pH is unitless. The fact that the pH > 7 means the solution is basic

pOH another way of expressing the acidity/basicity of a solution is pOH pOH = -log[OH], [OH] = 10-pOH pOHwater = -log[10-7] = 7 need to know the [OH] concentration to find pOH pOH < 7 is basic; pOH > 7 is acidic, pOH = 7 is neutral Tro, Chemistry: A Molecular Approach

pH and pOH Complete the Table Tro, Chemistry: A Molecular Approach

pH and pOH Complete the Table Tro, Chemistry: A Molecular Approach

Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution Tro, Chemistry: A Molecular Approach

pK a way of expressing the strength of an acid or base is pK pKa = -log(Ka), Ka = 10-pKa pKb = -log(Kb), Kb = 10-pKb the stronger the acid, the smaller the pKa larger Ka = smaller pKa because it is the –log Tro, Chemistry: A Molecular Approach

Finding the pH of a Strong Acid there are two sources of H3O+ in an aqueous solution of a strong acid – the acid and the water for the strong acid, the contribution of the water to the total [H3O+] is negligible shifts the Kw equilibrium to the left so far that [H3O+]water is too small to be significant except in very dilute solutions, generally < 1 x 10-4 M for a monoprotic strong acid [H3O+] = [HAcid] for polyprotic acids, the other ionizations can generally be ignored 0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00 Tro, Chemistry: A Molecular Approach

Finding the pH of a Weak Acid there are also two sources of H3O+ in and aqueous solution of a weak acid – the acid and the water however, finding the [H3O+] is complicated by the fact that the acid only undergoes partial ionization calculating the [H3O+] requires solving an equilibrium problem for the reaction that defines the acidity of the acid HAcid + H2O  Acid + H3O+ Tro, Chemistry: A Molecular Approach

Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution @ 25°C Write the reaction for the acid with water Construct an ICE table for the reaction Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0 HNO2 + H2O  NO2 + H3O+ [HNO2] [NO2-] [H3O+] initial change equilibrium [HNO2] [NO2-] [H3O+] initial 0.200 ≈ 0 change equilibrium since no products initially, Qc = 0, and the reaction is proceeding forward Tro, Chemistry: A Molecular Approach

Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution @ 25°C represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HNO2] [NO2-] [H3O+] initial 0.200 change equilibrium x +x +x x x 0.200 x Tro, Chemistry: A Molecular Approach

Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution @ 25°C Ka for HNO2 = 4.6 x 10-4 determine the value of Ka from Table 15.5 since Ka is very small, approximate the [HNO2]eq = [HNO2]init and solve for x [HNO2] [NO2-] [H3O+] initial 0.200 ≈ 0 change -x +x equilibrium x 0.200 x Tro, Chemistry: A Molecular Approach

Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution @ 25°C Ka for HNO2 = 4.6 x 10-4 check if the approximation is valid by seeing if x < 5% of [HNO2]init [HNO2] [NO2-] [H3O+] initial 0.200 ≈ 0 change -x +x equilibrium x x = 9.6 x 10-3 the approximation is valid Tro, Chemistry: A Molecular Approach

Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution @ 25°C Ka for HNO2 = 4.6 x 10-4 substitute x into the equilibrium concentration definitions and solve [HNO2] [NO2-] [H3O+] initial 0.200 ≈ 0 change -x +x equilibrium 0.190 0.0096 [HNO2] [NO2-] [H3O+] initial 0.200 ≈ 0 change -x +x equilibrium 0.200-x x x = 9.6 x 10-3 Tro, Chemistry: A Molecular Approach

Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution @ 25°C Ka for HNO2 = 4.6 x 10-4 substitute [H3O+] into the formula for pH and solve [HNO2] [NO2-] [H3O+] initial 0.200 ≈ 0 change -x +x equilibrium 0.190 0.0096 Tro, Chemistry: A Molecular Approach

Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution @ 25°C Ka for HNO2 = 4.6 x 10-4 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HNO2] [NO2-] [H3O+] initial 0.200 ≈ 0 change -x +x equilibrium 0.190 0.0096 though not exact, the answer is reasonably close Tro, Chemistry: A Molecular Approach

Practice - What is the pH of a 0 Practice - What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2? (Ka = 1.4 x 10-5 @ 25°C) Tro, Chemistry: A Molecular Approach

Practice - What is the pH of a 0 Practice - What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2? Write the reaction for the acid with water Construct an ICE table for the reaction Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0 HC6H4NO2 + H2O  C6H4NO2 + H3O+ [HA] [A-] [H3O+] initial 0.012 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach

Practice - What is the pH of a 0 Practice - What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2? HC6H4NO2 + H2O  C6H4NO2 + H3O+ represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HA] [A-] [H3O+] initial 0.012 change equilibrium x +x +x x x 0.012 x Tro, Chemistry: A Molecular Approach

Practice - What is the pH of a 0 Practice - What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 10-5 @ 25°C HC6H4NO2 + H2O  C6H4NO2 + H3O+ determine the value of Ka since Ka is very small, approximate the [HA]eq = [HA]init and solve for x [HA] [A2-] [H3O+] initial 0.012 ≈ 0 change -x +x equilibrium x 0.012 x Tro, Chemistry: A Molecular Approach

the approximation is valid Practice - What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 10-5 @ 25°C Ka for HC6H4NO2 = 1.4 x 10-5 check if the approximation is valid by seeing if x < 5% of [HC6H4NO2]init [HA] [A2-] [H3O+] initial 0.012 ≈ 0 change -x +x equilibrium x x = 4.1 x 10-4 the approximation is valid Tro, Chemistry: A Molecular Approach

Practice - What is the pH of a 0 Practice - What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 10-5 @ 25°C [HA] [A2-] [H3O+] initial 0.012 ≈ 0 change -x +x equilibrium 0.012-x x substitute x into the equilibrium concentration definitions and solve x = 4.1 x 10-4 Tro, Chemistry: A Molecular Approach

Practice - What is the pH of a 0 Practice - What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 10-5 @ 25°C [HA] [A2-] [H3O+] initial 0.012 ≈ 0 change -x +x equilibrium 0.00041 substitute [H3O+] into the formula for pH and solve Tro, Chemistry: A Molecular Approach

Practice - What is the pH of a 0 Practice - What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 10-5 @ 25°C check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HA] [A2-] [H3O+] initial 0.012 ≈ 0 change -x +x equilibrium 0.00041 the values match Tro, Chemistry: A Molecular Approach

Ex 15.7 Find the pH of 0.100 M HClO2(aq) solution @ 25°C Write the reaction for the acid with water Construct an ICE table for the reaction Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0 HClO2 + H2O  ClO2 + H3O+ [HClO2] [ClO2-] [H3O+] initial 0.100 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach

Ex 15.7 Find the pH of 0.100 M HClO2(aq) solution @ 25°C represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HClO2] [ClO2-] [H3O+] initial 0.100 ≈ 0 change -x +x equilibrium 0.100-x x Tro, Chemistry: A Molecular Approach

Ex 15.7 Find the pH of 0.100 M HClO2(aq) solution @ 25°C Ka for HClO2 = 1.1 x 10-2 determine the value of Ka from Table 15.5 since Ka is very small, approximate the [HClO2]eq = [HClO2]init and solve for x [HClO2] [ClO2-] [H3O+] initial 0.100 ≈ 0 change -x +x equilibrium 0.100-x x Tro, Chemistry: A Molecular Approach

Ex 15.7 Find the pH of 0.100 M HClO2(aq) solution @ 25°C Ka for HClO2 = 1.1 x 10-2 [HClO2] [ClO2-] [H3O+] initial 0.100 ≈ 0 change -x +x equilibrium 0.100-x x check if the approximation is valid by seeing if x < 5% of [HNO2]init x = 3.3 x 10-2 the approximation is invalid Tro, Chemistry: A Molecular Approach

Ex 15.7 Find the pH of 0.100 M HClO2(aq) solution @ 25°C Ka for HClO2 = 1.1 x 10-2 if the approximation is invalid, solve for x using the quadratic formula Tro, Chemistry: A Molecular Approach

Ex 15.7 Find the pH of 0.100 M HClO2(aq) solution @ 25°C Ka for HClO2 = 1.1 x 10-2 substitute x into the equilibrium concentration definitions and solve [HClO2] [ClO2-] [H3O+] initial 0.100 ≈ 0 change -x +x equilibrium 0.072 0.028 [HClO2] [ClO2-] [H3O+] initial 0.100 ≈ 0 change -x +x equilibrium 0.100-x x x = 0.028 Tro, Chemistry: A Molecular Approach

Ex 15.7 Find the pH of 0.100 M HClO2(aq) solution @ 25°C Ka for HClO2 = 1.1 x 10-2 [HClO2] [ClO2-] [H3O+] initial 0.100 ≈ 0 change -x +x equilibrium 0.072 0.028 substitute [H3O+] into the formula for pH and solve Tro, Chemistry: A Molecular Approach

Ex 15.7 Find the pH of 0.100 M HClO2(aq) solution @ 25°C Ka for HClO2 = 1.1 x 10-2 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HClO2] [ClO2-] [H3O+] initial 0.100 ≈ 0 change -x +x equilibrium 0.072 0.028 the answer matches Tro, Chemistry: A Molecular Approach

Ex 15. 8 - What is the Ka of a weak acid if a 0 Ex 15.8 - What is the Ka of a weak acid if a 0.100 M solution has a pH of 4.25? Use the pH to find the equilibrium [H3O+] Write the reaction for the acid with water Construct an ICE table for the reaction Enter the initial concentrations and [H3O+]equil HA + H2O  A + H3O+ [HA] [A-] [H3O+] initial 0.100 ≈ 0 change equilibrium 5.6E-05 [HA] [A-] [H3O+] initial change equilibrium Tro, Chemistry: A Molecular Approach

Ex 15. 8 - What is the Ka of a weak acid if a 0 Ex 15.8 - What is the Ka of a weak acid if a 0.100 M solution has a pH of 4.25? HA + H2O  A + H3O+ fill in the rest of the table using the [H3O+] as a guide if the difference is insignificant, [HA]equil = [HA]initial substitute into the Ka expression and compute Ka [HA] [A-] [H3O+] initial 0.100 change equilibrium −5.6E-05 +5.6E-05 +5.6E-05 0.100  5.6E-05 0.100 5.6E-05 5.6E-05 Tro, Chemistry: A Molecular Approach

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 since [ionized acid]equil = [H3O+]equil Tro, Chemistry: A Molecular Approach

Ex 15.9 - What is the percent ionization of a 2.5 M HNO2 solution? Write the reaction for the acid with water Construct an ICE table for the reaction Enter the Initial Concentrations Define the Change in Concentration in terms of x Sum the columns to define the Equilibrium Concentrations HNO2 + H2O  NO2 + H3O+ [HNO2] [NO2-] [H3O+] initial change equilibrium [HNO2] [NO2-] [H3O+] initial 2.5 ≈ 0 change equilibrium x +x +x 2.5  x x x Tro, Chemistry: A Molecular Approach

Ex 15.9 - What is the percent ionization of a 2.5 M HNO2 solution? Ka for HNO2 = 4.6 x 10-4 determine the value of Ka from Table 15.5 since Ka is very small, approximate the [HNO2]eq = [HNO2]init and solve for x [HNO2] [NO2-] [H3O+] initial 2.5 ≈ 0 change -x +x equilibrium 2.5-x ≈2.5 x Tro, Chemistry: A Molecular Approach

Ex 15.9 - What is the percent ionization of a 2.5 M HNO2 solution? HNO2 + H2O  NO2 + H3O+ substitute x into the Equilibrium Concentration definitions and solve [HNO2] [NO2-] [H3O+] initial 2.5 ≈ 0 change -x +x equilibrium 0.034 x = 3.4 x 10-2 2.5  x x x Tro, Chemistry: A Molecular Approach

Ex 15.9 - What is the percent ionization of a 2.5 M HNO2 solution? HNO2 + H2O  NO2 + H3O+ Apply the Definition and Compute the Percent Ionization [HNO2] [NO2-] [H3O+] initial 2.5 ≈ 0 change -x +x equilibrium 0.034 since the percent ionization is < 5%, the “x is small” approximation is valid Tro, Chemistry: A Molecular Approach

Relationship Between [H3O+]equilibrium & [HA]initial increasing the initial concentration of acid results in increased H3O+ concentration at equilibrium increasing the initial concentration of acid results in decreased percent ionization this means that the increase in H3O+ concentration is slower than the increase in acid concentration Tro, Chemistry: A Molecular Approach

Why doesn’t the increase in H3O+ keep up with the increase in HA? the reaction for ionization of a weak acid is: HA(aq) + H2O(l)  A−(aq) + H3O+(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 [H3O+] in the dilute solution compared to the initial acid concentration this will result in a larger percent ionization

Finding the pH of Mixtures of Acids generally, you can ignore the contribution of the weaker acid to the [H3O+]equil for a mixture of a strong acid with a weak acid, the complete ionization of the strong acid provides more than enough [H3O+] to shift the weak acid equilibrium to the left so far that the weak acid’s added [H3O+] is negligible for mixtures of weak acids, 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 Tro, Chemistry: A Molecular Approach

Ex 15. 10 Find the pH of a mixture of 0. 150 M HF(aq) solution and 0 Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq) Write the reactions for the acids with water and determine their Kas If the Kas are sufficiently different, use the strongest acid to construct an ICE table for the reaction Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0 HF + H2O  F + H3O+ Ka = 3.5 x 10-4 HClO + H2O  ClO + H3O+ Ka = 2.9 x 10-8 H2O + H2O  OH + H3O+ Kw = 1.0 x 10-14 [HF] [F-] [H3O+] initial 0.150 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach

Ex 15. 10 Find the pH of a mixture of 0. 150 M HF(aq) solution and 0 Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq) represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HF] [F-] [H3O+] initial 0.150 change equilibrium x +x +x x x 0.150 x Tro, Chemistry: A Molecular Approach

Ex 15. 10 Find the pH of a mixture of 0. 150 M HF(aq) solution and 0 Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq) Ka for HF = 3.5 x 10-4 determine the value of Ka for HF since Ka is very small, approximate the [HF]eq = [HF]init and solve for x [HF] [F-] [H3O+] initial 0.150 ≈ 0 change -x +x equilibrium x 0.150 x Tro, Chemistry: A Molecular Approach

Ex 15. 10 Find the pH of a mixture of 0. 150 M HF(aq) solution and 0 Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq) Ka for HF = 3.5 x 10-4 check if the approximation is valid by seeing if x < 5% of [HF]init [HF] [F-] [H3O+] initial 0.150 ≈ 0 change -x +x equilibrium x x = 7.2 x 10-3 the approximation is valid Tro, Chemistry: A Molecular Approach

Ex 15. 10 Find the pH of a mixture of 0. 150 M HF(aq) solution and 0 Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq) Ka for HF = 3.5 x 10-4 substitute x into the equilibrium concentration definitions and solve [HF] [F-] [H3O+] initial 0.150 ≈ 0 change -x +x equilibrium 0.143 0.0072 [HF] [F-] [H3O+] initial 0.150 ≈ 0 change -x +x equilibrium 0.150-x x x = 7.2 x 10-3 Tro, Chemistry: A Molecular Approach

Ex 15. 10 Find the pH of a mixture of 0. 150 M HF(aq) solution and 0 Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq) Ka for HF = 3.5 x 10-4 substitute [H3O+] into the formula for pH and solve [HF] [F-] [H3O+] initial 0.150 ≈ 0 change -x +x equilibrium 0.143 0.0072 Tro, Chemistry: A Molecular Approach

Ex 15. 10 Find the pH of a mixture of 0. 150 M HF(aq) solution and 0 Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq) Ka for HF = 3.5 x 10-4 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HF] [F-] [H3O+] initial 0.150 ≈ 0 change -x +x equilibrium 0.143 0.0072 though not exact, the answer is reasonably close Tro, Chemistry: A Molecular Approach

Strong Bases the stronger the base, the more willing it is to accept H use water as the standard acid for strong bases, practically all molecules are dissociated into OH– or accept H’s strong electrolyte multi-OH strong bases completely dissociated [HO–] = [strong base] x (# OH) NaOH ® Na+ + OH- Tro, Chemistry: A Molecular Approach

Example 15. 11b – Calculate the pH at 25°C of a 0 Example 15.11b – Calculate the pH at 25°C of a 0.0015 M Sr(OH)2 solution and determine if the solution is acidic, basic, or neutral Given: Find: [Sr(OH)2] = 1.5 x 10-3 M pH Concept Plan: Relationships: [H3O+] [OH] pH [Sr(OH)2] [OH]=2[Sr(OH)2] Solution: [OH] = 2(0.0015) = 0.0030 M Check: pH is unitless. The fact that the pH > 7 means the solution is basic

Practice - Calculate the pH of a 0 Practice - Calculate the pH of a 0.0010 M Ba(OH)2 solution and determine if it is acidic, basic, or neutral Tro, Chemistry: A Molecular Approach

Practice - Calculate the pH of a 0 Practice - Calculate the pH of a 0.0010 M Ba(OH)2 solution and determine if it is acidic, basic, or neutral Ba(OH)2 = Ba2+ + 2 OH- therefore [OH-] = 2 x 0.0010 = 0.0020 = 2.0 x 10-3 M Kw = [H3O+][OH] [H3O+] = 1.00 x 10-14 2.0 x 10-3 = 5.0 x 10-12M pH = -log [H3O+] = -log (5.0 x 10-12) pH = 11.30 pH > 7 therefore basic Tro, Chemistry: A Molecular Approach

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 NH3 + H2O Û NH4+ + OH- Tro, Chemistry: A Molecular Approach

Tro, Chemistry: A Molecular Approach

Structure of Amines Tro, Chemistry: A Molecular Approach

Ex 15.12 Find the pH of 0.100 M NH3(aq) solution Write the reaction for the base with water Construct an ICE table for the reaction Enter the initial concentrations – assuming the [OH] from water is ≈ 0 NH3 + H2O  NH4+ + OH [NH3] [NH4+] [OH] initial change equilibrium [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change equilibrium since no products initially, Qc = 0, and the reaction is proceeding forward Tro, Chemistry: A Molecular Approach

Ex 15.12 Find the pH of 0.100 M NH3(aq) solution represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [NH3] [NH4+] [OH] initial 0.100 change equilibrium x +x +x x x 0.100 x Tro, Chemistry: A Molecular Approach

Ex 15.12 Find the pH of 0.100 M NH3(aq) solution Kb for NH3 = 1.76 x 10-5 determine the value of Kb from Table 15.8 since Kb is very small, approximate the [NH3]eq = [NH3]init and solve for x [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change -x +x equilibrium x 0.100 x Tro, Chemistry: A Molecular Approach

Ex 15.12 Find the pH of 0.100 M NH3(aq) solution Kb for NH3 = 1.76 x 10-5 check if the approximation is valid by seeing if x < 5% of [NH3]init [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change -x +x equilibrium x x = 1.33 x 10-3 the approximation is valid Tro, Chemistry: A Molecular Approach

Ex 15.12 Find the pH of 0.100 M NH3(aq) solution Kb for NH3 = 1.76 x 10-5 substitute x into the equilibrium concentration definitions and solve [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change -x +x equilibrium 0.099 1.33E-3 [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change -x +x equilibrium 0.100 x x x = 1.33 x 10-3 Tro, Chemistry: A Molecular Approach

Ex 15.12 Find the pH of 0.100 M NH3(aq) solution Kb for NH3 = 1.76 x 10-5 use the [OH-] to find the [H3O+] using Kw substitute [H3O+] into the formula for pH and solve [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change -x +x equilibrium 0.099 1.33E-3 Tro, Chemistry: A Molecular Approach

Ex 15.12 Find the pH of 0.100 M NH3(aq) solution Kb for NH3 = 1.76 x 10-5 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change -x +x equilibrium 0.099 1.33E-3 though not exact, the answer is reasonably close Tro, Chemistry: A Molecular Approach

Practice – Find the pH of a 0. 0015 M morphine solution, Kb = 1 Practice – Find the pH of a 0.0015 M morphine solution, Kb = 1.6 x 10-6 Tro, Chemistry: A Molecular Approach

Practice – Find the pH of a 0.0015 M morphine solution Write the reaction for the base with water Construct an ICE table for the reaction Enter the initial concentrations – assuming the [OH] from water is ≈ 0 B + H2O  BH+ + OH [B] [BH+] [OH] initial 0.0015 ≈ 0 change equilibrium since no products initially, Qc = 0, and the reaction is proceeding forward Tro, Chemistry: A Molecular Approach

Practice – Find the pH of a 0.0015 M morphine solution represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [B] [BH+] [OH] initial 0.0015 change equilibrium x +x +x x x 0.0015 x Tro, Chemistry: A Molecular Approach

Practice – Find the pH of a 0.0015 M morphine solution Kb for Morphine = 1.6 x 10-6 determine the value of Kb since Kb is very small, approximate the [B]eq = [B]init and solve for x [B] [BH+] [OH] initial 0.0015 ≈ 0 change -x +x equilibrium x 0.0015 x Tro, Chemistry: A Molecular Approach

Practice – Find the pH of a 0.0015 M morphine solution Kb for Morphine = 1.6 x 10-6 check if the approximation is valid by seeing if x < 5% of [B]init [B] [BH+] [OH] initial 0.0015 ≈ 0 change -x +x equilibrium x x = 4.9 x 10-5 the approximation is valid Tro, Chemistry: A Molecular Approach

Practice – Find the pH of a 0.0015 M morphine solution Kb for Morphine = 1.6 x 10-6 substitute x into the equilibrium concentration definitions and solve [B] [BH+] [OH] initial 0.0015 ≈ 0 change -x +x equilibrium 4.9E-5 [B] [BH+] [OH] initial 0.0015 ≈ 0 change -x +x equilibrium 0.0015 x x x = 4.9 x 10-5 Tro, Chemistry: A Molecular Approach

Practice – Find the pH of a 0.0015 M morphine solution Kb for Morphine = 1.6 x 10-6 use the [OH-] to find the [H3O+] using Kw substitute [H3O+] into the formula for pH and solve [B] [BH+] [OH] initial 0.0015 ≈ 0 change -x +x equilibrium 4.9E-5 Tro, Chemistry: A Molecular Approach

Practice – Find the pH of a 0.0015 M morphine solution Kb for Morphine = 1.6 x 10-6 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb [B] [BH+] [OH] initial 0.0015 ≈ 0 change -x +x equilibrium 4.9E-5 the answer matches the given Kb Tro, Chemistry: A Molecular Approach

Acid-Base Properties of 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 NaHCO3 solutions are basic Na+ is the cation of the strong base NaOH HCO3− is the conjugate base of the weak acid H2CO3 salts that contain cations that are the conjugate acid of a weak base and an anion of a strong acid are acidic NH4Cl solutions are acidic NH4+ is the conjugate acid of the weak base NH3 Cl− is the anion of the strong acid HCl Tro, Chemistry: A Molecular Approach

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) + H2O(l)  HA(aq) + OH−(aq) the stronger the acid is, the weaker the conjugate base is an anion that is the conjugate base of a strong acid is pH neutral Cl−(aq) + H2O(l)  HCl(aq) + OH−(aq) since HCl is a strong acid, this equilibrium lies practically completely to the left an anion that is the conjugate base of a weak acid is basic F−(aq) + H2O(l)  HF(aq) + OH−(aq) since HF is a weak acid, the position of this equilibrium favors the right Tro, Chemistry: A Molecular Approach

Ex 15.13 - Use the Table to Determine if the Given Anion Is Basic or Neutral NO3− the conjugate base of a strong acid, therefore neutral NO2− the conjugate base of a weak acid, therefore basic Tro, Chemistry: A Molecular Approach

Relationship between Ka of an Acid and Kb of Its Conjugate Base many reference books only give tables of Ka values because Kb values can be found from them when you add equations, you multiply the K’s

Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution Na+ is the cation of a strong base – pH neutral. The CHO2− is the anion of a weak acid – pH basic Write the reaction for the anion with water Construct an ICE table for the reaction Enter the initial concentrations – assuming the [OH] from water is ≈ 0 CHO2− + H2O  HCHO2 + OH [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach

Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x Calculate the value of Kb from the value of Ka of the weak acid from Table 15.5 substitute into the equilibrium constant expression [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change equilibrium x +x +x 0.100 x x x

Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution Kb for CHO2− = 5.6 x 10-11 since Kb is very small, approximate the [CHO2−]eq = [CHO2−]init and solve for x [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change -x +x equilibrium x 0.100 x Tro, Chemistry: A Molecular Approach

Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution Kb for CHO2− = 5.6 x 10-11 [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change -x +x equilibrium x check if the approximation is valid by seeing if x < 5% of [CHO2−]init x = 2.4 x 10-6 the approximation is valid Tro, Chemistry: A Molecular Approach

Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution Kb for CHO2− = 5.6 x 10-11 [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change -x +x equilibrium 2.4E-6 [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change -x +x equilibrium 0.100 −x x substitute x into the equilibrium concentration definitions and solve x = 2.4 x 10-6 Tro, Chemistry: A Molecular Approach

Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution Kb for CHO2− = 5.6 x 10-11 use the [OH-] to find the [H3O+] using Kw substitute [H3O+] into the formula for pH and solve [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change -x +x equilibrium 2.4E-6 Tro, Chemistry: A Molecular Approach

Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution Kb for CHO2− = 5.6 x 10-11 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change -x +x equilibrium 2.4E-6 though not exact, the answer is reasonably close Tro, Chemistry: A Molecular Approach

Polyatomic Cations as Weak Acids some cations can be thought of as the conjugate acid of a base others are the counterions of a strong base therefore, some cation can potentially be an acid MH+(aq) + H2O(l)  MOH(aq) + H3O+(aq) the stronger the base is, the weaker the conjugate acid is 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 NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq) since NH3 is a weak base, the position of this equilibrium favors the right Tro, Chemistry: A Molecular Approach

Metal Cations as Weak Acids cations of small, highly charged metals are weakly acidic alkali metal cations and alkali earth metal cations pH neutral cations are hydrated Al(H2O)63+(aq) + H2O(l)  Al(H2O)5(OH)2+ (aq) + H3O+(aq) Tro, Chemistry: A Molecular Approach

Ex 15.15 - Determine if the Given Cation Is Acidic or Neutral C5N5NH2+ the conjugate acid of a weak base, therefore acidic Ca2+ the counterion of a strong base, therefore neutral Cr3+ a highly charged metal ion, therefore acidic Tro, Chemistry: A Molecular Approach

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(NO3)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(C2H3O2)2 KNO2 Tro, Chemistry: A Molecular Approach

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 NH4Cl 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(NO3)3 Tro, Chemistry: A Molecular Approach

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 NH4F since HF is a stronger acid than NH4+, Ka of NH4+ is larger than Kb of the F−; therefore the solution will be acidic Tro, Chemistry: A Molecular Approach

Ex 15.16 - Determine whether a solution of the following salts is acidic, basic, or neutral SrCl2 Sr2+ is the counterion of a strong base, pH neutral Cl− is the conjugate base of a strong acid, pH neutral solution will be pH neutral AlBr3 Al3+ is a small, highly charged metal ion, weak acid solution will be acidic CH3NH3NO3 CH3NH3+ is the conjugate acid of a weak base, acidic NO3− is the conjugate base of a strong acid, pH neutral Tro, Chemistry: A Molecular Approach

Ex 15.16 - Determine whether a solution of the following salts is acidic, basic, or neutral NaCHO2 Na+ is the counterion of a strong base, pH neutral CHO2− is the conjugate base of a weak acid, basic solution will be basic NH4F NH4+ is the conjugate acid of a weak base, acidic F− is the conjugate base of a weak acid, basic Ka(NH4+) > Kb(F−); solution will be acidic Tro, Chemistry: A Molecular Approach

Polyprotic Acids since polyprotic acids ionize in steps, each H has a separate Ka Ka1 > Ka2 > Ka3 generally, the difference in Ka 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 H2SO4  use [H2SO4] as the [H3O+] for the second ionization [A2-] = Ka2 as long as the second ionization is negligible Tro, Chemistry: A Molecular Approach

Ex 15.18 Find the pH of 0.0100 M H2SO4(aq) solution @ 25°C H2SO4 + H2O  HSO4 + H3O+ Write the reactions for the acid with water Construct an ICE table for the reaction Enter the initial concentrations – assuming the [HSO4−] and [H3O+] is ≈ [H2SO4] HSO4 + H2O  SO42 + H3O+ [HSO4 ] [SO42 ] [H3O+] initial 0.0100 change equilibrium Tro, Chemistry: A Molecular Approach

Ex 15.18 Find the pH of 0.0100 M H2SO4(aq) solution @ 25°C represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HSO4 ] [SO42 ] [H3O+] initial 0.0100 change −x +x equilibrium 0.0100 −x x Tro, Chemistry: A Molecular Approach

Ex 15.18 Find the pH of 0.0100 M H2SO4(aq) solution @ 25°C Ka for HSO4− = 0.012 expand and solve for x using the quadratic formula Tro, Chemistry: A Molecular Approach

Ex 15.18 Find the pH of 0.0100 M H2SO4(aq) solution @ 25°C Ka for HSO4− = 0.012 substitute x into the equilibrium concentration definitions and solve [HSO4 ] [SO42 ] [H3O+] initial 0.0100 change −x +x equilibrium 0.0055 0.0045 0.0145 [HSO4 ] [SO42 ] [H3O+] initial 0.0100 change −x +x equilibrium 0.0100 −x x x = 0.0045 Tro, Chemistry: A Molecular Approach

Ex 15.18 Find the pH of 0.0100 M H2SO4(aq) solution @ 25°C Ka for HSO4− = 0.012 substitute [H3O+] into the formula for pH and solve [HSO4 ] [SO42 ] [H3O+] initial 0.0100 change −x +x equilibrium 0.0055 0.0045 0.0145 Tro, Chemistry: A Molecular Approach

Ex 15.7 Find the pH of 0.100 M HClO2(aq) solution @ 25°C Ka for HSO4− = 0.012 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HSO4 ] [SO42 ] [H3O+] initial 0.0100 change −x +x equilibrium 0.0055 0.0045 0.0145 the answer matches Tro, Chemistry: A Molecular Approach

Strengths of Binary Acids the more d+ H-X d- 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 H-C < H-N < H-O < H-F binary acid strength increases down the column H-F < H-Cl < H-Br < H-I Tro, Chemistry: A Molecular Approach

Strengths of Oxyacids, H-O-Y the more electronegative the Y atom, the stronger the acid helps weakens the H-O bond the more oxygens attached to Y, the stronger the acid further weakens and polarizes the H-O bond Tro, Chemistry: A Molecular Approach

Lewis Acid - Base Theory electron sharing electron donor = Lewis Base = nucleophile must have a lone pair of electrons electron acceptor = Lewis Acid = electrophile electron deficient when Lewis Base gives electrons from lone pair to Lewis Acid, a covalent bond forms between the molecules Nucleophile: + Electrophile  Nucleophile:Electrophile product called an adduct other acid-base reactions also Lewis Tro, Chemistry: A Molecular Approach

Example - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile OH H C H  + OH-1  OH H C H  + OH-1  Electrophile Nucleophile •• • Tro, Chemistry: A Molecular Approach

Practice - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile BF3 + HF  CaO + SO3  KI + I2  Tro, Chemistry: A Molecular Approach

Practice - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile BF3 + HF  H+1BF4-1 CaO + SO3  Ca+2SO4-2 KI + I2  KI3 F B F H F •• + -1 H+1 Nuc Elec O S O •• Ca+2 O -2 + -2 Ca+2 Elec Nuc I I K+1 I -1 •• + Elec Nuc K+1 I I I -1 Tro, Chemistry: A Molecular Approach

What Is Acid Rain? natural rain water has a pH of 5.6 naturally slightly acidic due mainly to CO2 rain water with a pH lower than 5.6 is called acid rain acid rain is linked to damage in ecosystems and structures Tro, Chemistry: A Molecular Approach

What Causes Acid Rain? many natural and pollutant gases dissolved in the air are nonmetal oxides CO2, SO2, NO2 nonmetal oxides are acidic CO2 + H2O  H2CO3 2 SO2 + O2 + 2 H2O  2 H2SO4 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 weather patterns may cause rain to be acidic in regions other than where the nonmetal oxide is produced Tro, Chemistry: A Molecular Approach

pH of Rain in Different Regions

Sources of SO2 from Utilities

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 affecting aquatic life dissolving and leaching more minerals from soil making it difficult for trees Tro, Chemistry: A Molecular Approach

Acid Rain Legislation 1990 Clean Air Act attacks acid rain force utilities to reduce SO2 result is acid rain in northeast stabilized and beginning to be reduced Tro, Chemistry: A Molecular Approach

Damage from Acid Rain