Chapter 15 Acids and Bases

Chapter 15 Acids and Bases
Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Chapter 15 Acids and Bases Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

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 Tro, Chemistry: A Molecular Approach, 2/e

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, 2/e

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 Tro, Chemistry: A Molecular Approach, 2/e

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, 2/e

Common Acids Tro, Chemistry: A Molecular Approach, 2/e

Structures of Acids Binary acids have acid hydrogens attached to a nonmetal atom HCl, HF Tro, Chemistry: A Molecular Approach, 2/e

Structure of Acids Oxy acids have acid hydrogens attached to an oxygen atom H2SO4, HNO3 Tro, Chemistry: A Molecular Approach, 2/e

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, 2/e

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, 2/e

Common Bases Tro, Chemistry: A Molecular Approach, 2/e

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, 2/e

Indicators 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 Tro, Chemistry: A Molecular Approach, 2/e

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, 2/e

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, 2/e

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, H3O+ H+ + H2O  H3O+ there are also minor amounts of H+ with multiple water molecules, H(H2O)n+ Tro, Chemistry: A Molecular Approach, 2/e

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, 2/e

Problems with Arrhenius Theory
Does not explain why molecular substances, such as NH3, dissolve in water to form basic solutions – even though they do not contain OH– ions Does not explain how some ionic compounds, such as Na2CO3 or Na2O, dissolve in water to form basic solutions – even though they do not contain OH– ions Does not explain why molecular substances, such as 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, 2/e

Another Definition: Brønsted-Lowry Acid-Base Theory
Brønsted and Lowry redefined acids and bases based on what happens in a reaction. Any reaction that involves H+ being transferred 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, but many more do as well Tro, Chemistry: A Molecular Approach, 2/e

Brønsted-Lowry Theory
In a Brønsted-Lowry acid–base reaction, an H+ is transferred 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 gives an H+ to the base molecule H–A + :B  :A– + H–B+ Tro, Chemistry: A Molecular Approach, 2/e

Brønsted-Lowry Acids HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)
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 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, 2/e

Brønsted-Lowry Bases NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq) base acid
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 NH3 dissolves in water, the NH3(aq) is the base 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, 2/e

A Warning! Because chemists know common bonding patterns, we often do not draw lone pair electrons on our structures. You need to be able to recognize when an atom in a molecule has lone pair electrons and when it doesn’t! Tro, Chemistry: A Molecular Approach, 2/e

Practice – Draw structures of the following that include lone pairs of electrons
HClO HCO3− Tro, Chemistry: A Molecular Approach, 2/e

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) + 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, 2/e

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, 2/e

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 its conjugate acid; and the original acid becomes its conjugate base Tro, Chemistry: A Molecular Approach, 2/e

Brønsted-Lowry Acid–Base Reactions
H–A :B  :A– + H–B+ acid base conjugate conjugate base acid HCHO H2O  CHO2– + H3O+ acid base conjugate conjugate base acid H2O NH3  HO– + NH4+ acid base conjugate conjugate base acid Tro, Chemistry: A Molecular Approach, 2/e

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, 2/e

Practice – Write the formula for the conjugate acid of the following
H2O H3O+ NH3 NH4+ CO32− HCO3− H2PO41− H3PO4 H2O NH3 CO32− H2PO41− Tro, Chemistry: A Molecular Approach, 2/e

Practice – Write the formula for the conjugate base of the following
H2O HO− NH3 NH2− CO32− because CO32− does not have an H, it cannot be an acid H2PO41− HPO42− H2O NH3 CO32− H2PO41− Tro, Chemistry: A Molecular Approach, 2/e

Example 15.1a: Identify the Brønsted-Lowry acids and bases, and their conjugates, in the reaction
H2SO H2O  HSO4– + H3O+ When the H2SO4 becomes HSO4, it loses an H+ so H2SO4 must be the acid and HSO4 its conjugate base When the H2O becomes H3O+, it accepts an H+ so H2O must be the base and H3O+ its conjugate acid H2SO H2O  HSO4– + H3O+ acid base conjugate conjugate base acid Tro, Chemistry: A Molecular Approach, 2/e

Example 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 accepts an H+ so HCO3 must be the base and H2CO3 its conjugate acid When the H2O becomes OH, it donats 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, 2/e

Practice – Identify the Brønsted-Lowry acid, base, conjugate acid, and conjugate base in the following reaction HSO4−(aq) HCO3−(aq)  SO42−(aq) H2CO3(aq) Conjugate Base Conjugate Acid Acid Base Tro, Chemistry: A Molecular Approach, 2/e

HBr + H2O  Br− + H3O+ HBr HSO4− HSO4− + H2O  SO42− + H3O+
Practice—Write the equations for the following reacting with water and acting as a monoprotic acid & label the conjugate acid and base HBr + H2O  Br− + H3O+ Acid Base Conj Conj. base acid HBr HSO4− HSO4− + H2O  SO42− + H3O+ Acid Base Conj Conj. base acid Tro, Chemistry: A Molecular Approach, 2/e

Practice—Write the equations for the following reacting with water and acting as a monoprotic-accepting base and label the conjugate acid and base I− CO32− I− H2O  HI OH− Base Acid Conj Conj. acid base CO32− + H2O  HCO3− + OH− Base Acid Conj Conj. acid base Tro, Chemistry: A Molecular Approach, 2/e

Comparing Arrhenius Theory and Brønsted-Lowry Theory
HCl(aq)  H+(aq) + Cl−(aq) HF(aq)  H+(aq) + F−(aq) NaOH(aq)  Na+(aq) + OH−(aq) NH4OH(aq)  NH4+(aq) + OH−(aq) Brønsted–Lowry theory HCl(aq) + H2O(l)  Cl−(aq) + H3O+(aq) HF(aq) + H2O(l)  F−(aq) + H3O+(aq) NaOH(aq)  Na+(aq) + OH−(aq) NH3(aq) + H2O(l)  NH4+(aq) + OH−(aq) Tro, Chemistry: A Molecular Approach, 2/e

Arrow Conventions  in these notes
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, 2/e

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, 2/e

Strong Acids The stronger the acid, the more willing it is to donate H
we use water as the standard base to donate H to Strong acids donate practically all their H’s 100% ionized in water strong electrolyte [H3O+] = [strong acid] [X] means the molarity of X HCl ® H+ + Cl− HCl + H2O ® H3O+ + Cl− 0.10 M HCl = 0.10 M H3O+ Tro, Chemistry: A Molecular Approach, 2/e

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− 0.10 M HF ≠ 0.10 M H3O+ Tro, Chemistry: A Molecular Approach, 2/e

Strong & Weak Acids Tro, Chemistry: A Molecular Approach, 2/e

Tro, Chemistry: A Molecular Approach, 2/e

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− + H3O+ Base: + H2O  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 Tro, Chemistry: A Molecular Approach, 2/e

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+ > H2O > OH−; NH4+ > NH3 > NH2− trend in base strength opposite Tro, Chemistry: A Molecular Approach, 2/e

Acid Ionization Constant, Ka
Acid strength measured by the size of the equilibrium constant when reacts with H2O HAcid + H2O  Acid− + H3O+ The equilibrium constant for this reaction is called the acid ionization constant, Ka larger Ka = stronger acid Tro, Chemistry: A Molecular Approach, 2/e

Tro, Chemistry: A Molecular Approach, 2/e

Autoionization of Water
Water is actually an extremely weak electrolyte therefore there must be a few ions present About 2 out of every 1 billion 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–] = 25 °C Tro, Chemistry: A Molecular Approach, 2/e

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 aka the Dissociation Constant of Water [H3O+] x [OH–] = Kw = 1.00 x 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, 2/e

Acidic and Basic Solutions
All aqueous solutions contain both H3O+ and OH– ions Neutral solutions have equal [H3O+] and [OH–] [H3O+] = [OH–] = 1.00 x 10−7 Acidic solutions have a larger [H3O+] than [OH–] [H3O+] > 1.00 x 10−7; [OH–] < 1.00 x 10−7 Basic solutions have a larger [OH–] than [H3O+] [H3O+] < 1.00 x 10−7; [OH–] > 1.00 x 10−7 Tro, Chemistry: A Molecular Approach, 2/e

Practice – Complete the table [H+] vs. [OH−]
Tro, Chemistry: A Molecular Approach, 2/e

Practice – Complete the table [H+] vs. [OH−]
Acid Base [H+] − − − − − − −13 10−14 OH− H+ [OH−]10−14 10−13 10− − − − − − 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, 2/e

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] Conceptual Plan: Relationships: [H3O+] [OH] Solution: Check: the units are correct; the fact that the [H3O+] < [OH] means the solution is basic Tro, Chemistry: A Molecular Approach, 2/e

Practice – Determine the [H3O+] when the [OH−] = 2.5 x 10−9 M

Practice – Determine the [H3O+] when the [OH−] = 2.5 x 10−9 M
Given: Find: [OH] = 2.5 x 10−9 M [H3O+] Conceptual Plan: Relationships: [OH] [H3O+] Solution: Check: the units are correct; the fact that the [H3O+] > [OH] means the solution is acidic Tro, Chemistry: A Molecular Approach, 2/e

Measuring Acidity: pH The acidity or basicity of a solution is often expressed as pH pH = −log[H3O+] exponent on 10 with a positive sign pHwater = −log[10−7] = 7 need to know the [H3O+] concentration to find pH pH < 7 is acidic; pH > 7 is basic, pH = 7 is neutral [H3O+] = 10−pH Tro, Chemistry: A Molecular Approach, 2/e

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) = … = 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 x 106) = 6.30 Tro, Chemistry: A Molecular Approach, 2/e

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 [H3O+] = 1 M, pH 14 is [OH–] = 1 M pH can be negative (very acidic) or larger than 14 (very alkaline) Tro, Chemistry: A Molecular Approach, 2/e

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 Conceptual Plan: Relationships: [H3O+] [OH] pH Solution: Check: pH is unitless; the fact that the pH > 7 means the solution is basic Tro, Chemistry: A Molecular Approach, 2/e

Practice – Determine the pH @ 25 ºC of a solution that has [OH−] = 2
Practice – Determine the 25 ºC of a solution that has [OH−] = 2.5 x 10−9 M Tro, Chemistry: A Molecular Approach, 2/e

Practice – Determine the pH @ 25 ºC of a solution that has [OH−] = 2
Practice – Determine the 25 ºC of a solution that has [OH−] = 2.5 x 10−9 M Given: Find: [OH] = 2.5 x 10−9 M pH Conceptual Plan: Relationships: [H3O+] [OH] pH Solution: Check: pH is unitless; the fact that the pH < 7 means the solution is acidic Tro, Chemistry: A Molecular Approach, 2/e

Practice – Determine the [OH−] of a solution with a pH of 5.40

Practice – Determine the [OH−] of a solution with a pH of 5.40
Given: Find: pH = 5.40 [OH−], M Conceptual Plan: Relationships: [H3O+] pH [OH] Solution: Check: because the pH < 7, [OH−] should be less than 1 x 10−7; and it is Tro, Chemistry: A Molecular Approach, 2/e

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 pH + pOH = 14.0 Tro, Chemistry: A Molecular Approach, 2/e

pH and pOH Practice – Complete the table

pH and pOH Practice – Complete the table
Acid Base pH [H+] − − − − − − −13 10−14 OH− H+ [OH−]10−14 10−13 10− − − − − − pOH Tro, Chemistry: A Molecular Approach, 2/e

Relationship between pH and pOH
pH + pOH = 14.00 at 25 °C you can use pOH to find pH of a solution Tro, Chemistry: A Molecular Approach, 2/e

Example: Calculate the pH at 25 °C when the [OH] = 1
Example: 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 Conceptual Plan: Relationships: pOH [OH] pH Solution: Check: pH is unitless; the fact that the pH > 7 means the solution is basic Tro, Chemistry: A Molecular Approach, 2/e

Practice – Determine the pOH @ 25 ºC of a solution that has [H3O+] = 2
Practice – Determine the 25 ºC of a solution that has [H3O+] = 2.5 x 10−9 M Tro, Chemistry: A Molecular Approach, 2/e

Practice – Determine the pOH @ 25 ºC of a solution that has [H3O+] = 2
Practice – Determine the 25 ºC of a solution that has [H3O+] = 2.5 x 10−9 M Given: Find: [H3O+] = 2.5 x 10−9 M pOH Conceptual Plan: Relationships: pH [H3O+] pOH Solution: Check: pH is unitless; the fact that the pH < 7 means the solution is acidic Tro, Chemistry: A Molecular Approach, 2/e

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 The stronger the base, the smaller the pKb larger Kb = smaller pKb Tro, Chemistry: A Molecular Approach, 2/e

[H3O+] and [OH−] in a Strong Acid or Strong Base Solution
There are two sources of H3O+ 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 [H3O+] or [OH−] is negligible the [H3O+]acid shifts the Kw equilibrium so far that [H3O+]water is too small to be significant except in very dilute solutions, generally < 1 x 10−4 M Tro, Chemistry: A Molecular Approach, 2/e

Finding pH of a Strong Acid or Strong Base Solution
For a monoprotic strong acid [H3O+] = [HAcid] for polyprotic acids, the other ionizations can generally be ignored for H2SO4, the second ionization cannot be ignored 0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00 For a strong ionic base, [OH−] = (number OH−)x[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 Tro, Chemistry: A Molecular Approach, 2/e

Finding the pH of a Weak Acid
There are also two sources of H3O+ in an 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, 2/e

Example 15.6: Find the pH of 0.200 M HNO2(aq) solution @ 25 °C
HNO2 + H2O  NO2 + H3O+ 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] [NO2−] [H3O+] initial 0.200 ≈ 0 change equilibrium [HNO2] [NO2−] [H3O+] initial change equilibrium because no products initially, Qc = 0, and the reaction is proceeding forward Tro, Chemistry: A Molecular Approach, 2/e

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 HNO2 + H2O(l)  NO2−(aq) + H3O+(aq) Tro, Chemistry: A Molecular Approach, 2/e

Ka for HNO2 = 4.6 x 10−4 determine the value of Ka from Table 15.5 because 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, 2/e

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, 2/e

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, 2/e

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, 2/e

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, 2/e

Practice – What is the pH of a 0
Practice – What is the pH of a M solution of nicotinic acid, HC6H4NO2? (Ka = 1.4 x 25 °C) Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 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, 2/e

Practice – What is the pH of a M solution of nicotinic acid, HC6H4NO2? 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 HC6H4NO2 + H2O  C6H4NO2 + H3O+ [HA] [A−] [H3O+] initial 0.012 change equilibrium x +x +x x x 0.012 x Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a M solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 25 °C HC6H4NO2 + H2O  C6H4NO2 + H3O+ determine the value of Ka because Ka is very small, approximate the [HA]eq = [HA]init and solve for x [HA] [A−] [H3O+] initial 0.012 ≈ 0 change −x +x equilibrium x 0.012 x Tro, Chemistry: A Molecular Approach, 2/e

the approximation is valid
Practice – What is the pH of a M solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 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] [A−] [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, 2/e

Practice – What is the pH of a M solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 25 °C substitute x into the equilibrium concentration definitions and solve [HA] [A−] [H3O+] initial 0.012 ≈ 0 change −x +x equilibrium 0.012−x x x = 4.1 x 10−4 Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a M solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 25 °C substitute [H3O+] into the formula for pH and solve [HA] [A−] [H3O+] initial 0.012 ≈ 0 change −x +x equilibrium Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a M solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 25 °C check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HA] [A−] [H3O+] initial 0.012 ≈ 0 change −x +x equilibrium the values match Tro, Chemistry: A Molecular Approach, 2/e

Example 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, 2/e

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, 2/e

Ka for HClO2 = 1.1 x 10−2 determine the value of Ka from Table 15.5 because 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, 2/e

Ka for HClO2 = 1.1 x 10−2 check if the approximation is valid by seeing if x < 5% of [HNO2]init [HClO2] [ClO2−] [H3O+] initial 0.100 ≈ 0 change −x +x equilibrium 0.100−x x x = 3.3 x 10−2 the approximation is invalid Tro, Chemistry: A Molecular Approach, 2/e

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, 2/e

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 0.100−x x [HClO2] [ClO2−] [H3O+] initial 0.100 ≈ 0 change −x +x equilibrium 0.072 0.028 x = 0.028 Tro, Chemistry: A Molecular Approach, 2/e

Ka for HClO2 = 1.1 x 10−2 substitute [H3O+] into the formula for pH and solve [HClO2] [ClO2−] [H3O+] initial 0.100 ≈ 0 change −x +x equilibrium 0.072 0.028 Tro, Chemistry: A Molecular Approach, 2/e

Example 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, 2/e

Example 15. 8: What is the Ka of a weak acid if a 0
Example 15.8: What is the Ka of a weak acid if a 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 change equilibrium [HA] [A−] [H3O+] initial 0.100 ≈ 0 change equilibrium 5.6E-05 Tro, Chemistry: A Molecular Approach, 2/e

Example 15. 8: What is the Ka of a weak acid if a 0
Example 15.8: What is the Ka of a weak acid if a 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, 2/e

Practice – What is the Ka of nicotinic acid, HC6H4NO2, if a 0
Practice – What is the Ka of nicotinic acid, HC6H4NO2, if a M solution of nicotinic acid has a pH of 3.40? Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the Ka of nicotinic acid, HC6H4NO2, if a M solution of nicotinic acid has a pH of 3.40? 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 change equilibrium [HA] [A−] [H3O+] initial 0.012 ≈ 0 change equilibrium 4.0E-04 Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the Ka of nicotinic acid, HC6H4NO2, if a M solution of nicotinic acid has a pH of 3.40? 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 + H2O  A + H3O+ [HA] [A−] [H3O+] initial 0.012 change equilibrium −4.0E-04 +4.0E-04 +4.0E-04 0.012  4.0E-04 4.0E-04 4.0E-04 0.012 Tro, Chemistry: A Molecular Approach, 2/e

Chapter 15 Acids and Bases

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