Chapter 15 Acids and Bases.

Chapter 15 Acids and Bases

15.2 The Nature of Acids and Bases
Acids have A sour taste The ability to dissolve many metals The ability to turn litmus paper RED The ability to neutralize bases

Some Acid Structures

Carboxylic Acid Group

Bases have A bitter taste A slippery feel The ability to turn litmus paper blue The ability to neutralize acids (also known as alkalis) singular: alkali

15.3 Definitions of Acids and Bases
The Arrhenius Definition Acid: a substance that produces H+ ions in aqueous solutions Base: a substance that produces OH- ions in aqueous solutions HCl ionizes in water, producing H+ and Cl– ions NaOH dissociates in water, producing Na+ and OH– ions

Hydronium Ion H+ react with water molecules to produce complex ions, mainly hydronium ion, H3O+ H+ + H2O  H3O+

Arrhenius Acid-Base reaction
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 H+ + OH-  H2O 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)

The Bronsted-Lowry Definition
Acid : Proton (H+) donor Base: Proton (H+) acceptor 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+ 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+ In a Brønsted-Lowry acid-base reaction, the acid molecule gives an H+ to the base molecule H–A + :B  :A– + H–B+ HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq) Acid Base NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq) Base Acid

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!

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)

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

Conjugate Acid-Base 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.

In the reaction H2O + NH3  HO– + NH4+
H2O and OH– constitute an acid/conjugate base pair NH3 and NH4+ constitute a base/conjugate acid pair

Practice – Write the formula for the conjugate acid of the following
H2O H3O+ NH3 NH4+ CO32− HCO3− H2PO41− H3PO4 H2O NH3 CO32− H2PO41−

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−

EXAMPLE 15.1 Identifying Brønsted–Lowry Acids and Bases
and Their Conjugates In each reaction, identify the Brønsted–Lowry acid, the Brønsted–Lowry base, the conjugate acid, and the conjugate base. (a) Since H2SO4 donates a proton to H2O in this reaction, it is the acid (proton donor). After H2SO4 donates the proton, it becomes HSO4–, the conjugate base. Since H2O accepts a proton, it is the base (proton acceptor). After H2O accepts the proton it becomes H3O+, the conjugate acid. SOLUTION (b) Since H2O donates a proton to HCO3– in this reaction, it is the acid (proton donor). After H2O donates the proton, it becomes OH–, the conjugate base. Since HCO3– accepts a proton, it is the base (proton acceptor). After HCO3– accepts the proton it becomes H2CO3, the conjugate acid. FOR PRACTICE 15.1 In each reaction, identify the Brønsted–Lowry acid, the Brønsted–Lowry base, the conjugate acid, and the conjugate base. © 2011 Pearson Education, Inc.

HSO4−(aq) + HCO3−(aq)  SO42−(aq) + H2CO3(aq)
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

HBr + H2O  Br− + H3O+ 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

I− + H2O  HI + OH− CO32− + H2O  HCO3− + OH− Base Acid Conj. Conj.
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

15.4 Acid Strength and the Acid Ionization Constant (Ka)
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, 

Strong Acids HCl ® H+ + Cl− HCl + H2O ® H3O+ + Cl−
0.10 M HCl = 0.10 M H3O+ 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

Weak Acids HF Û H+ + F− HF + H2O Û H3O+ + F− 0.10 M HF ≠ 0.10 M H3O+
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]

Strong & Weak Acids 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

Acid Ionization Constant, Ka
Acid strength measured by the size of the equilibrium constant when reacts with H2O, or disassociates. The equilibrium constant for this reaction is called the acid ionization constant, Ka larger Ka = stronger acid The stronger the acid, the weaker the conjugate base. The weaker the acid, the stronger the conjugate base. The stronger the base, the weaker the conjugate acid. The weaker the base, the stronger the conjugate acid.

15.5 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

Ion Product of Water [H3O+] x [OH–] = Kw = 1.00 x 10−14 @ 25 °C
The product of the H3O+ and OH– concentrations in WATER 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 of [OH-][H3O+] stays constant inversely proportional

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

EXAMPLE 15.2 Using Kw in Calculations
Calculate [OH–] at 25 C for each solution and determine whether it is acidic, basic, or neutral. (a) [H3O+] = 7.5  10–5 M (b) [H3O+] = 1.5  10–9 M (c) [H3O+] = 1.0  10–7 M EXAMPLE 15.2 Using Kw in Calculations (a) To find [OH–] use the ion product constant. Substitute the given value for [H3O+] and solve the equation for [OH–]. Since [H3O+] > [OH–], the solution is acidic. SOLUTION (b) Substitute the given value for [H3O+] and solve the acid ionization equation for [OH–]. Since [H3O+] < [OH–], the solution is basic. (c) Substitute the given value for [H3O+] and solve the acid ionization equation for [OH–]. Since [H3O+] = 1.0  10–7 and [OH–] = 1.0  10–7, the solution is neutral. FOR PRACTICE 15.2 Calculate [H3O+] at 25 C for each solution and determine whether it 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 © 2011 Pearson Education, Inc.

Measuring Acidity: pH pH = −log[H3O+]
The acidity or basicity of a solution is 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

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)

EXAMPLE 15.3 Calculating pH from [H3O+] or [OH–]
Calculate the pH of each solution at 25 C and indicate whether it is acidic or basic. (a) [H3O+] = 1.8  10–4 M (b) [OH–] = 1.3  10–2 M (a) To calculate pH, substitute the given [H3O+] into the pH equation. Since pH < 7, this solution is acidic. SOLUTION (b) First use Kw to find [H3O+] from [OH–]. Then substitute [H3O+] into the pH expression to find pH. Since pH > 7, this solution is basic. © 2011 Pearson Education, Inc.

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

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

EXAMPLE 15.4 Calculating [H3O+] from pH
Calculate the H3O+ concentration for a solution with a pH of 4.80. To find the [H3O+] from pH, start with the equation that defines pH. Substitute the given value of pH and then solve for [H3O+]. Since the given pH value was reported to two decimal places, the [H3O+] is written to two significant figures. (Remember that 10log x = x (see Appendix I). Some calculators use an inv log key to represent this function.) SOLUTION © 2011 Pearson Education, Inc.

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

pOH pOH = −log[OH], [OH] = 10−pOH pH + pOH = 14.0
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

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

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 basic

pKa pKa = - log Ka The pKa of a weak acid is a way to quantify its strength. The smaller the pKa, the stronger the acid. Chlorous acid, Ka = 1.1 x 10-2 (pKa 1.96) Formic acid, Ka = 1.8 x (pKa 3.74) Chlorous acid is stronger than formic acid. Both are weak acids.

15.6 Finding the [H3O+] & pH of Strong and Weak Acid Solutions
Strong acids: Weak acids: Ka for weak acid is usually in the range of 10-2 to 10-10, so water contribution of hydronium ion is not significant and we focus on the weak acid.

The Approximation Technique (only works if K or Ka is very small)
K or Ka has to be very small (reactants decrease by an insignificant amount). THE APPROXIMATION SAYS: Initial concentration – x = Initial concentration Initial concentration + x = Initial concentration x calculated with the approximation has to be smaller than 5% of the initial concentration!!!! ALWAYS CHECK!!!! Otherwise, approximation is invalid and the calculations have to be performed without approximation.

Finding the pH (or [H3O+]) of a Weak Acid Solution EXAMPLE 15.5
PROCEDURE FOR... Finding the pH (or [H3O+]) of a Weak Acid Solution EXAMPLE 15.5 Finding the [H3O+] of a Weak Acid Solution Find the [H3O+] of a M HCN solution. To solve these types of problems, follow the procedure outlined below. 1. Write 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 [H3O+] concentration is listed as approximately zero because the autoionization of water produces a negligibly small amount of H3O+). 2. Represent the change in the concentration of H3O+ 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. continued… 3. Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x. © 2011 Pearson Education, Inc.

Therefore the approximation is valid.
4. Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the acid ionization constant (Ka). 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 Ka expression and solve for x. Confirm that the x is small approximation is valid by computing the ratio of x to the number it was subtracted from in the approximation. The ratio should be less than (or 5%). Therefore the approximation is valid. 5. Determine the [H3O+] concentration from the computed value of x and compute the pH if necessary. 6. Check your answer by substituting the computed equilibrium values into the acid ionization expression. The computed value of Ka should match the given value of Ka. Note that rounding errors and the x is small approximation could result in a difference in the least significant digit when comparing values of Ka. Since the computed value of Ka matches the given value, the answer is valid. © 2011 Pearson Education, Inc.

Finding the pH (or [H3O+]) of a Weak Acid Solution EXAMPLE 15.6
PROCEDURE FOR... Finding the pH (or [H3O+]) of a Weak Acid Solution EXAMPLE 15.6 Finding the pH of a Weak Acid Solution To solve these types of problems, follow the procedure outlined below. Find the pH of a M HNO2 solution. 1. Write 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 [H3O+] concentration is listed as approximately zero because the autoionization of water produces a negligibly small amount of H3O+). 2. Represent the change in the concentration of H3O+ 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. continued… 3. Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x. © 2011 Pearson Education, Inc.

Therefore the approximation is valid (but barely so).
4. Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the acid ionization constant (Ka). 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 Ka expression and solve for x. Confirm that the x is small approximation is valid by computing 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). 5. Determine the [H3O+] concentration from the computed value of x and compute the pH if necessary. 6. Check your answer by substituting the computed equilibrium values into the acid ionization expression. The computed value of Ka should match the given value of Ka. Note that rounding errors and the x is small approximation could result in a difference in the least significant digit when comparing values of Ka. Since the computed value of Ka matches the given value, the answer is valid. © 2011 Pearson Education, Inc.

EXAMPLE 15.7 Finding the pH of a Weak Acid Solution in Cases Where the x is small Approximation Does Not Work Find the pH of a M HClO2 solution. 1. Write 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 H3O+ concentration is listed as approximately zero. Although a little H3O+ is present from the autoionization of water, this amount is negligibly small compared to the amount of H3O+.) SOLUTION 2. Represent the change in [H3O+] with the variable x. Define the changes in the concentrations of the other reactants and products in terms of x. 3. Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x. continued… © 2011 Pearson Education, Inc.

Therefore, the x is small approximation is not valid.
4. Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the acid ionization constant (Ka). Make the x is small approximation and substitute the value of the acid ionization constant (from Table 15.5) into the Ka expression. Solve for x. Check to see if the x is small approximation is valid by computing 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. 4a. If 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. continued… © 2011 Pearson Education, Inc.

5. Determine the H3O+ concentration from the calculated value of x and calculate the pH (if necessary). 6. Check your answer by substituting the calculated equilibrium values into the acid ionization expression. The calculated value of Ka should match the given value of Ka. Note that rounding errors could result in a difference in the least significant digit when comparing values of Ka. Since the calculated value of Ka matches the given value, the answer is valid. © 2011 Pearson Education, Inc.

EXAMPLE 15.8 Finding the Equilibrium Constant from pH
Use the equilibrium concentration of H3O+ 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 equal because the amount that ionizes is usually very small compared to the initial concentration. Substitute the equilibrium concentrations into the expression for Ka and calculate its value. A M weak acid (HA) solution has a pH of Find Ka for the acid. Use the given pH to find the equilibrium concentration of [H3O+]. 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. SOLUTION

EXAMPLE 15.9 Finding the Percent Ionization of a Weak Acid
Use the definition of percent ionization to calculate it. (Since the percent ionization is less than 5%, the x is small approximation is valid.) Find the percent ionization of a 2.5 M HNO2 solution. To find the percent ionization, you must find the equilibrium concentration of H3O+. Follow the procedure in Example 15.5, shown in condensed form here. SOLUTION © 2011 Pearson Education, Inc.

Calculating [H+] or pH of Acid Mixtures
When a strong acid is mixed with a weak acid, strong acid production of [H+] is much higher compared to weak acid. Thus the weak acid’s contribution of [H+] is negligible. [H+] and pH can be calculated from only strong acid’s contribution. When a mixture of two weak acids exist, calculate the contribution of the weak acid with the higher Ka. (if the Ka’s differ by more than a factor of several hundred). Water also disassociates to produce [H+] but the Kw is only 1 x and is negligible compared to the production by the acids.

EXAMPLE 15.10 Mixtures of Weak Acids
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 M in HF and M in HClO. The three possible sources of H3O+ ions are HF, HClO, and H2O. 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 [H3O+] is by far the greatest. We can therefore simply calculate the [H3O+] formed by HF and neglect the other two potential sources of H3O+. SOLUTION continued… © 2011 Pearson Education, Inc.

Therefore the approximation is valid (though barely so).
Substitute the expressions for the equilibrium concentrations into the expression for the acid ionization constant (Ka). 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 Ka 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%). Determine the H3O+ concentration from the calculated value of x and find the pH. Therefore the approximation is valid (though barely so). © 2011 Pearson Education, Inc.

15.7 Base Solutions NaOH ® Na+ + OH− STRONG BASES
For ionic bases, practically all units are dissociated into OH– or accept H’s strong electrolyte multi-OH strong bases completely dissociated [HO–] = [strong base] x (# OH)

Weak Bases NH3 + H2O Û NH4+ + OH−
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 (instead of Ka, use Kb)

:Base + H2O  OH− + H:Base+
Kb Base strength measured by the size of the equilibrium constant when react with H2O :Base + H2O  OH− + H:Base+ The equilibrium constant is called the base ionization constant, Kb larger Kb = stronger base pKb = - log Kb

EXAMPLE 15.11 Finding the [OH–] and pH of a Strong Base Solution
What is the OH– concentration and pH in each solution? (a) M KOH (b) M Sr(OH)2 (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 Kw to find [H3O+]. Then substitute [H3O+] into the pH expression to find the pH. SOLUTION (b) Since Sr(OH)2 is a strong base, it completely dissociates into 1 mol of Sr2+ 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 Kw to find [H3O+]. Substitute [H3O+] into the pH expression to find the pH. © 2011 Pearson Education, Inc.

EXAMPLE 15.12 Finding the [OH–] and pH of a Weak Base Solution
Find the [OH–] and pH of a M NH3 solution. 1. Write 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 we 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.) SOLUTION 2. Represent 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. continued… 3. Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x. © 2011 Pearson Education, Inc.

Therefore the approximation is valid.
4. Substitute 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 Kb 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. 5. Determine the OH– concentration from the calculated value of x. Use the expression for Kw to find [H3O+]. Substitute [H3O+] into the pH equation to find pH. © 2011 Pearson Education, Inc.

15.8 The Acid-Base Properties of Ions and Salts
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) An anion that is the conjugate base of a weak acid is basic. F−(aq) + H2O(l)  HF(aq) + OH−(aq)

Example 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 C2H3O2−

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

EXAMPLE 15.14 Determining the pH of a Solution Containing an Anion Acting as a Base
Find the pH of a M NaCHO2 solution. The salt completely dissociates into Na+(aq) and CHO2–(aq) and the Na+ ion has no acid or base properties. 1. Since 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. SOLUTION 2. Represent 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. continued… 3. Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x. © 2011 Pearson Education, Inc.

Find Kb from Ka (for the conjugate acid).
Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for Kb. In many cases, you can make the approximation that x is small (as discussed in Chapter 14). Substitute the value of Kb into the Kb 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. Determine the OH– concentration from the calculated value of x. Use the expression for Kw to find [H3O+]. Substitute [H3O+] into the pH equation to find pH. © 2011 Pearson Education, Inc.

Polyatomic Cations as Weak Acids
Some cations can be thought of as the conjugate acid of a weak base. others are the counter-ions of a strong base. Therefore, some cations can potentially be acidic. MH+(aq) + H2O(l)  MOH(aq) + H3O+(aq) The stronger the base is, the weaker the conjugate acid is. a cation that is the counter-ion 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) because NH3 is a weak base, the position of this equilibrium favors the right

Metal Cations as Weak Acids
Cations of small, highly charged metals are weakly acidic. ( Al3+ or Fe3+ ) alkali metal cations (Group 1) and alkali earth metal (Group 2) cations are pH neutral cations are hydrated Al(H2O)63+(aq) + H2O(l)  Al(H2O)5(OH)2+ (aq) + H3O+(aq)

Example 15.15: Determine if the given cation Is acidic or neutral
C5N5NH2+ the conjugate acid of the weak base pyridine, therefore acidic Ca2+ the counter-ion of the strong base Ca(OH)2, therefore neutral Cr3+ a highly charged metal ion, therefore acidic

15.9 Polyprotic Acids Because polyprotic acids ionize in steps, each H has a separate Ka. Generally, the difference in Ka values is great enough so that the other ionization does not happen to a large enough extent to affect the pH.

Ka1 Ka2 Ka3

EXAMPLE 15.17 Finding the pH of a Polyprotic Acid Solution
Find the pH of a M ascorbic acid (H2C6H6O6) solution. To find the pH, you must find the equilibrium concentration of H3O+. 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 H3O+ and can be ignored. Follow the procedure in Example 15.6, shown in condensed form here. Use for ascorbic acid from Table 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 H3O+ concentration. SOLUTION The approximation is valid. Therefore, © 2011 Pearson Education, Inc.

EXAMPLE 15.18 Dilute H2SO4 Solutions
Find the pH of a M sulfuric acid (H2SO4) 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 H2SO4 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 M, you can’t neglect the H3O+ contribution from the second step, as you can for other polyprotic acids. You must calculate the H3O+ contributions from both steps. SOLUTION The [H3O+] that results from the first ionization step is M (because the first step is strong). To determine the [H3O+] formed by the second step, prepare an ICE table for the second step in which the initial concentration of H3O+ is M. The initial concentration of HSO4– must also be M (due to the stoichiometry of the ionization reaction). continued… © 2011 Pearson Education, Inc.

Solve the quadratic equation using the quadratic formula.
Substitute the expressions for the equilibrium concentrations (from the table above) into the expression for Ka2. 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 Ka2 and multiply out the expression to arrive at the standard quadratic form. 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 = Determine the H3O+ concentration from the calculated value of x and calculate the pH. Notice that the second step produces almost half as much H3O+ as the first step—an amount that must not be neglected. This will always be the case with dilute H2SO4 solutions. © 2011 Pearson Education, Inc.

15.10 Acid Strength and Molecular Structure
Acid is a proton donor, must lose H+. Acid strength can be judged by the ability to lose H+. The more d+ H─X d− polarized the bond, the more acidic the bond. If bond is d- H─X d+ like H—Li, it is not acidic. If bond is nonpolar like H—C, it is not acidic either. The stronger the H─X bond, the weaker the acid. Binary acid strength increases to the right across a period. (Increasing electronegativity from C to F) acidity: H─C < H─N < H─O < H─F Binary acid strength increases down the column. (Size is bigger, bond length is longer, bond is weaker). acidity: H─F < H─Cl < H─Br < H─I

Oxyacid (H—O—Y)Strength
The more electronegative Y is, the more polarized (hence easier to lose H+) H—O is, the more acidic it is. The more number of Oxygen bound to Y, the more polarized H—O is, the more acidic it is.

Practice: By Acidity (Least to Most) H3PO4 HNO3 H3PO3 H3AsO3
H3AsO3 < H3PO3 < H3PO4 < HNO3 HCl HBr H2S H2S < HCl < HBr

15.11 Lewis Acids and Bases Lewis Acid : Electron Pair Acceptor (has empty orbital space for more electrons, or can move electrons to create empty orbital space) Examples: Molecules with incomplete octets, and cations Lewis Base: Electron Pair Donor (has lone pair(s) of electrons) Lewis Base Lewis Acid

Chapter 15 Acids and Bases.

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