Acid-Base Equilibria Chapter 16.

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Acid-Base Equilibria Chapter 16

Conjugate Acid-Base Pairs
Eqn 16.7 p 670

Conjugate Acid-Base Pairs
Eqn 16.8 p 671

Fig 16.4 Relative strengths of some conjugate acid-base pairs
The stronger the acid… the weaker its conjugate base.

The Ion Product of Water
Kc = [H+][OH-] [H2O] H2O (l) H+ (aq) + OH- (aq) [H2O] = 55.6 M = constant Kc[H2O] = Kw = [H+][OH-] Ion-product constant (Kw) - the product of the molar concentrations of H+ and OH- ions at a particular temperature. Solution is: [H+] = [OH-] neutral At 25°C: Kw = [H+][OH-] = 1.0 x 10-14 [H+] > [OH-] acidic [H+] < [OH-] basic

pH – A Measure of Acidity
pH = −log [H+] pH = −log [H3O+] Solution is: At 25°C neutral [H+] = [OH-] [H+] = 1 x 10-7 pH = 7 acidic [H+] > [OH-] [H+] > 1 x 10-7 pH < 7 basic [H+] < [OH-] [H+] < 1 x 10-7 pH > 7 pH [H+]

Fig 16.5 H+ concentrations and pH of common substances
pOH = -log [OH-] [H+][OH-] = Kw = 1.0 x 10-14 −log [H+] – log [OH-] = 14.00 pH + pOH = 14.00

The pH of rainwater collected in a certain region of the northeastern United States on a particular day was What is the H+ ion concentration of the rainwater? pH = -log [H+] [H+] = 10-pH = = 1.5 x 10-5 M The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood? pH + pOH = 14.00 pOH = -log [OH-] = -log (2.5 x 10-7) = 6.60 pH = – pOH = – 6.60 = 7.40

Other “p” Functions The “p” in pH tells us to take the negative base - 10 logarithm of the quantity (in this case, hydronium ions) Some similar examples: pOH = -log [OH−] pKw = -log Kw pCl = -log [Cl−]

How Do We Measure pH? For less accurate measurements: Litmus paper
“Red” paper turns blue above ~pH = 8 “Blue” paper turns red below ~pH = 5 Indicator:

How Do We Measure pH? Fig 16.6 Digital pH meter
For more accurate measurements: pH meter, which measures the voltage in the solution Fig Digital pH meter

Strong Electrolyte – 100% dissociation
Strong Acids and Bases Strong Electrolyte – 100% dissociation Strong Acids and Strong Bases are strong electrolytes (p 130) HCl (aq) + H2O (l) H3O+ (aq) + Cl− (aq) NaOH (aq) Na+ (aq) + OH− (aq) Table 4.2

Weak Acids Weak electrolytes - only partially ionized in aqueous solution Weak Acids are weak electrolytes HF (aq) + H2O (l) H3O+ (aq) + F- (aq) HNO2 (aq) + H2O (l) H3O+ (aq) + NO2- (aq) Most acidic substances are weak acids

Weak Acids (HA) and Acid Ionization Constants
HA (aq) + H2O (l) H3O+ (aq) + A- (aq) HA (aq) H+ (aq) + A- (aq) Ka = [H+][A-] [HA] Ka ≡ acid ionization constant weak acid strength Ka

Table 16.2 Some Weak Acids in Water at 25 °C
16.5

Calculating Ka from the pH
Sample Exercise p 682 The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is Calculate Ka for formic acid at this temperature. HCOOH (aq) H+ (aq) + HCOO- (aq) Ka = [H+][HCOO-] [HCOOH] pH = -log [H+] 2.38 = -log [H+] -2.38 = log [H+] = 10log [H+] = [H+] 4.2  10-3 = [H+] = [HCOO-]

Calculating Ka from the pH
4.2  10-3 = [H+] = [HCOO-] HCOOH (aq) H+ (aq) + HCOO- (aq) [HCOOH], M [H3O+], M [HCOO-], M Initially 0.10 Change - 4.2  10-3 + 4.2  10-3 At Equilibrium  10-3 = = 0.10 4.2  10-3 Ka = [H+][HCOO-] [HCOOH] [4.2  10-3] [4.2  10-3] [0.10] = 1.8  10-4 =

For a monoprotic acid HA:
Concentration ionized Original concentration x 100% Percent ionization = For a monoprotic acid HA: Percent ionization = [H+] [HA]0 x 100% [HA]0 = initial concentration Fig 16.9 The more dilute the acid, the greater the percent ionization:

What is the pH of a 0.50 M HF solution (at 25°C)?
Ka = [H+][F-] [HF] = 6.8 x 10-4 HF (aq) H+ (aq) + F- (aq) HF (aq) H+ (aq) + F- (aq) Initial (M) 0.50 0.00 0.00 Change (M) -x +x +x Equilibrium (M) x x x Ka = x2 x = 6.8 x 10-4 If [HF] > 100 Ka 0.50 – x  0.50 Ka  x2 0.50 = 6.8 x 10-4 x2 = 3.40 x 10-4 x = M [H+] = [F-] = M pH = -log [H+] = 1.72 [HF] = 0.50 – x = 0.48 M

When can I use the approximation?
If [HF] > 100 Ka Then 0.50 – x  0.50 Let’s determine error introduced: 0.018 M 0.50 M x 100% = 3.6% Less than 5% Approximation ok. x = 0.018 What is the pH of a 0.05 M HF solution (at 25°C)? Ka  x2 0.05 = 6.8 x 10-4 x = M 0.006 M 0.05 M x 100% = 12% More than 5% Approximation not ok. Must solve for x exactly using quadratic equation...

Solving weak acid ionization problems:
Identify the major species that can affect the pH. In most cases, the autoionization of water can be ignored. Ignore [OH¯] because it is determined by [H+]. Use ICE table to express the equilibrium concentrations in terms of single unknown x. Write Ka in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly. Calculate concentrations of all species and/or pH of the solution.

What is the pH of a 0.122 M monoprotic acid whose
Ka is 5.7 x 10-3? HA (aq) H+ (aq) + A- (aq) Initial (M) 0.122 0.00 0.00 Change (M) -x +x +x Equilibrium (M) x x x Ka = x2 x = 5.7 x 10-3 Is [HF] > 100 Ka ? NO!! Approximation not ok.

Ka = x2 x = 5.7 x 10-3 x X 10-3 x – 6.95 X 10-4 = 0 -b ± b2 – 4ac 2a x = ax2 + bx + c =0 x = x = HA (aq) H+ (aq) + A- (aq) Initial (M) Change (M) Equilibrium (M) 0.122 0.00 -x +x x x [H+] = x = M pH = -log[H+] = 1.625

Polyprotic Acids Have more than one ionizable proton
If difference between the Ka1 and subsequent Ka values > 103, the pH generally depends only on the first dissociation.