University Chemistry — Year 1 · Bachelor Year 1

10Acid–Base Equilibria and the Predominant-Reaction Method

Vinegar dissolves the limescale of a kettle; so does lemon juice, a little faster. Blood keeps its pH between narrow limits whatever we eat; a fizzy drink is acidic because of the carbon dioxide dissolved in it. All these are acid–base equilibria in water, and a chemist faced with a solution containing several acids and bases — a mixture, a buffer, a polyprotic acid — needs a reliable way to find its pH and the concentrations of its species. This chapter builds that method: a scale of strengths, diagrams of predominance, and the predominant-reaction method, which reduces any mixture to the one reaction that matters.

You already know

Book 1 (grade 12) defined Brønsted acids and bases, the acidity constant $K_a$ and $\mathrm{p}K_a$ , the ionic product of water and the pH of strong acids and bases, and drew predominance diagrams. The reaction quotient, the equilibrium constant and the activities of solutes ( $c/c^\circ$ ) and of the solvent (1) were defined in Chapter 7; in this chapter $c^\circ = 1\,\mathrm{mol}/\mathrm{L}$ is not written in quotients.

Limescale, calcium carbonate, in a kettle. The weak acids of vinegar and lemon juice react with the carbonate ion and dissolve it.

10.1 Brønsted couples and the pH

Definition 10.1 (Brønsted acid and base, couple)

A Brønsted acid is a species able to give a proton $\ce{H+}$ ; a Brønsted base is a species able to accept one. An acid HA and the base A that it becomes by losing a proton form an acid–base couple HA/A, related by the formal half-equation $\mathrm{HA} \rightleftharpoons \mathrm{A} + \ce{H+}$ . In water the proton is never free: it is carried by a water molecule as $\ce{H3O+}$ .

Definition 10.2 (Ampholyte)

An ampholyte (an amphoteric species) is the base of one couple and the acid of another: water ( $\ce{H3O+}$ / $\ce{H2O}$ and $\ce{H2O}$ / $\ce{OH-}$ ), the hydrogencarbonate ion ( $\ce{CO2,H2O}$ / $\ce{HCO3-}$ and $\ce{HCO3-}$ / $\ce{CO3^{2-}}$ ), the dihydrogenphosphate ion.

Definition 10.3 (Autoprotolysis, ionic product, pH)

Water undergoes autoprotolysis, $\ce{2H2O <=> H3O+ + OH-}$ , of constant $K_e = [\ce{H3O+}][\ce{OH-}]$ , the ionic product of water; at $25\,{}^{\circ}\mathrm{C}$ , $K_e = 10^{-14.00}$ , so $\mathrm{p}K_e = 14.00$ . The pH of a solution is $\mathrm{pH} = -\log a(\ce{H3O+}) \approx -\log[\ce{H3O+}]$ . A solution is neutral when $[\ce{H3O+}] = [\ce{OH-}]$ , that is at $\mathrm{pH} = \mathrm{p}K_e/2 = 7.00$ at $25\,{}^{\circ}\mathrm{C}$ .

10.2 Strength: $K_a$ , p $K_a$ and the levelling effect

Definition 10.4 (Acidity constant)

The acidity constant of a couple HA/A is the constant of the reaction of the acid with water,

\mathrm{HA} + \ce{H2O} \rightleftharpoons \mathrm{A} + \ce{H3O+}, \qquad K_a = \frac{[\mathrm{A}][\ce{H3O+}]}{[\mathrm{HA}]}, \qquad \mathrm{p}K_a = -\log K_a .

The larger $K_a$ (the smaller $\mathrm{p}K_a$ ), the stronger the acid and the weaker its conjugate base.

Definition 10.5 (Strong and weak acids and bases)

An acid is strong in water if its reaction with water is total: no molecule HA remains (hydrochloric, nitric, perchloric acids, the first acidity of sulfuric acid); otherwise it is weak. A base is strong if its reaction with water is total, giving $\ce{OH-}$ (hydroxides, the oxide ion, alkoxides), and weak otherwise.

Definition 10.6 (Levelling effect)

In water, every acid stronger than $\ce{H3O+}$ reacts totally to give $\ce{H3O+}$ , and every base stronger than $\ce{OH-}$ totally to give $\ce{OH-}$ : their strengths cannot be distinguished. This is the levelling effect of the solvent. Weak acids and bases in water have $0 < \mathrm{p}K_a < 14$ , the couples of water itself ( $\mathrm{p}K_a(\ce{H3O+}/\ce{H2O}) = 0$ and $\mathrm{p}K_a(\ce{H2O}/\ce{OH-}) = 14$ ) bounding the scale.

couple	p $K_a$	couple	p $K_a$
$\ce{HSO4-}$ / $\ce{SO4^{2-}}$	1.99	$\ce{CO2,H2O}$ / $\ce{HCO3-}$	6.37
$\ce{H3PO4}$ / $\ce{H2PO4-}$	2.15	$\ce{H2PO4-}$ / $\ce{HPO4^{2-}}$	7.21
$\ce{HF}$ / $\ce{F-}$	3.16	$\ce{HClO}$ / $\ce{ClO-}$	7.55
$\ce{HCOOH}$ / $\ce{HCOO-}$	3.74	$\ce{NH4+}$ / $\ce{NH3}$	9.25
$\ce{CH3COOH}$ / $\ce{CH3COO-}$	4.76	$\ce{C6H5OH}$ / $\ce{C6H5O-}$	9.99
$\ce{C6H5NH3+}$ / $\ce{C6H5NH2}$	4.60	$\ce{HCO3-}$ / $\ce{CO3^{2-}}$	10.33
$\ce{C6H5COOH}$ / $\ce{C6H5COO-}$	4.21	$\ce{CH3NH3+}$ / $\ce{CH3NH2}$	10.66

10.3 Predominance and distribution diagrams

Definition 10.7 (Predominance and distribution diagrams)

For a couple HA/A, $\mathrm{pH} = \mathrm{p}K_a + \log([\mathrm{A}]/[\mathrm{HA}])$ : the acid predominates ( $[\mathrm{HA}] > [\mathrm{A}]$ ) for $\mathrm{pH} < \mathrm{p}K_a$ , the base for $\mathrm{pH} > \mathrm{p}K_a$ . The predominance diagram is the pH axis divided at each $\mathrm{p}K_a$ into the domains where each form predominates. The distribution diagram gives the fraction of each form, $[\text{form}]/C$ , as a function of pH.

Proposition 10.8 (The Henderson relation and the fractions)

For a couple HA/A of total concentration $C = [\mathrm{HA}] + [\mathrm{A}]$ ,

\mathrm{pH} = \mathrm{p}K_a + \log\frac{[\mathrm{A}]}{[\mathrm{HA}]}, \qquad \frac{[\mathrm{HA}]}{C} = \frac{1}{1 + 10^{\mathrm{pH} - \mathrm{p}K_a}}, \qquad \frac{[\mathrm{A}]}{C} = \frac{1}{1 + 10^{\mathrm{p}K_a - \mathrm{pH}}} .

One unit of pH away from $\mathrm{p}K_a$ , the minor form is below 10 % of the total; two units away, below 1 %.

Proof. Take the logarithm of $K_a = [\mathrm{A}][\ce{H3O+}]/[\mathrm{HA}]$ . Then $[\mathrm{A}]/[\mathrm{HA}] = 10^{\mathrm{pH}-\mathrm{p}K_a}$ , and dividing $[\mathrm{HA}]$ by $[\mathrm{HA}] + [\mathrm{A}]$ gives the fractions. At $\mathrm{pH} = \mathrm{p}K_a + 1$ , $[\mathrm{HA}]/C = 1/11 < 10\,\%$ ; at $+2$ , $1/101$ . ∎

Distribution diagrams of ethanoic acid (left) and phosphoric acid (right), and the predominance diagram of phosphoric acid. Each crossing of two neighbouring curves lies at a pK_a; the ampholytes H2PO4- and HPO42- are almost the only form at the middle of their domains. — Distribution diagrams of ethanoic acid (left) and phosphoric acid (right), and the predominance diagram of phosphoric acid. Each crossing of two neighbouring curves lies at a $\mathrm{p}K_a$ ; the ampholytes $\ce{H2PO4-}$ and $\ce{HPO4^{2-}}$ are almost the only form at the middle of their domains.

10.4 The predominant-reaction method

Proposition 10.9 (Constant of an acid–base reaction)

The reaction of the acid $\mathrm{HA}_1$ of a couple with the base $\mathrm{A}_2$ of another, $\mathrm{HA}_1 + \mathrm{A}_2 \rightleftharpoons \mathrm{A}_1 + \mathrm{HA}_2$ , has the constant

K = \frac{K_{a1}}{K_{a2}} = 10^{\,\mathrm{p}K_{a2} - \mathrm{p}K_{a1}} .

It is favourable ( $K > 1$ ) when the acid is stronger than the acid that forms ( $\mathrm{p}K_{a1} < \mathrm{p}K_{a2}$ ), and quantitative (in the sense of Method 7.23) when the difference of $\mathrm{p}K_a$ exceeds about 4.

Proof. The reaction is the sum of $\mathrm{HA}_1 + \ce{H2O} \rightleftharpoons \mathrm{A}_1 + \ce{H3O+}$ ( $K_{a1}$ ) and of the reverse of $\mathrm{HA}_2 + \ce{H2O} \rightleftharpoons \mathrm{A}_2 + \ce{H3O+}$ ( $1/K_{a2}$ ); Proposition 7.19 gives $K = K_{a1}/K_{a2}$ . ∎

Definition 10.10 (Predominant reaction, equivalent solution)

In a solution containing several acids and bases, the predominant reaction is the reaction of largest constant between the strongest acid and the strongest base present. When it is quantitative it is treated as total, and the solution is replaced by the equivalent solution: the solution obtained once this reaction is complete, which has the same pH as the real one.

Method 10.11 (The predominant-reaction method)

To find the pH and the composition of a solution:

list the species introduced (and water) and place their couples on a $\mathrm{p}K_a$ scale, acids on one side, bases on the other;
find the strongest acid and the strongest base present;
if the constant of their reaction is large ( $K > 10^4$ , or $\Delta\mathrm{p}K_a > 4$ ), treat the reaction as total and write the equivalent solution; go back to step 2;
when the predominant reaction has a small constant, it fixes the pH: write its progress table and solve $Q = K$ , usually with the hypothesis that its extent is small;
check the hypotheses: the minor species are indeed minor, and the other reactions (including the autoprotolysis of water) are negligible.

A scale of pK_a (drawn with pK_a increasing upwards; values on the right). An acid reacts favourably with any base placed above it on the scale: here ethanoic acid with ammonia, K = 109.25 - 4.76 = 104.49. — A scale of $\mathrm{p}K_a$ (drawn with p $K_a$ increasing upwards; values on the right). An acid reacts favourably with any base placed above it on the scale: here ethanoic acid with ammonia, $K = 10^{9.25 - 4.76} = 10^{4.49}$ .

Proposition 10.12 (pH of simple solutions)

At concentration $C$ (in $\mathrm{mol}/\mathrm{L}$ ), with $\mathrm{p}C = -\log C$ :

strong acid: $\mathrm{pH} = \mathrm{p}C$ , valid for $\mathrm{pH} < 6.5$ ;
weak acid, little dissociated: $\mathrm{pH} = \frac12(\mathrm{p}K_a + \mathrm{p}C)$ , valid if $\mathrm{pH} < \mathrm{p}K_a - 1$ ;
weak base, little protonated: $\mathrm{pH} = \frac12(\mathrm{p}K_e + \mathrm{p}K_a - \mathrm{p}C)$ , valid if $\mathrm{pH} > \mathrm{p}K_a + 1$ ;
ampholyte with $\mathrm{p}K_{a2} - \mathrm{p}K_{a1} > 4$ : $\mathrm{pH} = \frac12(\mathrm{p}K_{a1} + \mathrm{p}K_{a2})$ , whatever $C$ (if not too dilute).

Proof. Strong acid: $\ce{HA + H2O -> A- + H3O+}$ is total, $[\ce{H3O+}] = C$ , water’s own contribution ( $10^{-7}$ ) being negligible if $C > 10^{-6.5}$ . Weak acid: the predominant reaction is $\mathrm{HA} + \ce{H2O} \rightleftharpoons \mathrm{A} + \ce{H3O+}$ with extent $x = [\ce{H3O+}] = [\mathrm{A}]$ ; if $x \ll C$ , $K_a = x^2/C$ , $\mathrm{pH} = \frac12(\mathrm{p}K_a + \mathrm{p}C)$ ; the hypothesis $[\mathrm{A}] < [\mathrm{HA}]/10$ is $\mathrm{pH} < \mathrm{p}K_a - 1$ . Weak base: the same with $\mathrm{A} + \ce{H2O} \rightleftharpoons \mathrm{HA} + \ce{OH-}$ , of constant $K_e/K_a$ . Ampholyte HA $^-$ : the predominant reaction is $2\,\mathrm{HA}^- \rightleftharpoons \mathrm{H_2A} + \mathrm{A}^{2-}$ , so $[\mathrm{H_2A}] = [\mathrm{A}^{2-}]$ ; multiplying $K_{a1} = [\mathrm{HA}^-]h/[\mathrm{H_2A}]$ by $K_{a2} = [\mathrm{A}^{2-}]h/[\mathrm{HA}^-]$ gives $h^2 = K_{a1}K_{a2}$ . ∎

pH of ethanoic acid solutions against pC, computed exactly from the charge balance. Concentrated, the acid is little dissociated and follows 1/2( pK_a + pC); diluted below about 10-6 mol/L it is fully dissociated and behaves as a strong acid; very dilute, the pH tends to 7. — pH of ethanoic acid solutions against $\mathrm{p}C$ , computed exactly from the charge balance. Concentrated, the acid is little dissociated and follows $\frac12(\mathrm{p}K_a + \mathrm{p}C)$ ; diluted below about $10^{-6}$ mol/L it is fully dissociated and behaves as a strong acid; very dilute, the pH tends to 7.

Example 10.13 (Ethanoic acid and ammonia)

Mix $0.10\,\mathrm{mol}$ of ethanoic acid and $0.10\,\mathrm{mol}$ of ammonia in $1.0\,\mathrm{L}$ . The strongest acid is $\ce{CH3COOH}$ , the strongest base $\ce{NH3}$ : $\ce{CH3COOH + NH3 <=> CH3COO- + NH4+}$ , $K = 10^{9.25 - 4.76} = 10^{4.49}$ , quantitative. The equivalent solution contains $0.10\,\mathrm{mol}/\mathrm{L}$ of $\ce{CH3COO-}$ and of $\ce{NH4+}$ . Its predominant reaction, $\ce{NH4+ + CH3COO- <=> NH3 + CH3COOH}$ , has the small constant $10^{-4.49}$ and keeps $[\ce{NH3}] = [\ce{CH3COOH}]$ , whence, as for an ampholyte, $\mathrm{pH} = \frac12(4.76 + 9.25) = 7.01$ .

10.5 Polyprotic acids and buffers

Definition 10.14 (Polyprotic acid)

A polyprotic acid can give several protons in succession, each step with its own constant: phosphoric acid $\ce{H3PO4}$ ( $\mathrm{p}K_a$ = 2.15, 7.21, 12.34), carbonic acid ( $\ce{CO2,H2O}$ ; 6.37, 10.33), sulfuric acid (first acidity strong, second 1.99). The successive $\mathrm{p}K_a$ increase: removing a proton from a species that is already negative costs more.

Example 10.15 (Phosphoric acid at $0.10\,\mathrm{mol}/\mathrm{L}$ )

The predominant reaction is the first acidity. With $x = [\ce{H3O+}]$ , $x^2/(0.10 - x) = 10^{-2.15}$ ; here $x$ is not small compared with $C$ , so the quadratic $x^2 + K_{a1}x - K_{a1}C = 0$ is solved: $x = 0.0233\,\mathrm{mol}/\mathrm{L}$ , $\mathrm{pH} = 1.63$ . The second acidity ( $\mathrm{p}K_{a2} = 7.21$ ) contributes nothing at this pH.

Definition 10.16 (Buffer solution)

A buffer solution is a solution whose pH changes little when a small amount of acid or base is added, or when it is diluted. The usual buffer is a mixture of a weak acid and its conjugate base in comparable amounts; its pH is close to the $\mathrm{p}K_a$ of the couple.

Proposition 10.17 (Why a buffer resists)

For a buffer of $n_a$ mol of acid and $n_b$ mol of base, $\mathrm{pH} = \mathrm{p}K_a + \log(n_b/n_a)$ does not depend on the volume. Adding $\delta$ mol of a strong acid changes it by

\Delta\mathrm{pH} = \log\frac{n_b - \delta}{n_a + \delta} - \log\frac{n_b}{n_a} \approx -\frac{\delta}{\ln 10}\left(\frac{1}{n_a} + \frac{1}{n_b}\right),

which is smallest for $n_a = n_b$ at given total amount.

Proof. The strong acid reacts totally with the base: $n_b \to n_b - \delta$ , $n_a \to n_a + \delta$ . The derivative of $f(\delta) = \log(n_b - \delta) - \log(n_a + \delta)$ at 0 is $-\frac{1}{\ln10}\big(\frac{1}{n_b} + \frac{1}{n_a}\big)$ ; for $n_a + n_b$ fixed, $\frac1{n_a} + \frac1{n_b}$ is smallest when $n_a = n_b$ . ∎

Definition 10.18 (Degree of dissociation)

The degree of dissociation of a weak acid HA at analytical concentration $C$ is $\alpha = [\mathrm{A}]/C$ , the fraction of the acid dissociated by water; it satisfies Ostwald’s dilution law $\alpha^2C/(1 - \alpha) = K_a$ , and tends to 1 as $C \to 0$ .

In the lab — Making a buffer

A buffer is prepared in the laboratory by weighing the acid and its salt (for instance sodium dihydrogenphosphate and disodium hydrogenphosphate), or by adding a measured volume of sodium hydroxide solution to the acid while a pH meter, calibrated beforehand with two standard buffers, is read; the solution is then made up to the mark in a volumetric flask.

10.6 Exercises

Exercise 10.1 ★

Write the conjugate base of $\ce{H2SO4}$ , $\ce{HCO3-}$ , $\ce{NH4+}$ and $\ce{H2O}$ , and the conjugate acid of $\ce{NH3}$ , $\ce{HPO4^{2-}}$ , $\ce{OH-}$ and $\ce{H2O}$ . Which of these species are ampholytes?

Solution

Solution of Exercise 10.1.

Conjugate bases: $\ce{HSO4-}$ , $\ce{CO3^{2-}}$ , $\ce{NH3}$ , $\ce{OH-}$ . Conjugate acids: $\ce{NH4+}$ , $\ce{H2PO4-}$ , $\ce{H2O}$ , $\ce{H3O+}$ . Ampholytes: $\ce{HCO3-}$ , $\ce{H2O}$ , $\ce{HPO4^{2-}}$ (and $\ce{HSO4-}$ , base of sulfuric acid and acid of its second couple).

Exercise 10.2 ★

Using the table of $\mathrm{p}K_a$ , rank hydrofluoric acid, ethanoic acid, the ammonium ion and hypochlorous acid by strength, and give $K_a$ for each.

Solution

Solution of Exercise 10.2.

$\ce{HF}$ ( $K_a = 6.9 \times 10^{-4}$ ) $>$ $\ce{CH3COOH}$ ( $1.7 \times 10^{-5}$ ) $>$ $\ce{HClO}$ ( $2.8 \times 10^{-8}$ ) $>$ $\ce{NH4+}$ ( $5.6 \times 10^{-10}$ ).

Exercise 10.3 ★

Compute the pH of a $0.010\,\mathrm{mol}/\mathrm{L}$ solution of hydrochloric acid, and of a $0.010\,\mathrm{mol}/\mathrm{L}$ solution of sodium hydroxide, at $25\,{}^{\circ}\mathrm{C}$ .

Solution

Solution of Exercise 10.3.

Strong acid: $\mathrm{pH} = 2.00$ . Strong base: $[\ce{OH-}] = 10^{-2}$ , $\mathrm{pH} = 14.00 - 2.00 = 12.00$ .

Exercise 10.4 ★

Draw the predominance diagram of the couple $\ce{CH3COOH}$ / $\ce{CH3COO-}$ . Which form predominates in a soft drink of pH 3, in blood of pH 7.4? What fraction of the acid is in the acid form at pH 3.76?

Solution

Solution of Exercise 10.4.

$\ce{CH3COOH}$ below pH 4.76, $\ce{CH3COO-}$ above. At pH 3 the acid form predominates (98 %); in blood (pH 7.4) the ethanoate ion. At pH 3.76 the acid fraction is $1/(1 + 10^{-1}) = 0.91$ .

Exercise 10.5 ★★

Compute the pH of a $0.10\,\mathrm{mol}/\mathrm{L}$ solution of ammonia and check the hypothesis made.

Solution

Solution of Exercise 10.5.

Predominant reaction $\ce{NH3 + H2O <=> NH4+ + OH-}$ , $K = 10^{-(14.00 - 9.25)} = 10^{-4.75}$ . If little $\ce{NH3}$ reacts: $[\ce{OH-}] = \sqrt{K C} = 1.3 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ , $\mathrm{pH} = 11.13$ . Check: $[\ce{NH4+}] = 1.3\,\%$ of $C$ , and $\mathrm{pH} > \mathrm{p}K_a + 1$ : valid.

Exercise 10.6 ★★

Compute the pH of a $0.010\,\mathrm{mol}/\mathrm{L}$ ethanoic acid solution and the degree of dissociation; check the hypothesis.

Solution

Solution of Exercise 10.6.

$\mathrm{pH} = \frac12(4.76 + 2.00) = 3.38$ (the exact value is 3.39); $\alpha = 10^{-3.38}/0.010 = 0.042$ : 4 % dissociated, below 10 %, and $\mathrm{pH} < \mathrm{p}K_a - 1$ : valid.

Exercise 10.7 ★★

Write the predominant reaction, its constant and the pH of a solution obtained by mixing, in $1.0\,\mathrm{L}$ , $0.10\,\mathrm{mol}$ of hydrofluoric acid and $0.050\,\mathrm{mol}$ of sodium hydroxide.

Solution

Solution of Exercise 10.7.

$\ce{HF + OH- -> F- + H2O}$ , $K = 10^{14.00 - 3.16} = 10^{10.84}$ : total. The equivalent solution holds $0.050\,\mathrm{mol}/\mathrm{L}$ each of $\ce{HF}$ and $\ce{F-}$ , a buffer: $\mathrm{pH} = \mathrm{p}K_a = 3.16$ .

Exercise 10.8 ★★

A buffer contains $0.050\,\mathrm{mol}$ of ethanoic acid and $0.050\,\mathrm{mol}$ of sodium ethanoate in $1.0\,\mathrm{L}$ . Compute its pH, then the pH after adding $0.010\,\mathrm{mol}$ of hydrochloric acid. Compare with the same addition to $1.0\,\mathrm{L}$ of pure water.

Solution

Solution of Exercise 10.8.

$\mathrm{pH} = 4.76$ . The hydrochloric acid converts $0.010\,\mathrm{mol}$ of ethanoate into acid: $\mathrm{pH} = 4.76 + \log(0.040/0.060) = 4.58$ , a change of 0.18. In pure water the pH would fall from 7.00 to 2.00.

Exercise 10.9 ★★

Hydrochloric and nitric acids have the same strength in water, but not in pure ethanoic acid as solvent. Explain, and say what the strongest acid that can exist in water is.

Solution

Solution of Exercise 10.9.

In water both acids react totally to give $\ce{H3O+}$ : the solvent levels them. Ethanoic acid is a much weaker base than water, so in it the two acids react only partly, to different extents, and their strengths can be compared. The strongest acid that can exist in water is $\ce{H3O+}$ .

Exercise 10.10 ★★★

Compute the pH of a $0.10\,\mathrm{mol}/\mathrm{L}$ solution of sodium carbonate, and check that the second basicity can be neglected.

Solution

Solution of Exercise 10.10.

$\ce{CO3^{2-} + H2O <=> HCO3- + OH-}$ , $K = 10^{-(14.00 - 10.33)} = 10^{-3.67}$ ; $[\ce{OH-}] = \sqrt{10^{-3.67} \times 0.10} = 4.6 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ (4.6 % of $C$ ), $\mathrm{pH} = 11.67$ . The second basicity, $\ce{HCO3- + H2O <=> CO2 + H2O + OH-}$ , has $K = 10^{-7.63}$ : it gives $[\ce{CO2}] \approx 10^{-7.6}$ , negligible.

Exercise 10.11 ★★★

Prove Ostwald’s dilution law and compute the degree of dissociation of ethanoic acid at $C = 1.0 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ . Why does the approximation $\alpha \ll 1$ fail here?

Solution

Solution of Exercise 10.11.

$[\mathrm{A}] = [\ce{H3O+}] = \alpha C$ and $[\mathrm{HA}] = (1 - \alpha)C$ , so $K_a = \alpha^2C/(1 - \alpha)$ . With $C = 10^{-3}$ : $\alpha^2 + 0.0174\,\alpha - 0.0174 = 0$ , $\alpha = 0.12$ . The acid is 12 % dissociated: the hypothesis of a little-dissociated acid ( $\alpha < 10\,\%$ ) fails, and the quadratic must be solved.

Exercise 10.12 ★★★

A solution contains $0.010\,\mathrm{mol}/\mathrm{L}$ of hydrochloric acid and $0.10\,\mathrm{mol}/\mathrm{L}$ of ethanoic acid. Find its pH and the concentration of ethanoate ions; explain why the weak acid hardly dissociates.

Solution

Solution of Exercise 10.12.

Hydrochloric acid fixes $[\ce{H3O+}] = 0.010\,\mathrm{mol}/\mathrm{L}$ : pH 2.00. Then

[\ce{CH3COO-}] = \frac{K_a[\ce{CH3COOH}]}{[\ce{H3O+}]} = \frac{1.74 \times 10^{-5} \times 0.10}{0.010} = 1.7 \times 10^{-4}\,\mathrm{mol}/\mathrm{L},

negligible beside 0.010: the $\ce{H3O+}$ from the strong acid pushes the equilibrium of the weak acid back.

10.7 Problem: A Phosphate Buffer for the Cell-Culture Lab

Problem 10.1

Weekend problem — the three acidities of phosphoric acid, the pH of its three sodium salts, equivalent solutions, and the volume of sodium hydroxide that gives a pH 7.20 buffer

Cells grown in a laboratory need a medium at pH close to 7.2. Data at $25\,{}^{\circ}\mathrm{C}$ : $\mathrm{p}K_a$ of phosphoric acid 2.15, 7.21 and 12.34; $\mathrm{p}K_e = 14.00$ .

Part I — Phosphoric acid.

Write the three couples and their half-equations.
Draw the predominance diagram of phosphoric acid.
Which species predominates at pH 1, 5, 9 and 13?
Which species are ampholytes?
Why is each $\mathrm{p}K_a$ larger than the previous one?

Part II — Three solutions at $0.100\,\mathrm{mol}/\mathrm{L}$ .

For phosphoric acid, write the predominant reaction and show that its extent is not negligible compared with $C$ .
Compute the pH of the phosphoric acid solution.
For sodium dihydrogenphosphate $\ce{NaH2PO4}$ , write the predominant reaction and its constant.
Compute the pH of this solution.
Compute the pH of a solution of disodium hydrogenphosphate $\ce{Na2HPO4}$ .
Which of the three solutions is closest to the pH wanted?

Part III — Adding sodium hydroxide. A solution contains $0.100\,\mathrm{mol}$ of $\ce{H3PO4}$ ; sodium hydroxide is added, $n$ mol, total volume $1.00\,\mathrm{L}$ .

Write the reaction of the first $0.100\,\mathrm{mol}$ of hydroxide and its constant. Is it quantitative?
Describe the equivalent solution for $n = 0.100\,\mathrm{mol}$ and give its pH.
Same question for $n = 0.150\,\mathrm{mol}$ .
Same question for $n = 0.200\,\mathrm{mol}$ .
Sketch the pH as a function of $n$ from 0 to $0.300\,\mathrm{mol}$ .
Why is the solution of question 14 a good buffer?

Part IV — Preparing $1.00\,\mathrm{L}$ of pH 7.20 buffer. Sodium hydroxide is available at $1.00\,\mathrm{mol}/\mathrm{L}$ .

What ratio $[\ce{HPO4^{2-}}]/[\ce{H2PO4-}]$ gives pH 7.20?
Express the amounts of $\ce{H2PO4-}$ and $\ce{HPO4^{2-}}$ in terms of the amount $x$ of hydroxide added beyond the first $0.100\,\mathrm{mol}$ .
Compute $x$ .
By how much does the pH change if $1.0 \times 10^{-3}\,\mathrm{mol}$ of hydrochloric acid is added to this buffer? to $1.00\,\mathrm{L}$ of pure water?
Would the pH change if the buffer were diluted ten times? Explain.
Compute the volume of sodium hydroxide solution to add to $0.100\,\mathrm{mol}$ of phosphoric acid to obtain $1.00\,\mathrm{L}$ of buffer at pH 7.20.

Solution

Solution of Problem 10.1.

1. $\ce{H3PO4}$ / $\ce{H2PO4-}$ , $\ce{H2PO4-}$ / $\ce{HPO4^{2-}}$ , $\ce{HPO4^{2-}}$ / $\ce{PO4^{3-}}$ ; for instance $\ce{H3PO4 <=> H2PO4- + H+}$ . 2. Boundaries at 2.15, 7.21 and 12.34. 3. pH 1: $\ce{H3PO4}$ ; 5: $\ce{H2PO4-}$ ; 9: $\ce{HPO4^{2-}}$ ; 13: $\ce{PO4^{3-}}$ . 4. $\ce{H2PO4-}$ and $\ce{HPO4^{2-}}$ . 5. Each proton leaves an ion of higher negative charge, which holds the next proton more strongly. 6. $\ce{H3PO4 + H2O <=> H2PO4- + H3O+}$ ; the formula $\frac12(\mathrm{p}K_a + \mathrm{p}C) = 1.58$ violates $\mathrm{pH} < \mathrm{p}K_a - 1 = 1.15$ : about a fifth of the acid reacts. 7. $x^2 + K_{a1}x - K_{a1}C = 0$ with $K_{a1} = 10^{-2.15}$ : $x = 0.0233\,\mathrm{mol}/\mathrm{L}$ , $\mathrm{pH} = 1.63$ . 8. $\ce{2H2PO4- <=> H3PO4 + HPO4^{2-}}$ , $K = 10^{2.15 - 7.21} = 10^{-5.06}$ . 9. $\mathrm{pH} = \frac12(2.15 + 7.21) = 4.68$ . 10. $\mathrm{pH} = \frac12(7.21 + 12.34) = 9.78$ . 11. None: 1.63, 4.68 and 9.78; a mixture of two of them is needed. 12. $\ce{H3PO4 + OH- -> H2PO4- + H2O}$ , $K = 10^{14.00 - 2.15} = 10^{11.85}$ : quantitative. 13. $0.100\,\mathrm{mol}/\mathrm{L}$ of $\ce{H2PO4-}$ : pH 4.68. 14. The next $0.050\,\mathrm{mol}$ of hydroxide converts half of it: $0.050\,\mathrm{mol}/\mathrm{L}$ of $\ce{H2PO4-}$ and of $\ce{HPO4^{2-}}$ , pH 7.21. 15. $0.100\,\mathrm{mol}/\mathrm{L}$ of $\ce{HPO4^{2-}}$ : pH 9.78. 16. The pH rises from 1.63, jumps around $n = 0.100$ (to 4.68), rises slowly through 7.21 at $n = 0.150$ , jumps around $n = 0.200$ (to 9.78), passes 12.34 near $n = 0.250$ and reaches about 12.6 at $n = 0.300$ ( $0.100\,\mathrm{mol}/\mathrm{L}$ of $\ce{PO4^3-}$ : $x^2/(0.100 - x) = 10^{-1.66}$ , $x = 0.037$ , pH 12.57). 17. It contains equal amounts of the acid and the base of a couple of $\mathrm{p}K_a$ 7.21 (Proposition 10.17). 18. $10^{7.20 - 7.21} = 0.977$ . 19. $n(\ce{H2PO4-}) = 0.100 - x$ , $n(\ce{HPO4^{2-}}) = x$ . 20. $x/(0.100 - x) = 0.977$ , $x = 0.0494\,\mathrm{mol}$ . 21. The acid converts $0.0010\,\mathrm{mol}$ of $\ce{HPO4^{2-}}$ : $\mathrm{pH} = 7.21 + \log(0.0484/0.0516) = 7.18$ , a change of $-0.02$ . In pure water: from 7.00 to 3.00. 22. No: the pH depends on the ratio of the amounts, which dilution does not change (as long as the activities stay close to the concentrations). 23. $0.100 + 0.0494 = 0.149\,\mathrm{mol}$ of hydroxide: $149\,\mathrm{mL}$ of the $1.00\,\mathrm{mol}/\mathrm{L}$ solution, made up to $1.00\,\mathrm{L}$ .

Terms defined in this chapter

See all 852 terms in the glossary

10.1 Brønsted couples and the pH

10.2 Strength: KaK_aKa​, pKaK_aKa​ and the levelling effect

10.3 Predominance and distribution diagrams

10.4 The predominant-reaction method

10.5 Polyprotic acids and buffers

10.6 Exercises

10.7 Problem: A Phosphate Buffer for the Cell-Culture Lab

Terms defined in this chapter

10.2 Strength: $K_a$ , p $K_a$ and the levelling effect