6.4 Buffers, Salt Hydrolysis and Solubility Product
Salts are not always neutral, some solutions refuse to change their pH, and a few compounds barely dissolve at all. Salt hydrolysis and the pH of salt solutions, buffer solutions and the Henderson equation, and the solubility product with the common-ion effect and the prediction of precipitation.
Your blood has a pH of $7.40$, and it stays within about $0.05$ of that value whatever you eat or drink and however hard you exercise. If it fell to $7.0$ or rose to $7.8$ you would not survive. Lactic acid pours into the blood during a sprint, and yet the pH hardly moves: the blood is buffered. In a quite different corner of chemistry, a patient swallows a "barium meal" of barium sulphate before an X-ray, although barium ions are poisonous — because barium sulphate is so slightly soluble that almost no $\mathrm{Ba^{2+}}$ reaches the bloodstream. Both facts are explained by ionic equilibria.
This section completes the chapter. It first shows why solutions of many salts are acidic or basic (salt hydrolysis), then how a weak acid mixed with its salt makes a buffer whose pH is given by the Henderson equation. Finally it treats the equilibrium between a sparingly soluble salt and its ions, described by the solubility product $K_{sp}$, and uses it to calculate solubilities, to see the common-ion effect on solubility and to predict whether a precipitate will form.
Hydrolysis of Salts and the pH of Their Solutions
A salt is formed when an acid neutralises a base, but its solution is neutral only if neither of its ions reacts with water. The conjugate base of a weak acid (such as $\mathrm{CH_{3}COO^{-}}$) takes a proton from water and releases $\mathrm{OH^{-}}$; the conjugate acid of a weak base (such as $\mathrm{NH_{4}^{+}}$) gives a proton to water and releases $\mathrm{H_{3}O^{+}}$. Ions of strong acids and strong bases ($\mathrm{Cl^{-}}$, $\mathrm{NO_{3}^{-}}$, $\mathrm{Na^{+}}$, $\mathrm{K^{+}}$) do not react.
📘 Definition 6.14 — Salt hydrolysis and the hydrolysis constant
Salt hydrolysis is the reaction of the cation or anion of a salt with water, which changes the pH of the solution:
$$\begin{aligned}&\mathrm{CH_{3}COO^{-}(aq)+H_{2}O(l)}\\ &\quad\mathrm{\rightleftharpoons CH_{3}COOH(aq)+OH^{-}(aq)}\end{aligned}$$
$$\begin{aligned}&\mathrm{NH_{4}^{+}(aq)+H_{2}O(l)}\\ &\quad\mathrm{\rightleftharpoons NH_{3}(aq)+H_{3}O^{+}(aq)}\end{aligned}$$
Its equilibrium constant is the hydrolysis constant, $K_{h}=K_{w}/K_{a}$ for the anion of a weak acid and $K_{h}=K_{w}/K_{b}$ for the cation of a weak base (Rule 6.13). The fraction of the salt hydrolysed is the degree of hydrolysis, $h\approx\sqrt{K_{h}/c}$ when small.
📐 Rule 6.16 — pH of salt solutions at 298 K
For a salt of concentration $c$ (with $\mathrm{p}K_{a}$ of the weak acid and $\mathrm{p}K_{b}$ of the weak base involved):
Strong acid + strong base ($\mathrm{NaCl}$, $\mathrm{KNO_{3}}$): no hydrolysis, $\mathrm{pH}=7$.
Weak acid + weak base ($\mathrm{CH_{3}COONH_{4}}$): both ions hydrolyse, $\mathrm{pH}=7+\tfrac12(\mathrm{p}K_{a}-\mathrm{p}K_{b})$, independent of $c$: basic if $K_{b}\gt K_{a}$, acidic if $K_{a}\gt K_{b}$.
(Formulas 2–4 assume a small degree of hydrolysis.)
✏️ Example 1 — pH of salt solutions
Use $\mathrm{p}K_{a}(\mathrm{CH_{3}COOH})=4.74$, $\mathrm{p}K_{b}(\mathrm{NH_{3}})=4.74$, $K_{a}(\mathrm{HCN})=4.9\times10^{-10}$. Find the pH of (a) $0.10\ \mathrm{mol\,L^{-1}}$ $\mathrm{CH_{3}COONa}$, first from $K_{h}$ and then from Rule 6.16; (b) $0.10\ \mathrm{mol\,L^{-1}}$ $\mathrm{NH_{4}Cl}$; (c) $0.10\ \mathrm{mol\,L^{-1}}$ $\mathrm{CH_{3}COONH_{4}}$; (d) $0.010\ \mathrm{mol\,L^{-1}}$ NaCN, with its degree of hydrolysis.
(b) $\mathrm{pH}=7-\tfrac12(4.74-1.00)=5.13$ — acidic, the mirror image of (a).
(c) $\mathrm{pH}=7+\tfrac12(4.74-4.74)=7.00$: acetic acid and ammonia are equally weak, so the two hydrolyses cancel.
(d) $K_{h}=\dfrac{1.0\times10^{-14}}{4.9\times10^{-10}}=2.0\times10^{-5}$; $[\mathrm{OH^{-}}]=\sqrt{2.0\times10^{-5}\times0.010}=4.5\times10^{-4}$, $\mathrm{pOH}=3.35$, $\mathrm{pH}=10.65$; degree of hydrolysis $h=4.5\times10^{-4}/0.010=4.5\%$ — a solution of a cyanide is strongly alkaline.
Buffer Solutions
Add one drop of concentrated hydrochloric acid to a litre of pure water and the pH crashes from $7$ to about $3$. Add the same drop to a litre of a mixture of acetic acid and sodium acetate and the pH barely changes. The mixture contains a reservoir of a weak acid to neutralise added base and a reservoir of its conjugate base to neutralise added acid.
📘 Definition 6.15 — Buffer solution
A buffer solution resists a change in pH when a small amount of acid or base is added, or when it is diluted. An acidic buffer contains a weak acid and its salt with a strong base ($\mathrm{CH_{3}COOH}$ + $\mathrm{CH_{3}COONa}$, pH near $4.7$); a basic buffer contains a weak base and its salt with a strong acid ($\mathrm{NH_{3}}$ + $\mathrm{NH_{4}Cl}$, pH near $9.3$). Blood is buffered mainly by $\mathrm{H_{2}CO_{3}}$ / $\mathrm{HCO_{3}^{-}}$.
The buffer works because both members of the conjugate pair are present in large amounts:
Either way the strong acid or base is replaced by a weak one, and the ratio $[\mathrm{CH_{3}COO^{-}}]/[\mathrm{CH_{3}COOH}]$ — which fixes the pH — changes only a little. Taking logarithms of $[\mathrm{H^{+}}]=K_{a}[\mathrm{HA}]/[\mathrm{A^{-}}]$ (Section 6.3, Example 6) gives the buffer equation.
📐 Rule 6.17 — The Henderson equation
For a buffer of a weak acid HA and its conjugate base $\mathrm{A^{-}}$ (supplied by the salt),
$$\mathrm{pH}=\mathrm{p}K_{a}+\log\frac{[\text{conjugate base}]}{[\text{acid}]}$$
$$\mathrm{pH}=\mathrm{p}K_{a}+\log\frac{[\text{salt}]}{[\text{acid}]},$$
and for a weak base B with its salt,
$$\mathrm{pOH}=\mathrm{p}K_{b}+\log\frac{[\text{salt}]}{[\text{base}]}.$$
Valid when the acid (base) is weak and both concentrations are much larger than $[\mathrm{H^{+}}]$ and $[\mathrm{OH^{-}}]$, so that the equilibrium concentrations equal the amounts mixed. Because only the ratio appears, dilution leaves the pH unchanged. The buffer works best for $\mathrm{pH}=\mathrm{p}K_{a}\pm1$ (ratio between $0.1$ and $10$), and best of all at $\mathrm{pH}=\mathrm{p}K_{a}$, when $[\text{salt}]=[\text{acid}]$.
✏️ Example 2 — Designing a buffer
($\mathrm{p}K_{a}(\mathrm{CH_{3}COOH})=4.74$, $\mathrm{p}K_{b}(\mathrm{NH_{3}})=4.74$.) (a) Find the pH of a solution $0.10\ \mathrm{mol\,L^{-1}}$ in $\mathrm{CH_{3}COOH}$ and $0.20\ \mathrm{mol\,L^{-1}}$ in $\mathrm{CH_{3}COONa}$. (b) What ratio $[\mathrm{CH_{3}COO^{-}}]/[\mathrm{CH_{3}COOH}]$ gives a buffer of pH $5.00$? (c) What ratio $[\mathrm{NH_{4}^{+}}]/[\mathrm{NH_{3}}]$ gives pH $9.00$? (d) Why can an acetate buffer not be used to hold a pH of $7.0$?
(b) $\log(\text{ratio})=5.00-4.74=0.26$, ratio $=10^{0.26}=1.8$.
(c) $\mathrm{pOH}=14.00-9.00=5.00$; $\log\dfrac{[\mathrm{NH_{4}^{+}}]}{[\mathrm{NH_{3}}]}=5.00-4.74=0.26$, ratio $=1.8$.
(d) pH 7.0 would need a ratio of $10^{7.0-4.74}=180$: almost no acid would be left to neutralise added base. Choose a weak acid with $\mathrm{p}K_{a}$ close to 7 instead (for example $\mathrm{H_{2}PO_{4}^{-}}$, $\mathrm{p}K_{a_{2}}=7.21$).
✏️ Example 3 — Buffer action in numbers
$1.00\ \mathrm{L}$ of buffer contains $0.10\ \mathrm{mol}$ $\mathrm{CH_{3}COOH}$ and $0.10\ \mathrm{mol}$ $\mathrm{CH_{3}COONa}$ ($\mathrm{p}K_{a}=4.74$). Find the pH (a) of the buffer; (b) after adding $0.010\ \mathrm{mol}$ HCl; (c) after adding $0.010\ \mathrm{mol}$ NaOH instead. (d) Compare with adding $0.010\ \mathrm{mol}$ HCl to $1.00\ \mathrm{L}$ of pure water. (Ignore volume changes.)
(b) The HCl converts $0.010\ \mathrm{mol}$ acetate into acetic acid: acid $0.11$, salt $0.09\ \mathrm{mol}$. $\mathrm{pH}=4.74+\log\dfrac{0.09}{0.11}=4.74-0.09=4.65$.
(c) The NaOH converts $0.010\ \mathrm{mol}$ acid into acetate: acid $0.09$, salt $0.11$. $\mathrm{pH}=4.74+0.09=4.83$.
(d) In water $[\mathrm{H^{+}}]=0.010$, $\mathrm{pH}=2.00$: a fall of 5 units, against $0.09$ units for the buffer.
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — the titration curve of a weak acid
Not required by the CBSE Class 11 syllabus. It combines buffers and the hydrolysis of salts, which JEE Main, NEET and JEE Advanced all list, with acid–base titrations and indicators, which JEE Main and NEET include under practical chemistry.
When $0.10\ \mathrm{mol\,L^{-1}}$ acetic acid is titrated with $0.10\ \mathrm{mol\,L^{-1}}$ NaOH, the mixture before the end point is an acetate buffer. Halfway to the end point $[\text{salt}]=[\text{acid}]$, so $\mathrm{pH}=\mathrm{p}K_{a}=4.74$ — a neat way to measure $\mathrm{p}K_{a}$. At the equivalence point the flask holds only sodium acetate at $0.050\ \mathrm{mol\,L^{-1}}$ (the volume has doubled), so by Rule 6.16 $\mathrm{pH}=7+\tfrac12(4.74+\log0.050)=8.72$, not 7. An indicator that changes colour in the basic range, such as phenolphthalein (pH 8.3–10), must therefore be used; methyl orange (3.1–4.4) would change far too early.
Interactive 1 — Buffer Lab: Adding Acid and Base
One litre of solution. Choose the buffer pair and the amounts of weak acid (or base) and salt, then add strong acid or base with the buttons or the slider. The red curve is the buffer's pH against the amount added; the grey curve is the same additions made to pure water. The pH is computed exactly from the ionic equilibria, and the Henderson prediction is shown for comparison. Press ▶ Play to sweep from adding acid to adding base and back.
0.10 mol0.10 mol
0.000 mol
pH of buffer—
Henderson estimate—
pH of water + same addition—
ratio salt : acid now—
Solubility Equilibria and the Solubility Product
Salts differ enormously in solubility. As a rough guide, a salt is called soluble if more than $0.1\ \mathrm{mol}$ dissolves per litre, slightly soluble between $0.01$ and $0.1\ \mathrm{mol\,L^{-1}}$, and sparingly soluble below $0.01\ \mathrm{mol\,L^{-1}}$. In a saturated solution of a sparingly soluble ionic solid, the undissolved solid is in dynamic equilibrium with its ions — a heterogeneous equilibrium, so the solid does not appear in $K$ (Rule 6.6):
📘 Definition 6.16 — Solubility product and molar solubility
For a sparingly soluble salt $\mathrm{M}_{x}\mathrm{A}_{y}(s)\rightleftharpoons x\,\mathrm{M}^{y+}(aq)+y\,\mathrm{A}^{x-}(aq)$, the solubility product is
$$K_{sp}=[\mathrm{M}^{y+}]^{x}\,[\mathrm{A}^{x-}]^{y}$$
with the ion concentrations of the saturated solution at the given temperature. The molar solubility $S$ is the number of moles of the salt that dissolve per litre of saturated solution.
📐 Rule 6.18 — $K_{sp}$ and molar solubility in pure water
If $S\ \mathrm{mol\,L^{-1}}$ of $\mathrm{M}_{x}\mathrm{A}_{y}$ dissolves, $[\mathrm{M}^{y+}]=xS$ and $[\mathrm{A}^{x-}]=yS$, so
$$K_{sp}=x^{x}y^{y}S^{x+y}.$$
$\mathrm{AB}$ ($\mathrm{AgCl}$, $\mathrm{BaSO_{4}}$): $K_{sp}=S^{2}$. $\mathrm{AB_{2}}$ or $\mathrm{A_{2}B}$ ($\mathrm{PbI_{2}}$, $\mathrm{Mg(OH)_{2}}$, $\mathrm{Ag_{2}CrO_{4}}$): $K_{sp}=4S^{3}$. $\mathrm{AB_{3}}$: $27S^{4}$. $\mathrm{A_{2}B_{3}}$: $108S^{5}$. (Assumes the ions do not hydrolyse or form complexes.) Solubilities can be compared through $K_{sp}$ only for salts of the same formula type.
Salt
AgCl
AgBr
$\mathrm{BaSO_{4}}$
$\mathrm{PbI_{2}}$
$\mathrm{Mg(OH)_{2}}$
$\mathrm{Ag_{2}CrO_{4}}$
$K_{sp}$ (298 K)
$1.8\times10^{-10}$
$5.0\times10^{-13}$
$1.1\times10^{-10}$
$7.1\times10^{-9}$
$1.8\times10^{-11}$
$1.1\times10^{-12}$
✏️ Example 4 — From $K_{sp}$ to solubility and back
Use the table. (a) Find the molar solubility of AgCl and its solubility in $\mathrm{g\,L^{-1}}$ ($M=143.3\ \mathrm{g\,mol^{-1}}$). (b) Find the molar solubility of $\mathrm{Mg(OH)_{2}}$ and the pH of its saturated solution. (c) The molar solubility of $\mathrm{Ag_{2}CrO_{4}}$ is $6.5\times10^{-5}\ \mathrm{mol\,L^{-1}}$. Show that this agrees with $K_{sp}$. (d) $K_{sp}(\mathrm{Ag_{2}CrO_{4}})\lt K_{sp}(\mathrm{AgCl})$; which salt is more soluble in moles per litre?
💡 Show solution▶
(a) $S=\sqrt{1.8\times10^{-10}}=1.34\times10^{-5}\ \mathrm{mol\,L^{-1}}$; $1.34\times10^{-5}\times143.3=1.9\times10^{-3}\ \mathrm{g\,L^{-1}}$ — about 2 mg in a litre.
(d) $\mathrm{Ag_{2}CrO_{4}}$: $6.5\times10^{-5}\ \mathrm{mol\,L^{-1}}$, about five times the molar solubility of AgCl, despite its smaller $K_{sp}$ — the two salts are of different formula types.
The Common-Ion Effect on Solubility and Predicting Precipitation
If a saturated solution of AgCl is made in a solution that already contains chloride ions, the product $[\mathrm{Ag^{+}}][\mathrm{Cl^{-}}]$ is reached with far fewer silver ions: by Le Chatelier's principle the equilibrium $\mathrm{AgCl(s)\rightleftharpoons Ag^{+}(aq)+Cl^{-}(aq)}$ is pushed to the left, and less AgCl dissolves. The same idea tells us when a precipitate will appear on mixing two solutions.
📘 Definition 6.17 — Ionic product $Q_{sp}$
The ionic product $Q_{sp}$ of a salt is the expression for $K_{sp}$ evaluated with the ion concentrations actually present in a solution (for example just after mixing) — the reaction quotient of the dissolution equilibrium.
📐 Rule 6.19 — Precipitation and the common-ion effect on solubility
If $Q_{sp}\gt K_{sp}$ the solution is supersaturated and the salt precipitates until $Q_{sp}=K_{sp}$; if $Q_{sp}=K_{sp}$ it is just saturated; if $Q_{sp}\lt K_{sp}$ it is unsaturated and more solid can dissolve.
In a solution that already contains a common ion at concentration $c$ (much larger than the solubility), the solubility of $\mathrm{MA}$ falls to $S\approx K_{sp}/c$; for $\mathrm{MA_{2}}$ in a solution of $\mathrm{A^{-}}$, $S\approx K_{sp}/c^{2}$.
When two ions could precipitate with the same reagent, the salt whose $K_{sp}$ is reached first (at the lowest concentration of reagent) precipitates first.
The common-ion effect has many uses. Passing hydrogen chloride gas into a saturated solution of crude sodium chloride raises $[\mathrm{Cl^{-}}]$ so that pure NaCl crystallises while the more soluble impurities stay in solution. In qualitative analysis, the sulphides of group II are precipitated with $\mathrm{H_{2}S}$ in acid solution, where the common ion $\mathrm{H^{+}}$ keeps $[\mathrm{S^{2-}}]$ very low (Section 6.3, Example 5), so that only sulphides with very small $K_{sp}$ reach it; group IV sulphides need the higher $[\mathrm{S^{2-}}]$ of an ammoniacal solution. Group III hydroxides are precipitated by $\mathrm{NH_{4}OH}$ in the presence of $\mathrm{NH_{4}Cl}$, whose common ion $\mathrm{NH_{4}^{+}}$ lowers $[\mathrm{OH^{-}}]$ enough to keep $\mathrm{Mg(OH)_{2}}$ in solution.
✏️ Example 5 — The common-ion effect on solubility
(a) Find the molar solubility of AgCl in $0.10\ \mathrm{mol\,L^{-1}}$ NaCl and compare it with its solubility in water (Example 4). (b) Find the molar solubility of $\mathrm{PbI_{2}}$ in water and in $0.10\ \mathrm{mol\,L^{-1}}$ KI. (c) Find the molar solubility of $\mathrm{Mg(OH)_{2}}$ in a buffer of pH $9.00$.
💡 Show solution▶
(a) $[\mathrm{Cl^{-}}]=0.10+S\approx0.10$: $S=\dfrac{1.8\times10^{-10}}{0.10}=1.8\times10^{-9}\ \mathrm{mol\,L^{-1}}$, about $7500$ times smaller than $1.34\times10^{-5}$.
(b) water $4S^{3}=7.1\times10^{-9}$ → $S=\sqrt[3]{1.78\times10^{-9}}=1.2\times10^{-3}\ \mathrm{mol\,L^{-1}}$.
(b) in KI $[\mathrm{I^{-}}]\approx0.10$: $S=\dfrac{K_{sp}}{(0.10)^{2}}=7.1\times10^{-7}\ \mathrm{mol\,L^{-1}}$ — some 1700 times smaller.
(c) pH 9.00 → $[\mathrm{OH^{-}}]=1.0\times10^{-5}$ (fixed by the buffer). $S=[\mathrm{Mg^{2+}}]=\dfrac{1.8\times10^{-11}}{(1.0\times10^{-5})^{2}}=0.18\ \mathrm{mol\,L^{-1}}$: in this buffer $\mathrm{Mg(OH)_{2}}$ is quite soluble — the basis of keeping $\mathrm{Mg^{2+}}$ in solution with $\mathrm{NH_{4}Cl}$ present.
✏️ Example 6 — Will a precipitate form?
(a) $50.0\ \mathrm{mL}$ of $1.0\times10^{-4}\ \mathrm{mol\,L^{-1}}$ $\mathrm{AgNO_{3}}$ is mixed with $50.0\ \mathrm{mL}$ of $1.0\times10^{-4}\ \mathrm{mol\,L^{-1}}$ NaCl. Does AgCl precipitate? (b) Equal volumes of $2.0\times10^{-5}\ \mathrm{mol\,L^{-1}}$ $\mathrm{BaCl_{2}}$ and $2.0\times10^{-5}\ \mathrm{mol\,L^{-1}}$ $\mathrm{Na_{2}SO_{4}}$ are mixed. Does $\mathrm{BaSO_{4}}$ precipitate? (c) Silver nitrate is added slowly to a solution $0.010\ \mathrm{mol\,L^{-1}}$ in both $\mathrm{Cl^{-}}$ and $\mathrm{CrO_{4}^{2-}}$. Which precipitates first, and what is $[\mathrm{Cl^{-}}]$ left when red $\mathrm{Ag_{2}CrO_{4}}$ begins to appear?
💡 Show solution▶
(a) Mixing halves each concentration: $[\mathrm{Ag^{+}}]=[\mathrm{Cl^{-}}]=5.0\times10^{-5}$. $Q_{sp}=2.5\times10^{-9}\gt1.8\times10^{-10}$: yes, AgCl precipitates.
(b) $[\mathrm{Ba^{2+}}]=[\mathrm{SO_{4}^{2-}}]=1.0\times10^{-5}$; $Q_{sp}=1.0\times10^{-10}\lt1.1\times10^{-10}$: no precipitate — the mixture is just short of saturation.
(c) first $[\mathrm{Ag^{+}}]$ needed: AgCl, $\dfrac{1.8\times10^{-10}}{0.010}=1.8\times10^{-8}$; $\mathrm{Ag_{2}CrO_{4}}$, $\sqrt{\dfrac{1.1\times10^{-12}}{0.010}}=1.05\times10^{-5}\ \mathrm{mol\,L^{-1}}$. AgCl needs far less silver, so it precipitates first.
(c) left When $[\mathrm{Ag^{+}}]=1.05\times10^{-5}$, $[\mathrm{Cl^{-}}]=\dfrac{1.8\times10^{-10}}{1.05\times10^{-5}}=1.7\times10^{-5}\ \mathrm{mol\,L^{-1}}$ — over 99.8% of the chloride has already been removed. This is why chromate works as the indicator in titrations of chloride with silver nitrate.
Interactive 2 — Precipitation Predictor: $Q_{sp}$ against $K_{sp}$
Choose a salt and set the concentrations of its cation and anion in the mixture (log scales). The green line is the saturation boundary $Q_{sp}=K_{sp}$; above it (shaded) the salt precipitates, below it the solution is unsaturated. Points on the line are saturated solutions: the open circle is the salt dissolved in pure water, and moving along the line to the right shows how a common anion lowers the cation concentration that the solution can hold. Press ▶ Play to slowly add anion to a fixed amount of cation and watch the precipitate appear.
——
Q_sp—
K_sp—
verdict—
solubility in pure water—
Exercises
I
Salt hydrolysis ($298\ \mathrm{K}$; $\mathrm{p}K_{a}(\mathrm{CH_{3}COOH})=\mathrm{p}K_{b}(\mathrm{NH_{3}})=4.74$).
1 Predict whether aqueous solutions of $\mathrm{KNO_{3}}$, $\mathrm{Na_{2}CO_{3}}$, $\mathrm{NH_{4}NO_{3}}$, $\mathrm{CH_{3}COOK}$ and NaCN are acidic, basic or neutral.
2 Find the pH of $0.050\ \mathrm{mol\,L^{-1}}$ sodium acetate.
3 Find the pH of $0.020\ \mathrm{mol\,L^{-1}}$ ammonium chloride.
4 Find the hydrolysis constant of the acetate ion and its degree of hydrolysis in $0.10\ \mathrm{mol\,L^{-1}}$ sodium acetate.
5 Show that the pH of ammonium acetate solution is $7.00$ whatever its concentration.
6 Find the pH of ammonium cyanide solution ($\mathrm{p}K_{a}(\mathrm{HCN})=9.31$). Is it acidic or basic?
7 Write an equation to explain why washing soda ($\mathrm{Na_{2}CO_{3}}$) solution is alkaline.
8 Find the pH of $0.10\ \mathrm{mol\,L^{-1}}$ NaF ($K_{a}(\mathrm{HF})=3.5\times10^{-4}$).
5 $\mathrm{pH}=7+\tfrac12(\mathrm{p}K_{a}-\mathrm{p}K_{b})=7+\tfrac12(4.74-4.74)=7.00$; $c$ does not appear in the formula.
6 $\mathrm{pH}=7+\tfrac12(9.31-4.74)=9.28$: basic, because $\mathrm{NH_{3}}$ ($K_{b}=1.8\times10^{-5}$) is a stronger base than HCN ($K_{a}=4.9\times10^{-10}$) is an acid.
7 $\mathrm{CO_{3}^{2-}(aq)+H_{2}O(l)\rightleftharpoons HCO_{3}^{-}(aq)+OH^{-}(aq)}$: the carbonate ion, conjugate base of the weak acid $\mathrm{HCO_{3}^{-}}$, produces hydroxide ions.
1 What is a buffer solution? Give one acidic and one basic example.
2 Find the pH of a solution $0.20\ \mathrm{mol\,L^{-1}}$ in $\mathrm{CH_{3}COOH}$ and $0.10\ \mathrm{mol\,L^{-1}}$ in $\mathrm{CH_{3}COONa}$.
3 Find the pH of a solution $0.050\ \mathrm{mol\,L^{-1}}$ in $\mathrm{NH_{3}}$ and $0.10\ \mathrm{mol\,L^{-1}}$ in $\mathrm{NH_{4}Cl}$.
4 What ratio $[\mathrm{CH_{3}COO^{-}}]/[\mathrm{CH_{3}COOH}]$ gives pH $4.00$?
5 How many moles, and how many grams, of sodium acetate ($82.0\ \mathrm{g\,mol^{-1}}$) must be dissolved in $1.00\ \mathrm{L}$ of $0.10\ \mathrm{mol\,L^{-1}}$ acetic acid to give pH $5.20$?
6 $0.020\ \mathrm{mol}$ NaOH is added to $1.00\ \mathrm{L}$ of a buffer $0.10\ \mathrm{mol\,L^{-1}}$ in both $\mathrm{CH_{3}COOH}$ and $\mathrm{CH_{3}COONa}$. Find the new pH.
7 Why does diluting a buffer with an equal volume of water leave its pH almost unchanged?
8Applied. Blood plasma at pH $7.40$ is buffered by $\mathrm{H_{2}CO_{3}/HCO_{3}^{-}}$, with an effective $\mathrm{p}K_{a}$ of $6.10$ at body temperature. Find the ratio $[\mathrm{HCO_{3}^{-}}]/[\mathrm{H_{2}CO_{3}}]$.
💡 Show solutions▶
1 A solution that resists a change of pH when a little acid or base is added or when it is diluted. Acidic: $\mathrm{CH_{3}COOH}$ + $\mathrm{CH_{3}COONa}$; basic: $\mathrm{NH_{3}}$ + $\mathrm{NH_{4}Cl}$.
4 $\log(\text{ratio})=4.00-4.74=-0.74$, ratio $=0.18$.
5 $\log\dfrac{n}{0.10}=5.20-4.74=0.46$ → $n=0.10\times10^{0.46}=0.29\ \mathrm{mol}$, i.e. $0.29\times82.0=24\ \mathrm{g}$.
6 Acid $0.10-0.020=0.080$, salt $0.10+0.020=0.120\ \mathrm{mol}$: $\mathrm{pH}=4.74+\log1.5=4.74+0.18=4.92$.
7 Both concentrations are halved, so their ratio — and hence $\mathrm{pH}=\mathrm{p}K_{a}+\log(\text{ratio})$ — is unchanged.
8 $\log(\text{ratio})=7.40-6.10=1.30$, ratio $=10^{1.30}=20$: plasma holds about twenty times more hydrogencarbonate than carbonic acid — well suited to neutralising the acids produced by metabolism.
III
Solubility product and solubility (use the $K_{sp}$ table).
1 Write $K_{sp}$ expressions for $\mathrm{BaSO_{4}}$, $\mathrm{Ag_{2}CrO_{4}}$ and $\mathrm{Ca_{3}(PO_{4})_{2}}$.
2 Find the molar solubility of $\mathrm{BaSO_{4}}$ in water.
3 Find the molar solubility of $\mathrm{PbI_{2}}$, the concentration of iodide in its saturated solution, and its solubility in $\mathrm{g\,L^{-1}}$ ($461\ \mathrm{g\,mol^{-1}}$).
4 A saturated solution of AgBr contains $1.3\times10^{-4}\ \mathrm{g}$ per litre ($187.8\ \mathrm{g\,mol^{-1}}$). Find $K_{sp}$.
5 A salt $\mathrm{A_{2}B_{3}}$ has molar solubility $1.0\times10^{-5}\ \mathrm{mol\,L^{-1}}$. Find its $K_{sp}$.
6 Which has the larger molar solubility, AgCl or AgBr? Can you decide this from $K_{sp}$ alone? Could you for AgCl and $\mathrm{Ag_{2}CrO_{4}}$?
7 What mass of AgCl dissolves in $2.0\ \mathrm{L}$ of water at $298\ \mathrm{K}$?
8 Give the units of $K_{sp}$ for a salt of type AB and of type $\mathrm{AB_{2}}$.
6 AgCl ($1.34\times10^{-5}$ against $7.1\times10^{-7}\ \mathrm{mol\,L^{-1}}$). Yes: both are of type AB, so the larger $K_{sp}$ means the larger $S$. Not for AgCl and $\mathrm{Ag_{2}CrO_{4}}$, which have different formula types (Example 4(d)).
1 Find the molar solubility of AgCl in $0.010\ \mathrm{mol\,L^{-1}}$ NaCl.
2 Find the molar solubility of $\mathrm{BaSO_{4}}$ in $0.020\ \mathrm{mol\,L^{-1}}$ $\mathrm{Na_{2}SO_{4}}$.
3 Find the molar solubility of $\mathrm{PbI_{2}}$ in $0.050\ \mathrm{mol\,L^{-1}}$ $\mathrm{Pb(NO_{3})_{2}}$.
4 What minimum $[\mathrm{I^{-}}]$ is needed to start precipitating $\mathrm{PbI_{2}}$ from $0.10\ \mathrm{mol\,L^{-1}}$ $\mathrm{Pb^{2+}}$?
5 Equal volumes of $2.0\times10^{-4}\ \mathrm{mol\,L^{-1}}$ $\mathrm{BaCl_{2}}$ and $2.0\times10^{-4}\ \mathrm{mol\,L^{-1}}$ $\mathrm{Na_{2}SO_{4}}$ are mixed. Will $\mathrm{BaSO_{4}}$ precipitate?
6 A solution is $0.010\ \mathrm{mol\,L^{-1}}$ in $\mathrm{Mg^{2+}}$. Will $\mathrm{Mg(OH)_{2}}$ precipitate at pH $9.00$? At pH $10.00$? Find the pH at which precipitation just begins.
7 Explain why passing HCl gas into a saturated solution of impure sodium chloride gives crystals of pure NaCl.
8Applied. A barium meal is a suspension of $\mathrm{BaSO_{4}}$. Find $[\mathrm{Ba^{2+}}]$ in saturated $\mathrm{BaSO_{4}}$ solution, and explain why adding a little $\mathrm{Na_{2}SO_{4}}$ to the suspension (to make it $0.010\ \mathrm{mol\,L^{-1}}$ in sulphate) makes it even safer. By what factor does $[\mathrm{Ba^{2+}}]$ fall?
7 The added $\mathrm{Cl^{-}}$ (common ion) makes $[\mathrm{Na^{+}}][\mathrm{Cl^{-}}]$ exceed the solubility limit of NaCl, which crystallises; the impurities (such as $\mathrm{MgCl_{2}}$, $\mathrm{CaCl_{2}}$) are more soluble and their ions stay in solution.
8 In water $[\mathrm{Ba^{2+}}]=\sqrt{1.1\times10^{-10}}=1.05\times10^{-5}\ \mathrm{mol\,L^{-1}}$. With $[\mathrm{SO_{4}^{2-}}]=0.010$: $[\mathrm{Ba^{2+}}]=1.1\times10^{-8}\ \mathrm{mol\,L^{-1}}$, about 950 times lower — the common ion suppresses the dissolution of the poisonous barium.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET
E1.NEET The molar solubility $S$ of a sparingly soluble salt $\mathrm{MX_{2}}$ is related to its solubility product by (A) $K_{sp}=S^{2}$ (B) $K_{sp}=4S^{3}$ (C) $K_{sp}=2S^{3}$ (D) $K_{sp}=27S^{4}$
💡 Show solution▶
Idea $[\mathrm{M^{2+}}]=S$, $[\mathrm{X^{-}}]=2S$: $K_{sp}=S(2S)^{2}=4S^{3}$.
Answer(B).
E2.JEE MainNumerical answer. A buffer is $0.10\ \mathrm{mol\,L^{-1}}$ in $\mathrm{NH_{3}}$ and $0.010\ \mathrm{mol\,L^{-1}}$ in $\mathrm{NH_{4}Cl}$; $\mathrm{p}K_{b}(\mathrm{NH_{3}})=4.74$. Its pH at $298\ \mathrm{K}$, multiplied by 100 and rounded to the nearest integer, is ____.
💡 Show solution▶
Henderson $\mathrm{pOH}=4.74+\log\dfrac{0.010}{0.10}=4.74-1=3.74$, so $\mathrm{pH}=10.26$.
Answer1026.
E3.Board · 3 marks What is a buffer solution? Explain, with equations, how a mixture of acetic acid and sodium acetate resists a change in pH when a little HCl or NaOH is added, and find its pH when the two are present in equal concentrations ($\mathrm{p}K_{a}=4.74$).
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1 mark A buffer resists a change in pH on adding a small amount of acid or base or on dilution.
1 mark Added acid: $\mathrm{CH_{3}COO^{-}+H^{+}\rightarrow CH_{3}COOH}$; added base: $\mathrm{CH_{3}COOH+OH^{-}\rightarrow CH_{3}COO^{-}+H_{2}O}$. The ratio of salt to acid hardly changes.
1 mark $\mathrm{pH}=\mathrm{p}K_{a}+\log1=4.74$.
E4.JEE Main The solubility product of a metal hydroxide $\mathrm{M(OH)_{2}}$ is $4.0\times10^{-12}$ at $298\ \mathrm{K}$. The pH of its saturated solution is (A) $10.30$ (B) $9.70$ (C) $4.00$ (D) $10.00$
E5.JEE AdvancedOne or more options correct. At $298\ \mathrm{K}$: (A) AgCl is less soluble in $0.1\ \mathrm{mol\,L^{-1}}$ NaCl than in pure water; (B) an aqueous solution of $\mathrm{NH_{4}Cl}$ is acidic; (C) the pH of an acetate buffer changes greatly when it is diluted tenfold; (D) $K_{sp}$ of AgCl increases when NaCl is added to its saturated solution.
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(A) True — the common-ion effect ($S=K_{sp}/[\mathrm{Cl^{-}}]$).
(C) False: the ratio of salt to acid, and so the pH, is unchanged by dilution (until the solution is very dilute).
(D) False: $K_{sp}$ depends only on temperature; the solubility falls instead.
Answer(A), (B).
E6.Board · Case study Silver chloride ($K_{sp}=1.8\times10^{-10}$ at $298\ \mathrm{K}$) is one of the least soluble chlorides; this is used to test for chloride ions in water. (i) Write the expression for $K_{sp}$ of AgCl and find its molar solubility in pure water. (2 marks) (ii) A tap-water sample has $[\mathrm{Cl^{-}}]=1.0\times10^{-3}\ \mathrm{mol\,L^{-1}}$. After adding silver nitrate to make $[\mathrm{Ag^{+}}]=1.0\times10^{-4}\ \mathrm{mol\,L^{-1}}$, will a precipitate form? (1 mark) (iii) Explain why AgCl is even less soluble in sea water. (1 mark)
(ii) $Q_{sp}=(1.0\times10^{-4})(1.0\times10^{-3})=1.0\times10^{-7}\gt K_{sp}$: yes, a white precipitate forms.
(iii) Sea water contains much chloride (about $0.55\ \mathrm{mol\,L^{-1}}$); this common ion shifts $\mathrm{AgCl(s)\rightleftharpoons Ag^{+}+Cl^{-}}$ to the left and lowers the solubility.
🔬 Research–Teaching NexusResearch-tutored · optional, about 35 minutes
Research question. How does the carbonic acid–hydrogencarbonate buffer keep blood at pH $7.40$ during hard exercise, and why do the lungs matter as much as the buffer itself?
Why it matters. Sports scientists and doctors measure blood pH, $\mathrm{HCO_{3}^{-}}$ and dissolved $\mathrm{CO_{2}}$ ("blood gases") to diagnose acidosis in athletes, in diabetic patients and in intensive care. The same Henderson equation you used in this section is printed on the analysers used in hospitals around the world.
What to do.
In arterial plasma $[\mathrm{HCO_{3}^{-}}]\approx24\ \mathrm{mmol\,L^{-1}}$ and $[\mathrm{H_{2}CO_{3}}]$ (dissolved $\mathrm{CO_{2}}$) $\approx1.2\ \mathrm{mmol\,L^{-1}}$; the effective $\mathrm{p}K_{a}$ is $6.10$. Check that the Henderson equation gives pH $7.40$.
During a sprint, lactic acid adds about $5\ \mathrm{mmol\,L^{-1}}$ of $\mathrm{H^{+}}$, which reacts with $\mathrm{HCO_{3}^{-}}$. Find the new pH if the $\mathrm{H_{2}CO_{3}}$ formed stays in the blood (a "closed" system).
In reality faster breathing removes the extra $\mathrm{CO_{2}}$ and keeps $[\mathrm{H_{2}CO_{3}}]$ near $1.2\ \mathrm{mmol\,L^{-1}}$. Recalculate the pH for this "open" system.
Use Le Chatelier's principle on $\mathrm{H_{2}CO_{3}(aq)\rightleftharpoons CO_{2}(g)+H_{2}O(l)}$ to explain why breathing out $\mathrm{CO_{2}}$ raises the pH.
Compare: what would the same $5\ \mathrm{mmol\,L^{-1}}$ of $\mathrm{H^{+}}$ do to unbuffered water?
Tools: Rule 6.17, Interactive 1, Principle 6.10, a calculator.
Hand in: Your three pH values with working, and a paragraph explaining why blood is such an effective buffer only because the lungs and kidneys keep adjusting $[\mathrm{H_{2}CO_{3}}]$ and $[\mathrm{HCO_{3}^{-}}]$.
Going further. Find the normal range of arterial blood pH and of $[\mathrm{HCO_{3}^{-}}]$ used in clinical practice, and describe one condition that causes respiratory acidosis.
Instructor note: $6.10+\log(24/1.2)=6.10+1.30=7.40$. Closed: $[\mathrm{HCO_{3}^{-}}]=19$, $[\mathrm{H_{2}CO_{3}}]=6.2$ → $6.10+\log3.06=6.59$ (dangerous). Open: $6.10+\log(19/1.2)=7.30$. Removing $\mathrm{CO_{2}}$ pulls the equilibrium towards $\mathrm{CO_{2}}$, lowering $[\mathrm{H_{2}CO_{3}}]$ and raising the ratio. Unbuffered water: $[\mathrm{H^{+}}]=5\times10^{-3}$, pH $2.3$. Normal arterial pH $7.35$–$7.45$, $[\mathrm{HCO_{3}^{-}}]$ $22$–$26\ \mathrm{mmol\,L^{-1}}$; respiratory acidosis follows any failure to remove $\mathrm{CO_{2}}$ (e.g. lung disease, drug-induced slow breathing).
🧪 Self-check quiz
0/5 correct
Q1. An aqueous solution of sodium acetate is
Q2. For a buffer containing equal concentrations of a weak acid and its salt, the pH equals
Q3. If the molar solubility of $\mathrm{Ag_{2}CrO_{4}}$ is $S$, its $K_{sp}$ is
Q4. A precipitate forms when two solutions are mixed if
Q5. The molar solubility of AgCl in $0.10\ \mathrm{mol\,L^{-1}}$ NaCl ($K_{sp}=1.8\times10^{-10}$) is