Common Ion Effect, Salt Hydrolysis and Buffer Solutions

Adding an Ion That Is Already Present

Acetic acid in water settles at an ionisation equilibrium:

$$\mathrm{CH_3COOH(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + \mathrm{CH_3COO^-(aq)} \qquad K_a = 1.74 \times 10^{-5}$$

Sodium acetate is a strong electrolyte. It does not sit at an equilibrium at all; in water it comes apart completely:

$$\mathrm{CH_3COONa(aq)} \rightarrow \mathrm{Na^+(aq)} + \mathrm{CH_3COO^-(aq)}$$

Dissolve sodium acetate in the acetic acid solution and the beaker suddenly holds a large stock of acetate ion, a product of the first equilibrium. Acetate is the common ion. Two readings of what follows, and both give the same answer.

Le Chatelier's reading. The concentration of a product has been raised. The system shifts in the direction that consumes that product, which is the reverse direction. Acetate ions pick up hydronium ions, undissociated acetic acid is rebuilt, and $[\mathrm{H_3O^+}]$ drops. The acid ionises less than it did.

The $Q$ reading. Before the salt is added the mixture is at equilibrium, so $Q = K_a$. The moment acetate is dumped in, the numerator of

$$Q = \frac{[\mathrm{H_3O^+}][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]}$$

jumps while the denominator is untouched. Now $Q > K_a$, and a system with $Q$ above $K$ runs backwards until $Q$ falls back to $K_a$. It falls back by destroying $\mathrm{H_3O^+}$.

The same suppression happens from the other side. Add hydrochloric acid to acetic acid and the added $\mathrm{H_3O^+}$ is the common ion; the equilibrium again moves left and the acetic acid ionises less.

Key Point (Definition): The common ion effect is the suppression of the ionisation of a weak electrolyte caused by adding a strong electrolyte that supplies an ion already present in the ionisation equilibrium. It is Le Chatelier's principle applied to an ionic equilibrium.

$K_a$ is a constant at fixed temperature and adding a salt does not change it. What changes is the position of the equilibrium, and with it the degree of ionisation $\alpha$ and the pH. A common ion never alters $K_a$, $K_b$ or $K_w$.

[JEE/NEET] A common ion always pushes the weak electrolyte back towards its undissociated form. There is no case in which it increases ionisation.

How Large the Suppression Is

Take $0.05\ \mathrm{M}$ acetic acid, $K_a = 1.74 \times 10^{-5}$ at $298\ \mathrm{K}$.

On its own, Ostwald's dilution law gives

$$[\mathrm{H_3O^+}] = \sqrt{K_a c} = \sqrt{1.74 \times 10^{-5} \times 0.05} = 9.33 \times 10^{-4}\ \mathrm{M}, \qquad \mathrm{pH} = 3.03$$

$$\alpha = \frac{9.33 \times 10^{-4}}{0.05} = 1.87 \times 10^{-2} = 1.9\%$$

Now make the same solution $0.05\ \mathrm{M}$ in sodium acetate as well. Set up the equilibrium with $x$ as the amount of acid that ionises:

	$\mathrm{CH_3COOH}$	$\mathrm{H_3O^+}$	$\mathrm{CH_3COO^-}$
Initial (M)	$0.05$	$0$	$0.05$
Change (M)	$-x$	$+x$	$+x$
Equilibrium (M)	$0.05 - x$	$x$	$0.05 + x$

$$K_a = \frac{x(0.05 + x)}{0.05 - x}$$

The acid was already barely ionised, and the added acetate makes it ionise even less, so $x \ll 0.05$ and both $0.05 + x$ and $0.05 - x$ collapse to $0.05$:

$$1.74 \times 10^{-5} = \frac{x \times 0.05}{0.05} = x = [\mathrm{H_3O^+}], \qquad \mathrm{pH} = -\log(1.74 \times 10^{-5}) = 4.76$$

$$\alpha = \frac{1.74 \times 10^{-5}}{0.05} = 3.48 \times 10^{-4} = 0.035\%$$

The degree of ionisation has fallen from $1.9\%$ to $0.035\%$, a factor of about $54$. The pH has climbed by $1.73$ units, from $3.03$ to $4.76$.

The same arithmetic works for a weak base. Ammonia, $K_b = 1.77 \times 10^{-5}$:

$$\mathrm{NH_3(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{NH_4^+(aq)} + \mathrm{OH^-(aq)}$$

Ammonium chloride is fully ionised and floods the solution with $\mathrm{NH_4^+}$, the common ion. In $0.10\ \mathrm{M}$ ammonia alone, $[\mathrm{OH^-}] = \sqrt{1.77 \times 10^{-5} \times 0.10} = 1.33 \times 10^{-3}\ \mathrm{M}$ and $\mathrm{pH} = 11.12$. Make it $0.10\ \mathrm{M}$ in $\mathrm{NH_4Cl}$ and the same cancellation leaves $[\mathrm{OH^-}] = K_b = 1.77 \times 10^{-5}\ \mathrm{M}$, so $\mathrm{pOH} = 4.75$ and $\mathrm{pH} = 9.25$. The base is suppressed $75$-fold and the pH drops by $1.87$ units.

Two structural facts fall out of these calculations, and both matter later:

• When the weak electrolyte and its salt are at equal concentrations, $[\mathrm{H_3O^+}] = K_a$ exactly, so $\mathrm{pH} = \mathrm{p}K_a$. The acetic acid case landed on $4.76$ for that reason.
• A weak acid together with a large stock of its conjugate base behaves quite unlike the acid alone. That mixture is a buffer.

Which Ions React with Water

A salt is what an acid and a base leave behind after neutralisation. Dissolve it and it ionises completely; the question is what those ions then do to the water around them. Some simply become hydrated and stop there. Others attack water, taking a proton from it or handing one to it, and leave an excess of $\mathrm{H_3O^+}$ or $\mathrm{OH^-}$ behind. That second process is hydrolysis.

Key Point (Definition): Salt hydrolysis is the reaction of the cation or the anion of a salt with water to regenerate the parent weak acid or weak base, producing an excess of $\mathrm{OH^-}$ or $\mathrm{H_3O^+}$ and shifting the pH away from $7$.

Deciding which ion hydrolyses needs only one rule, and it is the conjugate rule from the previous section.

Anions. An anion hydrolyses if it is the conjugate base of a weak acid, because a weak acid has a conjugate base strong enough to pull a proton out of water. $\mathrm{CH_3COO^-}$, $\mathrm{CN^-}$, $\mathrm{F^-}$, $\mathrm{NO_2^-}$ and $\mathrm{CO_3^{2-}}$ all hydrolyse. $\mathrm{Cl^-}$, $\mathrm{Br^-}$, $\mathrm{I^-}$, $\mathrm{NO_3^-}$ and $\mathrm{ClO_4^-}$ come from strong acids; their conjugate bases are so feeble that they do nothing but get hydrated.

Cations. A cation hydrolyses if it is the conjugate acid of a weak base. $\mathrm{NH_4^+}$ and the anilinium and pyridinium ions do so. The cations of strong bases, $\mathrm{Na^+}$, $\mathrm{K^+}$, $\mathrm{Ca^{2+}}$ and $\mathrm{Ba^{2+}}$, are hydrated but never hydrolysed. Small highly charged metal ions such as $\mathrm{Al^{3+}}$ and $\mathrm{Fe^{3+}}$ are a separate case: their hydrated forms are acidic in their own right.

Read that rule backwards and it gives a fast prediction: the stronger partner wins. A salt of a strong acid and a weak base is acidic; a salt of a weak acid and a strong base is basic; a salt of two strong partners is neutral.

[Board] A one-line justification is expected with the prediction: name the ion that hydrolyses and write its equation with water.

The Four Combinations

1. Strong acid with strong base — $\mathrm{NaCl}$, $\mathrm{KBr}$, $\mathrm{KNO_3}$

$$\mathrm{NaCl(aq)} \rightarrow \mathrm{Na^+(aq)} + \mathrm{Cl^-(aq)}$$

$\mathrm{Na^+}$ comes from $\mathrm{NaOH}$ and $\mathrm{Cl^-}$ comes from $\mathrm{HCl}$. Neither reacts with water. Only the water equilibrium is left, and the solution is neutral: $\mathrm{pH} = 7$ at $298\ \mathrm{K}$.

2. Weak acid with strong base — $\mathrm{CH_3COONa}$, $\mathrm{NaCN}$, $\mathrm{KF}$, $\mathrm{NaNO_2}$

$$\mathrm{CH_3COONa(aq)} \rightarrow \mathrm{CH_3COO^-(aq)} + \mathrm{Na^+(aq)}$$

$$\mathrm{CH_3COO^-(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{CH_3COOH(aq)} + \mathrm{OH^-(aq)}$$

Acetic acid, $K_a = 1.74 \times 10^{-5}$, is weak, so the acetic acid formed stays largely unionised and the liberated $\mathrm{OH^-}$ is not mopped up. The solution is basic, $\mathrm{pH} > 7$. This is anionic hydrolysis.

3. Strong acid with weak base — $\mathrm{NH_4Cl}$, $\mathrm{NH_4NO_3}$, $\mathrm{(NH_4)_2SO_4}$

$$\mathrm{NH_4Cl(aq)} \rightarrow \mathrm{NH_4^+(aq)} + \mathrm{Cl^-(aq)}$$

$$\mathrm{NH_4^+(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{NH_3(aq)} + \mathrm{H_3O^+(aq)}$$

Ammonia, $K_b = 1.77 \times 10^{-5}$, is weak and holds the proton loosely enough that the reaction proceeds measurably, leaving excess $\mathrm{H_3O^+}$. The solution is acidic, $\mathrm{pH} < 7$. This is cationic hydrolysis.

4. Weak acid with weak base — $\mathrm{CH_3COONH_4}$, $\mathrm{NH_4CN}$, $\mathrm{HCOONH_4}$

$$\mathrm{CH_3COO^-(aq)} + \mathrm{NH_4^+(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{CH_3COOH(aq)} + \mathrm{NH_4OH(aq)}$$

Both ions hydrolyse at once, one making $\mathrm{OH^-}$ and the other making $\mathrm{H_3O^+}$. The result is a tug of war, and the winner is whichever parent is weaker. Hydrolysis here is extensive, and the pH sits close to $7$ but on the side of the stronger parent.

Salt type	Example	Ion that hydrolyses	Nature	pH at 298 K
Strong acid + strong base	$\mathrm{NaCl}$	neither	neutral	$= 7$
Weak acid + strong base	$\mathrm{CH_3COONa}$	anion, $\mathrm{CH_3COO^-}$	basic	$> 7$
Strong acid + weak base	$\mathrm{NH_4Cl}$	cation, $\mathrm{NH_4^+}$	acidic	$< 7$
Weak acid + weak base	$\mathrm{CH_3COONH_4}$	both ions	depends on $K_a$ vs $K_b$	near $7$

The Hydrolysis Constant and the pH Formulae

Hydrolysis is an equilibrium, so it has a constant. For the acetate ion:

$$\mathrm{A^-(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{HA(aq)} + \mathrm{OH^-(aq)} \qquad K_h = \frac{[\mathrm{HA}][\mathrm{OH^-}]}{[\mathrm{A^-}]}$$

That expression is exactly $K_b$ of the anion, so the conjugate relation gives it immediately:

$$K_h = \frac{K_w}{K_a}$$

Multiply top and bottom by $[\mathrm{H_3O^+}]$ to see it directly: $\dfrac{[\mathrm{HA}][\mathrm{OH^-}]}{[\mathrm{A^-}]} \times \dfrac{[\mathrm{H_3O^+}]}{[\mathrm{H_3O^+}]} = \dfrac{K_w}{K_a}$. For cationic hydrolysis the mirror result is $K_h = K_w/K_b$, and for a salt of a weak acid with a weak base $K_h = K_w/(K_a K_b)$.

A weaker parent acid means a smaller $K_a$ and a larger $K_h$: the weaker the acid, the more its salt hydrolyses.

Deriving the pH of a salt of a weak acid and a strong base. Let the salt concentration be $c$ and the degree of hydrolysis be $h$.

	$\mathrm{A^-}$	$\mathrm{HA}$	$\mathrm{OH^-}$
Initial (M)	$c$	$0$	$0$
Equilibrium (M)	$c(1-h)$	$ch$	$ch$

$$K_h = \frac{(ch)(ch)}{c(1-h)} = \frac{ch^2}{1-h} \approx ch^2 \quad (h \ll 1)$$

$$h = \sqrt{\frac{K_h}{c}} = \sqrt{\frac{K_w}{K_a c}}$$

$$[\mathrm{OH^-}] = ch = \sqrt{K_h c} = \sqrt{\frac{K_w c}{K_a}}$$

$$[\mathrm{H_3O^+}] = \frac{K_w}{[\mathrm{OH^-}]} = \sqrt{\frac{K_w K_a}{c}}$$

Take $-\log$ of both sides:

$$\mathrm{pH} = \tfrac{1}{2}\mathrm{p}K_w + \tfrac{1}{2}\mathrm{p}K_a + \tfrac{1}{2}\log c$$

$$\boxed{\ \mathrm{pH} = 7 + \tfrac{1}{2}\left(\mathrm{p}K_a + \log c\right) \ } \qquad \text{at } 298\ \mathrm{K}$$

Because $\log c$ is negative below $1\ \mathrm{M}$, dilution lowers the pH of such a salt towards $7$.

The identical derivation on $\mathrm{NH_4^+} + \mathrm{H_2O} \rightleftharpoons \mathrm{NH_3} + \mathrm{H_3O^+}$, with $K_h = K_w/K_b$, gives

$$\mathrm{pH} = 7 - \tfrac{1}{2}\left(\mathrm{p}K_b + \log c\right) \qquad \text{at } 298\ \mathrm{K}$$

For a salt of a weak acid with a weak base both ions hydrolyse, the concentration terms cancel between them, and the result carries no $c$ at all:

$$\mathrm{pH} = 7 + \tfrac{1}{2}\left(\mathrm{p}K_a - \mathrm{p}K_b\right) \qquad \text{at } 298\ \mathrm{K}$$

Key Point: For $\mathrm{CH_3COONH_4}$ with $\mathrm{p}K_a = 4.76$ and $\mathrm{p}K_b = 4.75$, $\mathrm{pH} = 7 + \frac{1}{2}(0.01) = 7.005$. The pH of a weak-acid-weak-base salt does not change on dilution, because concentration never enters the formula. The solution is basic if $\mathrm{p}K_a > \mathrm{p}K_b$ (acid weaker than base) and acidic if $\mathrm{p}K_a < \mathrm{p}K_b$.

[JEE Main] The three formulae differ only in signs and in which p-value appears. The safe route in an examination is to write the hydrolysis equation first, decide from it whether the solution must be acidic or basic, and only then substitute; a formula that contradicts the equation has been misremembered.

Buffer Solutions and How They Absorb Acid or Base

Add a little hydrochloric acid to a litre of pure water and the pH crashes by several units. Do the same to blood and the pH moves by a hundredth of a unit. Blood is buffered.

Key Point (Definition): A buffer solution resists change in pH when a small amount of acid or alkali is added to it, and on moderate dilution. It contains a weak acid together with its conjugate base, or a weak base together with its conjugate acid, both in appreciable concentration.

Two kinds are made in the laboratory.

Acidic buffer: a weak acid plus its salt with a strong base. $\mathrm{CH_3COOH}$ with $\mathrm{CH_3COONa}$ buffers around pH $4.76$.

Basic buffer: a weak base plus its salt with a strong acid. $\mathrm{NH_3}$ with $\mathrm{NH_4Cl}$ buffers around pH $9.25$.

The mechanism is a stockpile of two species that neutralise opposite intruders. In an acetate buffer, the salt supplies plenty of $\mathrm{CH_3COO^-}$ and the acid supplies plenty of $\mathrm{CH_3COOH}$.

Added acid meets the acetate ion:

$$\mathrm{CH_3COO^-(aq)} + \mathrm{H_3O^+(aq)} \rightarrow \mathrm{CH_3COOH(aq)} + \mathrm{H_2O(l)}$$

The invading $\mathrm{H_3O^+}$ is converted into unionised acetic acid, a weak acid that barely releases it again.

Added base meets the acetic acid:

$$\mathrm{CH_3COOH(aq)} + \mathrm{OH^-(aq)} \rightarrow \mathrm{CH_3COO^-(aq)} + \mathrm{H_2O(l)}$$

The invading $\mathrm{OH^-}$ is turned into water and acetate.

Either way the intruder is removed and only the ratio of salt to acid shifts. Since the pH depends on that ratio through a logarithm, a shift of ten or twenty per cent moves the pH by a few hundredths of a unit.

For a basic buffer the same two reactions run on the ammonia pair:

$$\mathrm{NH_3(aq)} + \mathrm{H_3O^+(aq)} \rightarrow \mathrm{NH_4^+(aq)} + \mathrm{H_2O(l)}$$

$$\mathrm{NH_4^+(aq)} + \mathrm{OH^-(aq)} \rightarrow \mathrm{NH_3(aq)} + \mathrm{H_2O(l)}$$

Buffer capacity is set by how many moles of each partner are present. Once the added acid has consumed all the acetate, the next drop behaves as it would in plain water. A buffer made from $0.5\ \mathrm{M}$ solutions absorbs five times as much acid as one made from $0.1\ \mathrm{M}$ solutions at the same pH.

A mixture of a strong acid and its salt, such as $\mathrm{HCl}$ with $\mathrm{NaCl}$, is not a buffer. $\mathrm{Cl^-}$ is too weak a base to accept a proton, so there is nothing to absorb added acid.

The Henderson-Hasselbalch Equation

Write the ionisation of the weak acid of the buffer:

$$\mathrm{HA(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + \mathrm{A^-(aq)} \qquad K_a = \frac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$$

Rearrange for the hydronium ion:

$$[\mathrm{H_3O^+}] = K_a \times \frac{[\mathrm{HA}]}{[\mathrm{A^-}]}$$

Take logarithms of both sides:

$$\log[\mathrm{H_3O^+}] = \log K_a + \log\frac{[\mathrm{HA}]}{[\mathrm{A^-}]}$$

Multiply throughout by $-1$, which turns $\log[\mathrm{H_3O^+}]$ into $-\mathrm{pH}$ and $\log K_a$ into $-\mathrm{p}K_a$, and inverts the fraction inside the last logarithm:

$$\mathrm{pH} = \mathrm{p}K_a + \log\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}$$

Two approximations turn this into a working formula. The acid is weak and its ionisation is further suppressed by the common ion, so $[\mathrm{HA}]$ is essentially the concentration of acid weighed out. Almost all the $\mathrm{A^-}$ comes from the fully ionised salt, so $[\mathrm{A^-}]$ is essentially the concentration of salt weighed out.

$$\mathrm{pH} = \mathrm{p}K_a + \log\frac{[\text{salt}]}{[\text{acid}]}$$

That is the Henderson-Hasselbalch equation. Three consequences follow at once.

Equimolar means $\mathrm{pH} = \mathrm{p}K_a$. When $[\text{salt}] = [\text{acid}]$ the ratio is $1$, $\log 1 = 0$, and the pH equals the $\mathrm{p}K_a$ of the acid. An equimolar acetic acid-sodium acetate buffer sits at $4.76$.

Dilution does not change the pH. Adding water divides salt and acid by the same factor, so the ratio, and with it the pH, is unchanged. Buffer capacity does fall, because both stocks shrink.

The useful range is $\mathrm{p}K_a \pm 1$. A ratio of $10 : 1$ gives $\mathrm{pH} = \mathrm{p}K_a + 1$, and $1 : 10$ gives $\mathrm{pH} = \mathrm{p}K_a - 1$. Outside that window one partner is so depleted that it can no longer soak up an intruder, and the buffering collapses.

For a basic buffer the same derivation applied to $K_b$ gives

$$\mathrm{pOH} = \mathrm{p}K_b + \log\frac{[\text{salt}]}{[\text{base}]}$$

and the pH follows from $\mathrm{pH} + \mathrm{pOH} = 14$ at $298\ \mathrm{K}$. Substituting $\mathrm{p}K_a + \mathrm{p}K_b = \mathrm{p}K_w$ converts it into the acid form written for the conjugate pair:

$$\mathrm{pH} = \mathrm{p}K_a + \log\frac{[\text{base}]}{[\text{conjugate acid}]}$$

where $\mathrm{p}K_a$ now belongs to the cation. For the ammonia system that $\mathrm{p}K_a$ is $9.25$, so an equimolar $\mathrm{NH_3}$-$\mathrm{NH_4Cl}$ mixture sits at pH $9.25$.

[JEE Main] The logarithm holds a ratio, so any unit cancels: moles, millimoles or molarity may be used, provided both partners are in the same volume and the same unit.

Designing a Buffer of a Required pH

The design problem is stated as: make one litre of buffer at a stated pH. The Henderson-Hasselbalch equation turns that into two decisions.

Step 1 — choose the acid. Pick a weak acid whose $\mathrm{p}K_a$ lies within one unit of the target pH, and as close to it as possible. The ratio then stays between $1:10$ and $10:1$, where the buffer has real capacity. A target of pH $4.5$ points to acetic acid, $\mathrm{p}K_a = 4.76$. A target of pH $7.2$ points to the dihydrogenphosphate ion, $\mathrm{p}K_a = 7.21$. A target of pH $9.2$ points to the ammonium ion, $\mathrm{p}K_a = 9.25$, so a base-plus-salt mixture is used.

Step 2 — fix the ratio.

$$\log\frac{[\text{salt}]}{[\text{acid}]} = \mathrm{pH} - \mathrm{p}K_a \qquad \Rightarrow \qquad \frac{[\text{salt}]}{[\text{acid}]} = 10^{(\mathrm{pH}\, -\, \mathrm{p}K_a)}$$

A target above the $\mathrm{p}K_a$ needs more salt than acid; a target below it needs more acid than salt.

Step 3 — set the absolute concentrations. The ratio fixes the pH; the concentrations fix the capacity, and totals of $0.1$ to $1\ \mathrm{M}$ are typical.

A buffer can also be made without weighing out any salt, by partial neutralisation. Add strong base amounting to less than one equivalent of the weak acid and it converts part of the acid into its salt in the same beaker. Half-neutralisation converts exactly half, making salt and acid equal, so the pH is $\mathrm{p}K_a$. This is why the half-way point of a weak-acid titration is a buffer and why the titration curve is flattest there.

Some $\mathrm{p}K_a$ values worth carrying:

Acid system	$\mathrm{p}K_a$ at 298 K	Buffer range
$\mathrm{HCOOH}$ / $\mathrm{HCOO^-}$	$3.74$	$2.74$ to $4.74$
$\mathrm{CH_3COOH}$ / $\mathrm{CH_3COO^-}$	$4.76$	$3.76$ to $5.76$
$\mathrm{H_2CO_3}$ / $\mathrm{HCO_3^-}$	$6.35$	$5.35$ to $7.35$
$\mathrm{H_2PO_4^-}$ / $\mathrm{HPO_4^{2-}}$	$7.21$	$6.21$ to $8.21$
$\mathrm{NH_4^+}$ / $\mathrm{NH_3}$	$9.25$	$8.25$ to $10.25$

[Board] A design answer must name the pair, quote the $\mathrm{p}K_a$, show the ratio calculation and state the concentrations chosen.

The Buffer That Keeps Blood at 7.4

Human blood plasma is held between pH $7.35$ and $7.45$. Enzymes lose their shape and their activity outside that band. A drop below $7.35$ is acidosis; a rise above $7.45$ is alkalosis. Both are medical emergencies at a fraction of a pH unit.

The main defence is the carbonic acid-hydrogencarbonate pair. Carbon dioxide from respiring tissue dissolves and hydrates:

$$\mathrm{CO_2(g)} \rightleftharpoons \mathrm{CO_2(aq)} \qquad \mathrm{CO_2(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_2CO_3(aq)}$$

$$\mathrm{H_2CO_3(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + \mathrm{HCO_3^-(aq)}$$

Acid produced by metabolism is absorbed by the hydrogencarbonate ion, and any base is absorbed by carbonic acid:

$$\mathrm{HCO_3^-(aq)} + \mathrm{H_3O^+(aq)} \rightarrow \mathrm{H_2CO_3(aq)} + \mathrm{H_2O(l)}$$

$$\mathrm{H_2CO_3(aq)} + \mathrm{OH^-(aq)} \rightarrow \mathrm{HCO_3^-(aq)} + \mathrm{H_2O(l)}$$

The acid of the pair is carbonic acid itself, $\mathrm{p}K_{a_1} = 6.35$, the value already listed in the table above. The Henderson-Hasselbalch equation then reads

$$\mathrm{pH} = 6.35 + \log\frac{[\mathrm{HCO_3^-}]}{[\mathrm{H_2CO_3}]}$$

and pH $7.4$ requires $\log$ of the ratio to be $7.4 - 6.35 = 1.05$, that is a ratio of $10^{1.05} \approx 11$. Blood accordingly keeps its hydrogencarbonate stock roughly eleven times its carbonic acid, well over on the base side.

An $11:1$ ratio is outside the $\mathrm{p}K_a \pm 1$ window, and a sealed bottle of this mixture would be a poor buffer. Blood escapes the limit because the system is open at both ends. The lungs blow off $\mathrm{CO_2}$, adjusting the denominator within seconds by changing the breathing rate; the kidneys retain or excrete $\mathrm{HCO_3^-}$, adjusting the numerator over hours. Both stocks are under active control, so blood absorbs far more acid than the equation alone would predict, and the lopsided ratio is deliberate: metabolism produces acid, so the reserve is kept on the base side.

The dihydrogenphosphate-hydrogenphosphate pair, $\mathrm{p}K_a = 7.21$, does the same job inside cells and in urine.

Key Point: Blood is buffered at pH $7.4$ by $\mathrm{H_2CO_3}$ and $\mathrm{HCO_3^-}$ in a ratio near $11:1$, with the lungs controlling $\mathrm{CO_2}$ and the kidneys controlling $\mathrm{HCO_3^-}$. Holding the breath raises dissolved $\mathrm{CO_2}$ and lowers blood pH; hyperventilation strips $\mathrm{CO_2}$ out and raises it.

Question 1: How far a common ion suppresses ionisation

For $0.05\ \mathrm{M}$ acetic acid, $K_a = 1.74 \times 10^{-5}$, find the degree of ionisation and the pH. Repeat for the same acid made $0.05\ \mathrm{M}$ in sodium acetate.

Answer:

Without the salt I use Ostwald's dilution law.

$\alpha = \sqrt{K_a/c} = \sqrt{1.74 \times 10^{-5}/0.05} = \sqrt{3.48 \times 10^{-4}} = 1.87 \times 10^{-2}$

$[\mathrm{H_3O^+}] = c\alpha = 0.05 \times 1.87 \times 10^{-2} = 9.35 \times 10^{-4}\ \mathrm{M}$, so $\mathrm{pH} = 3.03$.

With the salt, acetate starts at $0.05\ \mathrm{M}$. Letting $x$ be the acid that ionises, and using $x \ll 0.05$ on both sides,

$K_a = \dfrac{x(0.05 + x)}{0.05 - x} \approx \dfrac{x \times 0.05}{0.05} = x$

So $[\mathrm{H_3O^+}] = 1.74 \times 10^{-5}\ \mathrm{M}$ and $\mathrm{pH} = 4.76$.

$\alpha = 1.74 \times 10^{-5}/0.05 = 3.48 \times 10^{-4}$

Ans: Alone, $\alpha = 1.9\%$ and $\mathrm{pH} = 3.03$; with acetate, $\alpha = 0.035\%$ and $\mathrm{pH} = 4.76$. Ionisation is suppressed about $54$ times and the pH rises by $1.73$ units. Watch out: Ostwald's law cannot be used on the second solution. It assumes the only source of the anion is the acid itself, which is exactly what the added salt destroys.

Question 2: A common ion supplied by a strong acid

The ionisation constant of propanoic acid is $1.32 \times 10^{-5}$. Find its degree of ionisation in a $0.05\ \mathrm{M}$ solution and the pH. What is the degree of ionisation if the solution is also $0.01\ \mathrm{M}$ in $\mathrm{HCl}$?

Answer:

Alone: $\alpha = \sqrt{1.32 \times 10^{-5}/0.05} = \sqrt{2.64 \times 10^{-4}} = 1.62 \times 10^{-2}$.

$[\mathrm{H_3O^+}] = 0.05 \times 1.62 \times 10^{-2} = 8.12 \times 10^{-4}\ \mathrm{M}$, so $\mathrm{pH} = 3.09$.

With $\mathrm{HCl}$ present, the hydronium ion is the common ion. Hydrochloric acid is strong and fully ionised, and the propanoic acid contributes almost nothing beside it, so $[\mathrm{H_3O^+}] = 0.01\ \mathrm{M}$.

$K_a = \dfrac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]} = \dfrac{0.01 \times 0.05\alpha}{0.05} = 0.01\alpha$

$\alpha = \dfrac{1.32 \times 10^{-5}}{0.01} = 1.32 \times 10^{-3}$

Ans: $\alpha = 1.62 \times 10^{-2}$ and $\mathrm{pH} = 3.09$ alone; $\alpha = 1.32 \times 10^{-3}$ in $0.01\ \mathrm{M}$ $\mathrm{HCl}$, a suppression of about $12$ times.

Question 3: Ammonia with ammonium chloride

Calculate the pH of $0.10\ \mathrm{M}$ ammonia, $K_b = 1.77 \times 10^{-5}$, and the pH after enough $\mathrm{NH_4Cl}$ is dissolved in it to make the solution $0.10\ \mathrm{M}$ in ammonium ion.

Answer:

Alone: $[\mathrm{OH^-}] = \sqrt{K_b c} = \sqrt{1.77 \times 10^{-5} \times 0.10} = 1.33 \times 10^{-3}\ \mathrm{M}$.

$\mathrm{pOH} = -\log(1.33 \times 10^{-3}) = 2.88$, so $\mathrm{pH} = 14.00 - 2.88 = 11.12$.

With the salt, $[\mathrm{NH_4^+}] = 0.10\ \mathrm{M}$ and $[\mathrm{NH_3}] = 0.10\ \mathrm{M}$ to a good approximation.

$K_b = \dfrac{[\mathrm{NH_4^+}][\mathrm{OH^-}]}{[\mathrm{NH_3}]} = \dfrac{0.10 \times [\mathrm{OH^-}]}{0.10}$, so $[\mathrm{OH^-}] = K_b = 1.77 \times 10^{-5}\ \mathrm{M}$.

$\mathrm{pOH} = 4.75$ and $\mathrm{pH} = 9.25$.

Ans: $\mathrm{pH} = 11.12$ alone; $\mathrm{pH} = 9.25$ with the common ion. Watch out: For a base, the equimolar mixture gives $\mathrm{pOH} = \mathrm{p}K_b$, not $\mathrm{pH} = \mathrm{p}K_b$. Stopping at $4.75$ and calling it the pH is the standard slip.

Question 4: pH and degree of hydrolysis of sodium nitrite

The ionisation constant of nitrous acid is $4.5 \times 10^{-4}$. Calculate the pH of $0.04\ \mathrm{M}$ sodium nitrite and its degree of hydrolysis.

Answer:

$\mathrm{NaNO_2}$ is a salt of a weak acid with a strong base, so the nitrite ion hydrolyses:

$$\mathrm{NO_2^-(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{HNO_2(aq)} + \mathrm{OH^-(aq)}$$

$K_h = \dfrac{K_w}{K_a} = \dfrac{1.0 \times 10^{-14}}{4.5 \times 10^{-4}} = 2.22 \times 10^{-11}$

$h = \sqrt{K_h/c} = \sqrt{2.22 \times 10^{-11}/0.04} = \sqrt{5.56 \times 10^{-10}} = 2.36 \times 10^{-5}$

$[\mathrm{OH^-}] = ch = 0.04 \times 2.36 \times 10^{-5} = 9.43 \times 10^{-7}\ \mathrm{M}$

$\mathrm{pOH} = 6.03$, so $\mathrm{pH} = 7.97$.

The formula route agrees: $\mathrm{p}K_a = 3.35$, and $\mathrm{pH} = 7 + \frac{1}{2}(3.35 - 1.40) = 7.97$.

Ans: $\mathrm{pH} = 7.97$; $h = 2.4 \times 10^{-5}$, that is $0.0024\%$.

Question 5: An acid and its salt at the same concentration

The ionisation constant of chloroacetic acid is $1.35 \times 10^{-3}$. Find the pH of a $0.1\ \mathrm{M}$ solution of the acid and of a $0.1\ \mathrm{M}$ solution of its sodium salt.

Answer:

The acid first, with the usual check: $c/K_a = 0.1/(1.35 \times 10^{-3}) = 74$, far below the threshold of $400$, so $\alpha$ comes out well above 5 per cent and the square-root form is not allowed. I solve the quadratic.

$\dfrac{x^2}{0.1 - x} = 1.35 \times 10^{-3} \quad\Rightarrow\quad x^2 + 1.35 \times 10^{-3}x - 1.35 \times 10^{-4} = 0$

$x = \dfrac{-1.35 \times 10^{-3} + \sqrt{1.82 \times 10^{-6} + 5.40 \times 10^{-4}}}{2} = \dfrac{-1.35 \times 10^{-3} + 2.328 \times 10^{-2}}{2} = 1.10 \times 10^{-2}\ \mathrm{M}$

$\mathrm{pH} = -\log(1.10 \times 10^{-2}) = 1.96$

The salt hydrolyses through its anion. $\mathrm{p}K_a = -\log(1.35 \times 10^{-3}) = 2.87$.

$\mathrm{pH} = 7 + \tfrac{1}{2}(\mathrm{p}K_a + \log c) = 7 + \tfrac{1}{2}(2.87 - 1.00) = 7 + 0.94 = 7.94$

Ans: Acid, $\mathrm{pH} = 1.96$; salt, $\mathrm{pH} = 7.94$. Watch out: The shortcut $\sqrt{K_a c} = 1.16 \times 10^{-2}\ \mathrm{M}$ would have given $1.94$; with $c/K_a = 74$ it was never available, and the quadratic is what the chapter's own test demands. Chloroacetic acid is also strong enough that its salt hydrolyses only slightly, which is why $7.94$ sits nearer $7$ than the $8.88$ given by $0.1\ \mathrm{M}$ sodium acetate.

Question 6: A salt of a weak acid and a weak base

Given $\mathrm{p}K_a(\mathrm{CH_3COOH}) = 4.76$, $\mathrm{p}K_a(\mathrm{HCOOH}) = 3.74$ and $\mathrm{p}K_b(\mathrm{NH_4OH}) = 4.75$, find the pH of ammonium acetate solution and of ammonium formate solution.

Answer:

Both ions hydrolyse, so concentration drops out and

$$\mathrm{pH} = 7 + \tfrac{1}{2}(\mathrm{p}K_a - \mathrm{p}K_b)$$

Ammonium acetate: $\mathrm{pH} = 7 + \frac{1}{2}(4.76 - 4.75) = 7 + 0.005 = 7.005$.

Ammonium formate: $\mathrm{pH} = 7 + \frac{1}{2}(3.74 - 4.75) = 7 - 0.505 = 6.50$ to two decimals.

Formic acid is the stronger of the two parents, so its salt with a weak base is acidic.

Ans: Ammonium acetate, $\mathrm{pH} = 7.005$ (very nearly neutral); ammonium formate, $\mathrm{pH} = 6.50$ (acidic). Watch out: No concentration is given, and none is needed for this salt type.

Question 7: Working backwards to the base constant of pyridine

A $0.02\ \mathrm{M}$ solution of pyridinium hydrochloride is found to have $[\mathrm{H_3O^+}] = 3.36 \times 10^{-4}\ \mathrm{M}$. Calculate the ionisation constant of pyridine, and the pH of the solution.

Answer:

The salt comes from a strong acid and the weak base pyridine, so the pyridinium cation hydrolyses:

$$\mathrm{C_5H_5NH^+(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{C_5H_5N(aq)} + \mathrm{H_3O^+(aq)}$$

$\mathrm{pH} = -\log(3.36 \times 10^{-4}) = 3.47$

$h = \dfrac{3.36 \times 10^{-4}}{0.02} = 1.68 \times 10^{-2}$

$K_h = ch^2 = 0.02 \times (1.68 \times 10^{-2})^2 = 5.64 \times 10^{-6}$

$K_h = \dfrac{K_w}{K_b}$, so $K_b = \dfrac{1.0 \times 10^{-14}}{5.64 \times 10^{-6}} = 1.77 \times 10^{-9}$.

Ans: $K_b(\text{pyridine}) = 1.77 \times 10^{-9}$; $\mathrm{pH} = 3.47$ Watch out: $K_h$ here is $K_w/K_b$ because the cation is hydrolysing. Dividing by $K_a$ instead inverts the whole calculation.

Question 8: What a buffer actually saves you from

One litre of solution contains $0.10\ \mathrm{mol}$ acetic acid and $0.10\ \mathrm{mol}$ sodium acetate, $\mathrm{p}K_a = 4.76$. Find the pH, and the pH after adding $0.010\ \mathrm{mol}$ of $\mathrm{HCl}$ and, separately, $0.010\ \mathrm{mol}$ of $\mathrm{NaOH}$. Compare with adding the same acid to a litre of pure water.

Answer:

Initially the ratio is $1$, so $\mathrm{pH} = \mathrm{p}K_a = 4.76$.

Added $\mathrm{H_3O^+}$ converts acetate into acetic acid, mole for mole: salt $0.10 - 0.010 = 0.090\ \mathrm{mol}$, acid $0.10 + 0.010 = 0.110\ \mathrm{mol}$.

$\mathrm{pH} = 4.76 + \log\dfrac{0.090}{0.110} = 4.76 - 0.09 = 4.67$

Added $\mathrm{OH^-}$ converts acid into acetate: salt $0.110$, acid $0.090$.

$\mathrm{pH} = 4.76 + \log\dfrac{0.110}{0.090} = 4.76 + 0.09 = 4.85$

In pure water, $0.010\ \mathrm{mol}$ of $\mathrm{HCl}$ per litre gives $[\mathrm{H_3O^+}] = 0.010\ \mathrm{M}$ and $\mathrm{pH} = 2.00$.

Ans: Buffer moves from $4.76$ to $4.67$ or $4.85$, a change of $0.09$; water moves from $7.00$ to $2.00$, a change of $5.00$.

Question 9: A buffer made by partial neutralisation

$50.0\ \mathrm{mL}$ of $0.10\ \mathrm{M}$ ammonia is treated with $25.0\ \mathrm{mL}$ of $0.10\ \mathrm{M}$ $\mathrm{HCl}$. Calculate the pH of the mixture. $K_b = 1.77 \times 10^{-5}$.

Answer:

Millimoles: ammonia $50.0 \times 0.10 = 5.0$; $\mathrm{HCl}$ $25.0 \times 0.10 = 2.5$.

$$\mathrm{NH_3} + \mathrm{HCl} \rightarrow \mathrm{NH_4^+} + \mathrm{Cl^-}$$

The acid is limiting. It consumes $2.5\ \mathrm{mmol}$ of ammonia and produces $2.5\ \mathrm{mmol}$ of $\mathrm{NH_4^+}$, leaving $2.5\ \mathrm{mmol}$ of ammonia. Total volume is $75.0\ \mathrm{mL}$, so each is $0.033\ \mathrm{M}$.

Base and its conjugate acid are present in equal amounts, so

$[\mathrm{OH^-}] = K_b \times \dfrac{[\mathrm{NH_3}]}{[\mathrm{NH_4^+}]} = 1.77 \times 10^{-5}\ \mathrm{M}$

$[\mathrm{H_3O^+}] = \dfrac{1.0 \times 10^{-14}}{1.77 \times 10^{-5}} = 5.6 \times 10^{-10}\ \mathrm{M}$, so $\mathrm{pH} = 9.25$.

Ans: $\mathrm{pH} = 9.25$ Watch out: This is the half-neutralisation point, so the pH is fixed by $\mathrm{p}K_a$ of $\mathrm{NH_4^+}$ alone and the dilution to $75\ \mathrm{mL}$ cancels in the ratio.

Question 10: Designing a buffer to order

How much sodium acetate must be dissolved in one litre of $0.10\ \mathrm{M}$ acetic acid to give a buffer of pH $4.30$? Take $\mathrm{p}K_a = 4.76$ and $M(\mathrm{CH_3COONa}) = 82\ \mathrm{g\,mol^{-1}}$. Assume no volume change.

Answer:

$4.30$ lies within one unit of $4.76$, so acetic acid is a sound choice.

$\log\dfrac{[\text{salt}]}{[\text{acid}]} = \mathrm{pH} - \mathrm{p}K_a = 4.30 - 4.76 = -0.46$

$\dfrac{[\text{salt}]}{[\text{acid}]} = 10^{-0.46} = 0.347$

$[\text{salt}] = 0.347 \times 0.10 = 0.0347\ \mathrm{M}$

Mass $= 0.0347 \times 82 = 2.85\ \mathrm{g}$.

Ans: About $2.8\ \mathrm{g}$ of sodium acetate per litre. Watch out: The target is below the $\mathrm{p}K_a$, so the answer must contain less salt than acid. A ratio greater than $1$ here means the subtraction was done the wrong way round.

Question 11: The bicarbonate ratio in blood

Blood plasma is buffered by the $\mathrm{H_2CO_3}$-$\mathrm{HCO_3^-}$ pair, $\mathrm{p}K_{a_1} = 6.35$. Find the ratio needed for pH $7.4$, and the pH that results if illness halves that ratio.

Answer:

$\log\dfrac{[\mathrm{HCO_3^-}]}{[\mathrm{H_2CO_3}]} = \mathrm{pH} - \mathrm{p}K_{a_1} = 7.4 - 6.35 = 1.05$

$\dfrac{[\mathrm{HCO_3^-}]}{[\mathrm{H_2CO_3}]} = 10^{1.05} = 11.2$

Halved, the ratio is $5.6$, and $\mathrm{pH} = 6.35 + \log 5.6 = 6.35 + 0.75 = 7.10$.

Ans: A ratio near $11 : 1$ holds blood at $7.4$; halving it to about $5.6 : 1$ drops the pH to $7.10$, which is severe acidosis. Watch out: A drop of $0.3$ in pH looks small and is a doubling of $[\mathrm{H_3O^+}]$.