JEE Corner — Simultaneous Equilibria, Degree of Dissociation and Advanced Ionic Equilibrium

Reading Dissociation off a Density Measurement

A gas that dissociates in a closed vessel keeps its mass and gains molecules. Two moles of $\mathrm{NO_2}$ weigh exactly what the one mole of $\mathrm{N_2O_4}$ they came from weighed, but they occupy twice the volume at the same temperature and pressure. Anything that measures mass per mole therefore falls as dissociation proceeds, and measuring how far it has fallen measures the degree of dissociation.

Key Point (Definition): The degree of dissociation $\alpha$ is the fraction of the reactant originally taken that has dissociated when equilibrium is reached. It runs from $0$ to $1$; multiplied by 100 it is the percentage dissociation.

The quantity actually measured in the laboratory is the vapour density.

Key Point (Definition): The vapour density of a gas is the mass of a given volume of that gas divided by the mass of the same volume of hydrogen, both measured at the same temperature and pressure. For an ideal gas this equals $M/2$, where $M$ is the molar mass in $\mathrm{g\,mol^{-1}}$.

Two symbols are used throughout, and mixing them up is the commonest error in the topic. $D$ is the theoretical vapour density, calculated from the formula of the undissociated substance. $d$ is the observed vapour density, measured on the equilibrium mixture. Dissociation makes $d$ smaller than $D$.

The derivation

Write the dissociation so that one mole of reactant appears on the left, and let it produce $n$ moles of gaseous product in total:

$$\mathrm{A(g)} \rightleftharpoons n\,\mathrm{B(g)}$$

Step 1. Start with 1 mol of $\mathrm{A}$ and let a fraction $\alpha$ of it dissociate. At equilibrium there is $(1-\alpha)$ mol of $\mathrm{A}$ left and $n\alpha$ mol of product formed.

Step 2. Add the moles present at equilibrium:

$$n_{\text{total}} = (1-\alpha) + n\alpha = 1 + (n-1)\alpha$$

Step 3. Mass is conserved. The mass of the sample equals moles multiplied by molar mass, both before and after:

$$1 \times M_{\text{theoretical}} = \left[1 + (n-1)\alpha\right] \times M_{\text{observed}}$$

Step 4. Divide, and replace each molar mass by twice its vapour density, since the factor 2 cancels:

$$\frac{D}{d} = \frac{M_{\text{theoretical}}}{M_{\text{observed}}} = 1 + (n-1)\alpha$$

Step 5. Make $\alpha$ the subject:

$$\alpha = \frac{D - d}{(n-1)\,d} \qquad\text{equivalently}\qquad \alpha = \frac{M_{\text{theoretical}} - M_{\text{observed}}}{(n-1)\,M_{\text{observed}}}$$

The ratio $D/d$ is the gas-phase counterpart of the van't Hoff factor: it is the number of particles now present for every particle originally taken, $i = 1 + (n-1)\alpha$.

What $n$ actually counts

$n$ is the number of moles of gaseous product formed from one mole of the reactant. The equation must be rewritten with a coefficient of 1 on the reactant before $n$ is read off, even if fractions appear on the right.

Dissociation, written per mole of reactant	$n$	$\alpha$ in terms of $D$ and $d$
$\mathrm{PCl_5(g)} \rightleftharpoons \mathrm{PCl_3(g)} + \mathrm{Cl_2(g)}$	2	$(D-d)/d$
$\mathrm{N_2O_4(g)} \rightleftharpoons 2\mathrm{NO_2(g)}$	2	$(D-d)/d$
$\mathrm{NH_4Cl(g)} \rightleftharpoons \mathrm{NH_3(g)} + \mathrm{HCl(g)}$	2	$(D-d)/d$
$\mathrm{NH_3(g)} \rightleftharpoons \frac{1}{2}\mathrm{N_2(g)} + \frac{3}{2}\mathrm{H_2(g)}$	2	$(D-d)/d$
$\mathrm{SO_3(g)} \rightleftharpoons \mathrm{SO_2(g)} + \frac{1}{2}\mathrm{O_2(g)}$	$1.5$	$2(D-d)/d$
$\mathrm{N_2O_5(g)} \rightleftharpoons 2\mathrm{NO_2(g)} + \frac{1}{2}\mathrm{O_2(g)}$	$2.5$	$(D-d)/1.5d$

For the first four the formula collapses to the memorable $\alpha = (D-d)/d$, which is why so many students apply it everywhere. It is only correct when $n = 2$.

The pressure version of the same result

Nothing in the derivation depends on density as such — only on the mole number changing at constant mass. If the pure reactant would exert a pressure $p_0$ in the vessel at that temperature, and the equilibrium mixture exerts $P$, then at constant volume and temperature pressure is proportional to moles:

$$\frac{P}{p_0} = 1 + (n-1)\alpha$$

An observed total pressure and a calculated undissociated pressure give $\alpha$ exactly as a pair of densities do.

The conditions attached to the formula

Every step above used an assumption, and each one is a place where a question can be set.

1. Ideal gas behaviour. $D = M/2$ and $P \propto n$ both come from $pV = nRT$. Near condensation, or at high pressure, the relation drifts.
2. One dissociation only. If the product dissociates further, or two reactions run at once, the single $n$ in the formula no longer describes the mixture.
3. The sample must be entirely in the gas phase. A solid or liquid residue holds back mass that the calculation assumes is in the vapour.
4. $D$ must come from the formula mass of the undissociated species, and $d$ must be measured at the same temperature and pressure.
5. The vapour must not associate. If $d$ comes out greater than $D$, molecules are combining rather than splitting — acetic acid vapour dimerising through hydrogen bonds is the standard case — and a different relation applies. The sign of $D - d$ tells which process is happening before any arithmetic is done.

Reading $\alpha$ back out of a measurement

Four different measurements all lead to the same $\alpha$, and a question may supply any one of them.

Measurement given	Relation used
Theoretical and observed vapour density	$\alpha = \dfrac{D-d}{(n-1)d}$
Theoretical and observed molar mass	$\alpha = \dfrac{M_{\text{th}} - M_{\text{obs}}}{(n-1)M_{\text{obs}}}$
Undissociated and observed pressure, constant $V$ and $T$	$\alpha = \dfrac{P - p_0}{(n-1)p_0}$
Initial and final total moles, constant $V$ and $T$	$\alpha = \dfrac{n_{\text{final}} - n_{\text{initial}}}{(n-1)n_{\text{initial}}}$

All four are the single statement $1 + (n-1)\alpha$ = (particles now)/(particles taken), read through whichever property was measured. They differ in two places, not one. First the numerator: density and molar mass fall on dissociation while pressure and mole number rise, so the subtraction is written the other way round in the lower two rows. Second the denominator: it is always whichever quantity sits below the line in that ratio, which is the observed density or molar mass in the upper two rows but the initial pressure or mole number in the lower two.

[JEE Main] A question that gives a vapour density and asks for $K_p$ is really two questions joined: get $\alpha$ from the densities, then feed $\alpha$ into the $K_p$ expression of the next block.

$K_p$ in Terms of $\alpha$ and Total Pressure

Once $\alpha$ is known, the equilibrium constant follows without ever converting to concentrations. The route is always the same: moles at equilibrium, then mole fractions, then partial pressures as mole fraction multiplied by total pressure $P$, then substitution.

Case 1: $\mathrm{A(g)} \rightleftharpoons 2\mathrm{B(g)}$

This covers $\mathrm{N_2O_4} \rightleftharpoons 2\mathrm{NO_2}$, $\mathrm{Cl_2} \rightleftharpoons 2\mathrm{Cl}$ and $\mathrm{I_2} \rightleftharpoons 2\mathrm{I}$ — every case in which one mole of reactant splits into two moles of the same product. $\mathrm{PCl_5}$ does not belong here: its two products are different species, and it follows Case 2 below.

Step 1. Start with 1 mol of $\mathrm{A}$; at equilibrium, $\mathrm{A} = 1-\alpha$, $\mathrm{B} = 2\alpha$, total $= 1+\alpha$.

Step 2. Mole fractions are each species divided by the total:

$$x_{\mathrm{A}} = \frac{1-\alpha}{1+\alpha} \qquad x_{\mathrm{B}} = \frac{2\alpha}{1+\alpha}$$

Step 3. Partial pressure is mole fraction multiplied by the total pressure $P$:

$$p_{\mathrm{A}} = \frac{(1-\alpha)P}{1+\alpha} \qquad p_{\mathrm{B}} = \frac{2\alpha P}{1+\alpha}$$

Step 4. Substitute into $K_p = p_{\mathrm{B}}^{2}/p_{\mathrm{A}}$:

$$K_p = \frac{\left(\dfrac{2\alpha P}{1+\alpha}\right)^{2}}{\dfrac{(1-\alpha)P}{1+\alpha}} = \frac{4\alpha^{2}P^{2}}{(1+\alpha)^{2}} \times \frac{1+\alpha}{(1-\alpha)P}$$

Step 5. Cancel one factor of $(1+\alpha)$ and one of $P$, and use $(1+\alpha)(1-\alpha) = 1-\alpha^{2}$:

$$\boxed{K_p = \frac{4\alpha^{2}P}{1-\alpha^{2}}}$$

Step 6. Inverting for $\alpha$: cross-multiplying gives $K_p - K_p\alpha^{2} = 4\alpha^{2}P$, so

$$\alpha = \sqrt{\frac{K_p}{K_p + 4P}}$$

Case 2: $\mathrm{A(g)} \rightleftharpoons \mathrm{B(g)} + \mathrm{C(g)}$

This is the $\mathrm{PCl_5} \rightleftharpoons \mathrm{PCl_3} + \mathrm{Cl_2}$ pattern, and also $\mathrm{NH_4Cl}$ vapour and $\mathrm{COCl_2}$. The total is again $1+\alpha$, but the two products each carry a coefficient of 1:

$$p_{\mathrm{B}} = p_{\mathrm{C}} = \frac{\alpha P}{1+\alpha} \qquad p_{\mathrm{A}} = \frac{(1-\alpha)P}{1+\alpha}$$

$$K_p = \frac{p_{\mathrm{B}}\,p_{\mathrm{C}}}{p_{\mathrm{A}}} = \frac{\alpha^{2}P^{2}}{(1+\alpha)^{2}} \times \frac{1+\alpha}{(1-\alpha)P} = \frac{\alpha^{2}P}{1-\alpha^{2}} \qquad \alpha = \sqrt{\frac{K_p}{K_p + P}}$$

The only difference from Case 1 is the factor 4, and it comes from squaring the coefficient 2 in $2\mathrm{B}$. Dropping it is the single most common slip in this calculation.

The standard cases collected

Equilibrium	Total moles	$K_p$	$\alpha$	$K_c$ (volume $V$, 1 mol taken)
$\mathrm{A} \rightleftharpoons 2\mathrm{B}$	$1+\alpha$	$\dfrac{4\alpha^{2}P}{1-\alpha^{2}}$	$\sqrt{\dfrac{K_p}{K_p+4P}}$	$\dfrac{4\alpha^{2}}{(1-\alpha)V}$
$\mathrm{A} \rightleftharpoons \mathrm{B}+\mathrm{C}$	$1+\alpha$	$\dfrac{\alpha^{2}P}{1-\alpha^{2}}$	$\sqrt{\dfrac{K_p}{K_p+P}}$	$\dfrac{\alpha^{2}}{(1-\alpha)V}$
$\mathrm{A} \rightleftharpoons 3\mathrm{B}$	$1+2\alpha$	$\dfrac{27\alpha^{3}P^{2}}{(1-\alpha)(1+2\alpha)^{2}}$	solved numerically	$\dfrac{27\alpha^{3}}{(1-\alpha)V^{2}}$
$2\mathrm{A} \rightleftharpoons \mathrm{B}+\mathrm{C}$	$1$ (unchanged)	$\dfrac{\alpha^{2}}{4(1-\alpha)^{2}}$	independent of $P$	same as $K_p$

The last row is worth a moment. When the number of gaseous moles does not change, $P$ cancels out of the expression entirely, and $\alpha$ is fixed by temperature alone. $\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$ and the dissociation of $\mathrm{HI}$ behave this way, which is why a vapour-density measurement on them shows nothing at all — $d$ equals $D$ however far the reaction has gone.

How $\alpha$ responds to pressure

Take $K_p = 4\alpha^{2}P/(1-\alpha^{2})$ and hold the temperature fixed, so $K_p$ is a constant. Raising $P$ must be paid for by lowering $\alpha^{2}/(1-\alpha^{2})$, which means lowering $\alpha$. Compression suppresses dissociation whenever the dissociation increases the gaseous mole number, which is exactly what Le Chatelier's principle predicts, now with a number attached.

When $\alpha$ is small, $1-\alpha^{2} \approx 1$ and

$$\alpha \approx \sqrt{\frac{K_p}{4P}} \quad\text{(Case 1)} \qquad \alpha \approx \sqrt{\frac{K_p}{P}} \quad\text{(Case 2)}$$

so $\alpha$ falls off as $P^{-1/2}$. Quadrupling the pressure roughly halves the degree of dissociation. That approximation needs $\alpha$ small — below about $0.1$ it is good to a few per cent, and at $\alpha = 0.5$ it is useless.

Adding an inert gas at constant volume changes neither the partial pressures nor $\alpha$: the extra gas raises the total pressure but every $p_i$ stays where it was. Adding an inert gas at constant total pressure forces the volume up, dilutes every reacting species, and pushes dissociation forward exactly as a genuine pressure reduction would.

[JEE Main] Recognise the pattern by counting the moles on the right of the equation written for one mole of reactant. Two moles gives the $1+\alpha$ denominator; the factor 4 appears only when both moles are the same species.

Simultaneous and Coupled Equilibria

Two equilibria that share a species cannot be solved separately. The shared species has one concentration in the vessel, and both equilibrium expressions must be satisfied by that single value at the same time.

The combination rule

If a third equation is obtained by adding two others, its equilibrium constant is the product of theirs at the same temperature:

$$\text{Reaction 3} = \text{Reaction 1} + \text{Reaction 2} \quad\Longrightarrow\quad K_3 = K_1 \times K_2$$

The reason is mechanical: writing the two expressions and multiplying them cancels the concentration of every species that appears as a product in one and a reactant in the other. Two consequences are used constantly.

Stepwise ionisation. For hydrogen sulphide,

$$\mathrm{H_2S} \rightleftharpoons \mathrm{H^+} + \mathrm{HS^-},\ K_{a_1} = 9.1 \times 10^{-8} \qquad \mathrm{HS^-} \rightleftharpoons \mathrm{H^+} + \mathrm{S^{2-}},\ K_{a_2} = 1.2 \times 10^{-13}$$

$$\mathrm{H_2S} \rightleftharpoons 2\mathrm{H^+} + \mathrm{S^{2-}} \qquad K_{\text{net}} = K_{a_1}K_{a_2} = 1.1 \times 10^{-20}$$

Rearranged, this is the master equation of qualitative analysis:

$$[\mathrm{S^{2-}}] = \frac{K_{a_1}K_{a_2}[\mathrm{H_2S}]}{[\mathrm{H^+}]^{2}}$$

The sulphide concentration is controlled by the square of the hydrogen ion concentration. Saturating a solution with $\mathrm{H_2S}$ holds $[\mathrm{H_2S}]$ near $0.1\ \mathrm{M}$; the acidity then sets $[\mathrm{S^{2-}}]$ anywhere over about six orders of magnitude. In $0.1\ \mathrm{M}$ HCl the sulphide ion concentration is only about $1.0 \times 10^{-19}\ \mathrm{M}$, while in $0.1\ \mathrm{M}$ $\mathrm{H_2S}$ alone it is near $1.2 \times 10^{-13}\ \mathrm{M}$.

Dissolution helped by complex formation. Silver chloride dissolves in ammonia because two equilibria are added:

$$\mathrm{AgCl(s)} \rightleftharpoons \mathrm{Ag^+} + \mathrm{Cl^-},\ K_{sp} = 1.8 \times 10^{-10}$$ $$\mathrm{Ag^+} + 2\mathrm{NH_3} \rightleftharpoons [\mathrm{Ag(NH_3)_2}]^{+},\ K_f = 1.6 \times 10^{7}$$ $$\mathrm{AgCl(s)} + 2\mathrm{NH_3} \rightleftharpoons [\mathrm{Ag(NH_3)_2}]^{+} + \mathrm{Cl^-},\ K = K_{sp}K_f = 2.9 \times 10^{-3}$$

A net constant of $10^{-3}$ is small but workable, so ammonia dissolves silver chloride. For silver iodide, $K_{sp} = 8.3 \times 10^{-17}$ gives a net constant near $1.3 \times 10^{-9}$, and ammonia leaves it untouched. The complexing agent removes $\mathrm{Ag^+}$, the shared species, and the solid dissolves to replace it.

Two solids sharing a gas

The classic gas-phase case has two solids decomposing in the same vessel, each giving a different gas plus a common one:

$$\mathrm{A(s)} \rightleftharpoons \mathrm{B(g)} + \mathrm{C(g)},\ K_{p_1} \qquad \mathrm{D(s)} \rightleftharpoons \mathrm{E(g)} + \mathrm{C(g)},\ K_{p_2}$$

Let $p_{\mathrm{B}} = x$ and $p_{\mathrm{E}} = y$. Every molecule of $\mathrm{C}$ came from one solid or the other, so $p_{\mathrm{C}} = x + y$. The two constants read

$$K_{p_1} = x(x+y) \qquad K_{p_2} = y(x+y)$$

Adding them gives $K_{p_1} + K_{p_2} = (x+y)^{2}$, so

$$p_{\mathrm{C}} = \sqrt{K_{p_1} + K_{p_2}} \qquad x = \frac{K_{p_1}}{p_{\mathrm{C}}} \qquad y = \frac{K_{p_2}}{p_{\mathrm{C}}}$$

$$P_{\text{total}} = x + y + p_{\mathrm{C}} = 2\sqrt{K_{p_1} + K_{p_2}}$$

The result holds only for this shape — both solids giving two gaseous moles, one of them common, and both solids still present at equilibrium. If either solid runs out, its equilibrium no longer exists and the expression fails.

What sharing does to the shared species

The shared species ends up at a higher concentration than either equilibrium alone would give, and every species that is not shared ends up lower. Solid $\mathrm{A}$ alone would give $p_{\mathrm{C}} = \sqrt{K_{p_1}}$; with $\mathrm{D}$ present as well, $p_{\mathrm{C}} = \sqrt{K_{p_1}+K_{p_2}}$, which is larger. Since $K_{p_1} = x\,p_{\mathrm{C}}$ is fixed, a larger $p_{\mathrm{C}}$ forces a smaller $x$.

Key Point: Any second source of a shared species pushes the first equilibrium backwards. The common ion effect in solution, the suppression of $\mathrm{PCl_5}$ dissociation by added chlorine, and the suppression of $\mathrm{H_2S}$ release by added ammonia are one principle wearing three costumes.

The gas-phase version is worth writing out. Heating $\mathrm{PCl_5}$ with extra $\mathrm{Cl_2}$ already present changes the mole table: starting from 1 mol $\mathrm{PCl_5}$ and $b$ mol $\mathrm{Cl_2}$, at equilibrium there are $(1-\alpha)$, $\alpha$ and $(b+\alpha)$ moles with a total of $(1+b+\alpha)$, and

$$K_p = \frac{\alpha(b+\alpha)}{(1-\alpha)(1+b+\alpha)}\,P$$

Setting $b = 0$ recovers $\alpha^{2}P/(1-\alpha^{2})$. For any $b > 0$ the same $K_p$ and $P$ give a smaller $\alpha$.

Water as the universal shared species

In every aqueous problem, $\mathrm{H_3O^+}$ and $\mathrm{OH^-}$ are shared between the solute's equilibrium and water's own autoionisation. The two are linked by $K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}]$, which is $1.0 \times 10^{-14}$ at 298 K and changes with temperature. Most of the time the solute swamps water and the coupling is ignored. When the solution is very dilute, or the acid very weak, it cannot be — which is the subject of the next block.

The Exact Treatment of a Weak Acid

The working formula $[\mathrm{H_3O^+}] = \sqrt{K_a c}$ is an approximation, and knowing when it fails is worth more marks than knowing the formula.

Where the approximation enters

For a monoprotic weak acid $\mathrm{HA}$ at initial concentration $c$, with $x$ the amount ionised per litre:

$$\mathrm{HA} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{A^-} \qquad K_a = \frac{x^{2}}{c - x}$$

Replacing $c - x$ by $c$ gives $x = \sqrt{K_a c}$. The exact statement is a quadratic:

$$x^{2} + K_a x - K_a c = 0 \qquad x = \frac{-K_a + \sqrt{K_a^{2} + 4K_a c}}{2}$$

The negative root is discarded because a concentration cannot be negative. In terms of the degree of ionisation $\alpha = x/c$, the same equation is Ostwald's dilution law without any approximation:

$$K_a = \frac{c\alpha^{2}}{1-\alpha} \qquad \alpha = \frac{-K_a + \sqrt{K_a^{2} + 4K_ac}}{2c}$$

and the familiar $\alpha = \sqrt{K_a/c}$ is what is left when $1-\alpha$ is set to 1.

The $c/K_a \ge 400$ test

The usual working rule is that the approximation is acceptable while $\alpha \le 0.05$, a 5% ionisation. Turning that into a test on the data:

Step 1. Accept the approximate $\alpha \approx \sqrt{K_a/c}$ for the purpose of the test itself.

Step 2. Impose $\alpha \le 0.05$, so $K_a/c \le 0.0025$.

Step 3. Invert: $c/K_a \ge 400$.

Key Point: Compute $c/K_a$ before anything else. If it is at least $400$, use $[\mathrm{H_3O^+}] = \sqrt{K_a c}$. If it is smaller, solve the quadratic — the approximation will overstate $[\mathrm{H_3O^+}]$.

The approximation always errs the same way. The exact $x$ satisfies $x = \sqrt{K_a(c-x)}$, which is smaller than $\sqrt{K_ac}$, so the approximate answer is always too high and the approximate pH always too low. The size of the error:

$c/K_a$	Exact $\alpha$	Approximate $\alpha$	Error in $[\mathrm{H_3O^+}]$	Error in pH
$10000$	$0.00995$	$0.01000$	$+0.5\%$	$0.002$
$400$	$0.0488$	$0.0500$	$+2.5\%$	$0.011$
$100$	$0.0951$	$0.1000$	$+5.1\%$	$0.022$
$10$	$0.270$	$0.316$	$+17\%$	$0.069$
$1$	$0.618$	$1.000$	$+62\%$	$0.21$

At $c/K_a = 400$ the pH is wrong in the second decimal place, by $0.011$, which no examination cares about. At $c/K_a = 1$ the approximation predicts complete ionisation of a weak acid, which is nonsense. The rule is a boundary between harmless and harmful, not between right and wrong.

Bringing water into the calculation

Two exact conditions govern the solution. Charge balance says the positive and negative charges must sum to zero:

$$[\mathrm{H_3O^+}] = [\mathrm{A^-}] + [\mathrm{OH^-}]$$

Mass balance says all the acid taken is somewhere:

$$c = [\mathrm{HA}] + [\mathrm{A^-}]$$

The $[\mathrm{OH^-}]$ term is what the ordinary treatment throws away. Keeping it, and writing $[\mathrm{A^-}] = K_a[\mathrm{HA}]/[\mathrm{H_3O^+}]$ and $[\mathrm{OH^-}] = K_w/[\mathrm{H_3O^+}]$:

$$[\mathrm{H_3O^+}] = \frac{K_a[\mathrm{HA}]}{[\mathrm{H_3O^+}]} + \frac{K_w}{[\mathrm{H_3O^+}]}$$

Multiplying through by $[\mathrm{H_3O^+}]$ and, where ionisation is slight, putting $[\mathrm{HA}] \approx c$:

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

That one extra term is all that is usually needed. Solving the two balances without any approximation gives a cubic,

$$[\mathrm{H_3O^+}]^{3} + K_a[\mathrm{H_3O^+}]^{2} - (K_ac + K_w)[\mathrm{H_3O^+}] - K_aK_w = 0$$

which is beyond what is ever required, but its existence is the reason the simpler forms are called approximations.

When water matters. Compare $K_ac$ with $K_w$. Water contributes about 1% when $K_ac \approx 10^{4}K_w$, and dominates when $K_ac < K_w$. Two situations trigger it: an extremely weak acid such as HCN or phenol, and an extremely dilute solution of any acid. A $10^{-5}\ \mathrm{M}$ solution of HCN has $K_ac = 4.9 \times 10^{-15}$, smaller than $K_w$; ignoring water gives $\mathrm{pH} = 7.15$ for an acid, which cannot be right.

The strong-acid version of the same correction, for an acid of concentration $C_a$ that ionises completely, is

$$[\mathrm{H_3O^+}] = \frac{C_a + \sqrt{C_a^{2} + 4K_w}}{2}$$

For $10^{-8}\ \mathrm{M}$ HCl this returns $1.05 \times 10^{-7}\ \mathrm{M}$ and $\mathrm{pH} = 6.98$, slightly acidic, rather than the impossible $\mathrm{pH} = 8$ from the naive calculation. Adding acid to water can never take the pH above the neutral value.

The temperature caveat

$K_w$ rises with temperature because the autoionisation of water is endothermic. At 298 K it is $1.0 \times 10^{-14}$, giving a neutral pH of exactly 7.00 and $\mathrm{pH} + \mathrm{pOH} = 14$. At 310 K, body temperature, $K_w$ is $2.7 \times 10^{-14}$, so neutral water has $\mathrm{pH} = 6.78$ and $\mathrm{pH} + \mathrm{pOH} = \mathrm{p}K_w = 13.57$. The water is still exactly neutral — $[\mathrm{H_3O^+}] = [\mathrm{OH^-}]$ — and its pH is not 7. Every statement in this chapter that attaches a number to neutrality carries "at 298 K" with it.

Buffer Capacity

A buffer resists pH change, but not without limit. Buffer capacity puts a number on the resistance.

Key Point (Definition): The buffer capacity $\beta$ is the number of moles of strong acid or strong base that must be added to one litre of the buffer to change its pH by one unit. Formally $\beta = \mathrm{d}n_b/\mathrm{d(pH)}$, where $n_b$ is moles of strong base added per litre. It is always positive, since adding base raises the pH and adding acid lowers it.

The expression and where its maximum lies

For a buffer made from a weak acid and its conjugate base with total concentration $C = [\mathrm{HA}] + [\mathrm{A^-}]$, differentiating the mass and charge balances gives

$$\beta = 2.303\left(\,[\mathrm{H_3O^+}] + [\mathrm{OH^-}] + C\,\frac{K_a[\mathrm{H_3O^+}]}{\left(K_a + [\mathrm{H_3O^+}]\right)^{2}}\right)$$

The first two terms are water's own contribution; they are negligible between about pH 2 and pH 12 and take over outside that range, which is why a strong acid at pH 1 is itself an excellent buffer. The third term is the buffer's own.

Finding where it peaks takes one substitution. Put $r = [\mathrm{H_3O^+}]/K_a$, so the buffer term becomes

$$C\,\frac{r}{(1+r)^{2}}$$

Step 1. Differentiate with respect to $r$:

$$\frac{\mathrm{d}}{\mathrm{d}r}\left[\frac{r}{(1+r)^{2}}\right] = \frac{(1+r)^{2} - 2r(1+r)}{(1+r)^{4}} = \frac{1-r}{(1+r)^{3}}$$

Step 2. Set the derivative to zero: $r = 1$.

Step 3. $r = 1$ means $[\mathrm{H_3O^+}] = K_a$, that is $\mathrm{pH} = \mathrm{p}K_a$.

Step 4. Substituting $r = 1$ gives the peak value $C/4$, so

$$\beta_{\max} = \frac{2.303\,C}{4} = 0.576\,C$$

Key Point: Buffer capacity is greatest when $\mathrm{pH} = \mathrm{p}K_a$, where the salt-to-acid ratio is $1{:}1$, and it is directly proportional to the total buffer concentration.

Why the $1{:}1$ mixture is the strongest

The Henderson-Hasselbalch equation, $\mathrm{pH} = \mathrm{p}K_a + \log\dfrac{[\text{salt}]}{[\text{acid}]}$, says the pH tracks the logarithm of a ratio. Adding base converts acid into salt: it subtracts from the denominator and adds the same amount to the numerator. A ratio near 1 is the least sensitive place on that logarithmic curve, because both quantities are large and the fractional change in each is small.

Numbers make it concrete. Take one litre of buffer with $C = 1.0\ \mathrm{M}$ and $\mathrm{p}K_a = 4.76$, and add $0.05$ mol of NaOH.

Starting mixture	Starting pH	Final ratio	Final pH	$\Delta \mathrm{pH}$	$\beta$ from this addition
$0.50$ acid, $0.50$ salt	$4.76$	$0.55/0.45$	$4.85$	$0.087$	$0.57$
$0.70$ acid, $0.30$ salt	$4.39$	$0.35/0.65$	$4.49$	$0.10$	$0.50$
$0.90$ acid, $0.10$ salt	$3.81$	$0.15/0.85$	$4.01$	$0.20$	$0.25$

The lopsided buffer shifts more than twice as far for the same addition. The last column is an average capacity over the interval rather than the instantaneous $\beta$ at the starting pH, and the two need not agree, because $\beta$ itself changes as the base goes in. For the two lopsided mixtures $\beta$ is climbing towards its peak as acid is converted, so the average comes out above the starting value — $0.50$ against $0.48$, and $0.25$ against $0.21$. The $1{:}1$ mixture starts at the peak, so its $\beta$ can only fall, and its average $0.57$ sits just below the starting $0.576$.

The useful range

At $\mathrm{pH} = \mathrm{p}K_a \pm 1$ the ratio is $10{:}1$ or $1{:}10$, and $r/(1+r)^{2}$ has fallen from $0.25$ to $10/121 = 0.083$ — almost exactly one third of the maximum. Beyond that the buffer is running out of one component and the capacity collapses. The working range of a buffer is $\mathrm{p}K_a \pm 1$, and an acid is chosen for a target pH by finding one whose $\mathrm{p}K_a$ is within a unit of it.

Dilution: the trap

Diluting a buffer tenfold does not change $[\text{salt}]/[\text{acid}]$, so the Henderson-Hasselbalch equation returns the same pH. The capacity, being proportional to $C$, falls to one tenth. A diluted buffer holds the same pH and defends it ten times more weakly. The pH-unchanged result also has a limit of its own: once the components fall to around $10^{-6}\ \mathrm{M}$, the water terms and the neglected $\pm[\mathrm{H_3O^+}]$ corrections in the mass balance matter, and the pH drifts towards 7.

Buffer capacity is also raised by having both components present in bulk rather than by any special ratio. A $0.01\ \mathrm{M}$ acetate buffer at $\mathrm{pH} = 4.76$ is destroyed by $0.006$ mol of acid per litre; a $1.0\ \mathrm{M}$ buffer at the same pH absorbs a hundred times as much for the same pH shift.

Choosing the acid for a target pH

Since the capacity peaks at $\mathrm{pH} = \mathrm{p}K_a$ and stays usable within a unit of it, designing a buffer starts by picking a conjugate pair whose $\mathrm{p}K_a$ brackets the required pH, and only then adjusting the ratio.

Buffer pair	$\mathrm{p}K_a$ at 298 K	Best working range
$\mathrm{HCOOH}/\mathrm{HCOO^-}$	$3.74$	$2.74 - 4.74$
$\mathrm{CH_3COOH}/\mathrm{CH_3COO^-}$	$4.76$	$3.76 - 5.76$
$\mathrm{H_2CO_3}/\mathrm{HCO_3^-}$	$6.35$	$5.35 - 7.35$
$\mathrm{H_2PO_4^-}/\mathrm{HPO_4^{2-}}$	$7.21$	$6.21 - 8.21$
$\mathrm{NH_4^+}/\mathrm{NH_3}$	$9.25$	$8.25 - 10.25$
$\mathrm{HCO_3^-}/\mathrm{CO_3^{2-}}$	$10.33$	$9.3 - 11.3$

A basic buffer is handled with the same table by converting: an ammonia buffer has $\mathrm{p}K_b = 4.75$ for the base, and the $\mathrm{p}K_a$ of its conjugate acid $\mathrm{NH_4^+}$ is $14 - 4.75 = 9.25$ at 298 K. Blood is held near $\mathrm{pH}\ 7.4$ by the carbonic acid pair even though $\mathrm{p}K_a$ is $6.35$, a full unit away, because the lungs and kidneys continuously restock both components — a working reminder that the $\pm 1$ rule describes a closed beaker, not an open system.

[JEE Main] Questions phrase this as "which buffer resists change best". Rank first by how close the pH is to $\mathrm{p}K_a$, then by total concentration. Both matter, and a highly concentrated buffer two pH units from its $\mathrm{p}K_a$ can still lose to a dilute one sitting exactly on it.

Titration Curves and the Choice of Indicator

A titration curve plots the pH of the flask against the volume of titrant added. Its shape is fixed by which of the two reagents is strong and which is weak, and the shape decides which indicator can be used.

Strong acid against strong base

Take $25.0\ \mathrm{mL}$ of $0.10\ \mathrm{M}$ HCl titrated with $0.10\ \mathrm{M}$ NaOH. The pH is set entirely by whichever reagent is in excess, since neither ion of the salt formed reacts with water.

NaOH added / mL	Species in excess	pH
$0.0$	$0.10\ \mathrm{M}\ \mathrm{H_3O^+}$	$1.00$
$24.0$	$\mathrm{H_3O^+}$, $2.0 \times 10^{-3}\ \mathrm{M}$	$2.69$
$24.9$	$\mathrm{H_3O^+}$, $2.0 \times 10^{-4}\ \mathrm{M}$	$3.70$
$25.0$	neither; NaCl solution	$7.00$
$25.1$	$\mathrm{OH^-}$, $2.0 \times 10^{-4}\ \mathrm{M}$	$10.30$
$50.0$	$\mathrm{OH^-}$, $3.3 \times 10^{-2}\ \mathrm{M}$	$12.52$

Between $24.9$ and $25.1\ \mathrm{mL}$ — about four drops — the pH climbs from $3.70$ to $10.30$. The equivalence pH is $7.00$ at 298 K, because $\mathrm{Na^+}$ and $\mathrm{Cl^-}$ are the ions of a strong base and a strong acid and neither hydrolyses. At another temperature the neutral pH is not 7 and neither is the equivalence point.

Weak acid against strong base

Take $25.0\ \mathrm{mL}$ of $0.10\ \mathrm{M}$ $\mathrm{CH_3COOH}$ ($K_a = 1.74 \times 10^{-5}$, $\mathrm{p}K_a = 4.76$) with $0.10\ \mathrm{M}$ NaOH. Four regions follow one another.

1. Before any base. A weak acid on its own: $[\mathrm{H_3O^+}] = \sqrt{K_ac} = 1.32 \times 10^{-3}$, $\mathrm{pH} = 2.88$. The curve starts far higher than the strong acid's $1.00$.
2. The buffer region. Between the first drop and the equivalence point the flask holds acid and its conjugate base together, and the pH follows the Henderson-Hasselbalch equation. The curve is at its flattest here.
3. The half-equivalence point, at $12.5\ \mathrm{mL}$. Exactly half the acid has been converted, so $[\text{salt}] = [\text{acid}]$, the logarithm vanishes and $\mathrm{pH} = \mathrm{p}K_a = 4.76$. This is the standard experimental route to $\mathrm{p}K_a$ — no concentrations need to be known, only the volume at which half the equivalence volume has been delivered.
4. The equivalence point, at $25.0\ \mathrm{mL}$. Every acid molecule has become acetate, at $0.05\ \mathrm{M}$ in the doubled volume. Acetate is a Bronsted base with $K_b = K_w/K_a = 5.7 \times 10^{-10}$, so $[\mathrm{OH^-}] = \sqrt{K_b c} = 5.34 \times 10^{-6}$ and $\mathrm{pH} = 8.73$. The equivalence pH is above 7 and it is not the point where pH equals 7.

Measured over the same $24.9$ to $25.1\ \mathrm{mL}$ window used for the strong acid, the jump runs from $7.1$ to $10.3$ — shorter than for the strong-strong case and displaced upwards.

Strong acid against weak base

Take $25.0\ \mathrm{mL}$ of $0.10\ \mathrm{M}$ $\mathrm{NH_3}$ ($K_b = 1.77 \times 10^{-5}$, $\mathrm{p}K_b = 4.75$) with $0.10\ \mathrm{M}$ HCl. The mirror image: the curve starts at $\mathrm{pH} = 11.12$, the half-equivalence point has $\mathrm{pOH} = \mathrm{p}K_b$ so $\mathrm{pH} = 9.25$, and at equivalence the flask holds $0.05\ \mathrm{M}$ $\mathrm{NH_4^+}$, an acid with $K_a = K_w/K_b = 5.6 \times 10^{-10}$. Then $[\mathrm{H_3O^+}] = 5.29 \times 10^{-6}$ and $\mathrm{pH} = 5.28$, below 7. Over the same $24.9$ to $25.1\ \mathrm{mL}$ window the jump runs down from $6.9$ to $3.7$.

Titration	pH at start	pH at half-equivalence	pH at equivalence (298 K)	Jump from $24.9$ to $25.1\ \mathrm{mL}$
Strong acid, strong base	$1.00$	no special value	$7.00$	$3.7$ to $10.3$
Weak acid, strong base	$2.88$	$\mathrm{p}K_a = 4.76$	above 7, $8.73$	$7.1$ to $10.3$
Strong acid, weak base	$11.12$	$14 - \mathrm{p}K_b = 9.25$	below 7, $5.28$	$6.9$ to $3.7$
Weak acid, weak base	intermediate	—	near 7 if $K_a \approx K_b$	no sharp jump

The weak-weak titration has no usable indicator end point at all, because there is no steep region for a colour change to sit in; conductometric titration is used instead.

How an indicator works

An acid-base indicator is itself a weak acid whose two forms differ in colour:

$$\mathrm{HIn} \rightleftharpoons \mathrm{H^+} + \mathrm{In^-} \qquad K_{\mathrm{In}} = \frac{[\mathrm{H^+}][\mathrm{In^-}]}{[\mathrm{HIn}]} \qquad \mathrm{pH} = \mathrm{p}K_{\mathrm{In}} + \log\frac{[\mathrm{In^-}]}{[\mathrm{HIn}]}$$

The eye stops seeing one colour when the other form reaches roughly a tenth of it, so the visible transition spans the ratios $1{:}10$ to $10{:}1$, that is $\mathrm{p}K_{\mathrm{In}} \pm 1$. Only a very small amount of indicator is added, so it does not consume a measurable quantity of the titrant.

Indicator	pH range	Colour in acid	Colour in base	$\mathrm{p}K_{\mathrm{In}}$
Thymol blue (first change)	$1.2 - 2.8$	red	yellow	$\approx 1.7$
Methyl orange	$3.1 - 4.4$	red	yellow	$\approx 3.7$
Bromophenol blue	$3.0 - 4.6$	yellow	blue	$\approx 4.0$
Methyl red	$4.2 - 6.3$	red	yellow	$\approx 5.1$
Bromothymol blue	$6.0 - 7.6$	yellow	blue	$\approx 7.0$
Phenol red	$6.8 - 8.4$	yellow	red	$\approx 7.9$
Phenolphthalein	$8.3 - 10.0$	colourless	pink	$\approx 9.1$
Thymolphthalein	$9.3 - 10.5$	colourless	blue	$\approx 9.9$
Alizarin yellow R	$10.1 - 12.0$	yellow	red	$\approx 11.0$

Key Point: Choose an indicator whose transition range lies wholly inside the steep part of the curve, so that the colour changes within one drop of the equivalence point. It is the position of the range, not a match with pH 7, that decides.

Applying that rule to the three cases: strong against strong takes methyl red or phenolphthalein outright, both ranges lying wholly inside $3.7$ to $10.3$, and methyl orange serves in practice as well — its range opens at $3.1$, a shade below the foot of the jump, but its change to yellow is complete by $4.4$, comfortably inside the steep region. Weak acid against strong base needs phenolphthalein or thymolphthalein; methyl orange would have finished its change to yellow by about $8\ \mathrm{mL}$, less than a third of the way to the equivalence point. Strong acid against weak base needs methyl red or methyl orange; phenolphthalein would lose its pink long before the equivalence point and report far too little acid.

Selective Precipitation

When one reagent can precipitate two or more ions in the same solution, the ion whose salt needs the smallest concentration of the added ion comes down first. Adding the reagent slowly separates them.

Setting the threshold

Precipitation begins the instant the ionic product reaches $K_{sp}$. For a $1{:}1$ salt $\mathrm{AgX}$ in a solution already containing $\mathrm{X^-}$,

$$[\mathrm{Ag^+}]_{\text{start}} = \frac{K_{sp}(\mathrm{AgX})}{[\mathrm{X^-}]}$$

Below this the solution is unsaturated and nothing appears; at it the solution is exactly saturated; above it solid separates until the product falls back to $K_{sp}$.

Take a litre holding $0.10\ \mathrm{M}$ each of $\mathrm{Cl^-}$, $\mathrm{Br^-}$ and $\mathrm{I^-}$, with silver nitrate solution added drop by drop.

Salt	$K_{sp}$ at 298 K	$[\mathrm{Ag^+}]$ needed with the halide at $0.10\ \mathrm{M}$	Order
$\mathrm{AgI}$	$8.3 \times 10^{-17}$	$8.3 \times 10^{-16}\ \mathrm{M}$	first
$\mathrm{AgBr}$	$5.0 \times 10^{-13}$	$5.0 \times 10^{-12}\ \mathrm{M}$	second
$\mathrm{AgCl}$	$1.8 \times 10^{-10}$	$1.8 \times 10^{-9}\ \mathrm{M}$	third

How clean the separation is

The useful question is not which comes first but how much of the first ion is still in solution when the second starts. Silver bromide begins when $[\mathrm{Ag^+}] = 5.0 \times 10^{-12}\ \mathrm{M}$. At that instant the iodide left over is fixed by the silver iodide equilibrium:

$$[\mathrm{I^-}] = \frac{K_{sp}(\mathrm{AgI})}{[\mathrm{Ag^+}]} = \frac{8.3 \times 10^{-17}}{5.0 \times 10^{-12}} = 1.7 \times 10^{-5}\ \mathrm{M}$$

That is $0.017\%$ of the original $0.10\ \mathrm{M}$: over $99.98\%$ of the iodide is already in the precipitate before any bromide is touched. Repeating the calculation at the moment silver chloride starts, $[\mathrm{Ag^+}] = 1.8 \times 10^{-9}\ \mathrm{M}$ leaves $[\mathrm{Br^-}] = 2.8 \times 10^{-4}\ \mathrm{M}$, or $0.28\%$ of the bromide.

Key Point: For two salts of the same formula type sharing a common ion, the separation is essentially complete when their solubility products differ by a factor of about $10^{3}$ or more. A smaller gap leaves the two precipitating together over much of the addition.

The comparison that must be made with care

Ranking salts by $K_{sp}$ alone works only when they are of the same formula type. Silver chromate has $K_{sp} = 1.1 \times 10^{-12}$, smaller than silver chloride's $1.8 \times 10^{-10}$, yet it is the more soluble salt: $s(\mathrm{Ag_2CrO_4}) = (K_{sp}/4)^{1/3} = 6.5 \times 10^{-5}\ \mathrm{M}$ against $s(\mathrm{AgCl}) = \sqrt{K_{sp}} = 1.3 \times 10^{-5}\ \mathrm{M}$. The chromate releases two silver ions per formula unit and its $K_{sp}$ carries a squared term, so the numbers are not on the same footing.

The safe procedure is always to compute the concentration of the added ion at which each salt starts. With $[\mathrm{Cl^-}] = [\mathrm{CrO_4^{2-}}] = 0.10\ \mathrm{M}$:

$$[\mathrm{Ag^+}]_{\mathrm{AgCl}} = \frac{1.8\times 10^{-10}}{0.10} = 1.8 \times 10^{-9}\ \mathrm{M} \qquad [\mathrm{Ag^+}]_{\mathrm{Ag_2CrO_4}} = \sqrt{\frac{1.1\times 10^{-12}}{0.10}} = 3.3 \times 10^{-6}\ \mathrm{M}$$

Silver chloride needs a thousand times less silver, so all the chloride precipitates before the first red speck of silver chromate appears. That is exactly the basis of Mohr's method, in which chromate is the indicator for a chloride titration.

Controlling an ion instead of adding it

Sulphide separations do not add sulphide directly; they control it through acidity, using the coupled equilibrium of the earlier block:

$$[\mathrm{S^{2-}}] = \frac{K_{a_1}K_{a_2}[\mathrm{H_2S}]}{[\mathrm{H^+}]^{2}}$$

In a solution saturated with $\mathrm{H_2S}$ and made $0.1\ \mathrm{M}$ in HCl, $[\mathrm{S^{2-}}]$ sits near $1.0 \times 10^{-19}\ \mathrm{M}$. With a metal ion at around $10^{-2}\ \mathrm{M}$ the ionic product is near $10^{-21}$, which exceeds $K_{sp}$ for CuS ($8 \times 10^{-37}$), CdS ($8 \times 10^{-27}$) and ZnS ($1.6 \times 10^{-24}$) but not for FeS ($6.3 \times 10^{-18}$) or MnS ($2.5 \times 10^{-13}$). Raising the acid to $0.3\ \mathrm{M}$ divides $[\mathrm{S^{2-}}]$ by nine, to about $1.1 \times 10^{-20}\ \mathrm{M}$, which pulls the ionic product down to about $10^{-22}$ — still above $K_{sp}$ for ZnS, so zinc does not return to solution on these numbers. What the extra acid buys is margin: it pushes the borderline sulphides closer to their saturation point while leaving the very insoluble ones untouched. Setting the sulphide groups of qualitative analysis apart is done the same way, by choosing the acidity rather than by adding sulphide, and the zinc group is held back not by more acid but by precipitating it separately from an alkaline medium, where $[\mathrm{S^{2-}}]$ is orders of magnitude higher. The same hydroxide-based logic separates $\mathrm{Fe^{3+}}$ from $\mathrm{Mg^{2+}}$: a buffered $\mathrm{pH}$ near 5 precipitates $\mathrm{Fe(OH)_3}$ ($K_{sp} = 1.0 \times 10^{-38}$) completely while leaving $\mathrm{Mg(OH)_2}$ ($K_{sp} = 8.9 \times 10^{-12}$) untouched.

[JEE/NEET] Every one of these problems reduces to a single sentence: compute the ionic product with the concentrations actually present after mixing, and compare it with $K_{sp}$. Dilution on mixing is the step most often skipped.