Ionic Equilibrium

Principles

• Solubility product principle: a sparingly soluble salt $A_xB_y$ precipitates only when its ionic product $[A^{y+}]^x[B^{x-}]^y$ exceeds its $K_{sp}$.
• Common ion effect: the ionisation of a weak electrolyte is suppressed when a strong electrolyte with a common ion is added.

In qualitative analysis the cations are divided into groups. For each group, the concentration of the precipitating ion is controlled with a common ion, so that the ionic product exceeds $K_{sp}$ only for the salts of that group.

Group-wise application

Group I (dil. HCl): $AgCl$, $PbCl_2$ and $Hg_2Cl_2$ have low $K_{sp}$ and precipitate. Other chlorides are soluble. A slight excess of $Cl^-$ (common ion) ensures complete precipitation. $PbCl_2$ is fairly soluble, so some $Pb^{2+}$ passes into group II.

Group II ($H_2S$ + dil. HCl):

$$H_2S \rightleftharpoons 2H^+ + S^{2-}$$

$H^+$ from HCl suppresses the ionisation of $H_2S$, so $[S^{2-}]$ is low. Only sulphides of very low $K_{sp}$ (CuS, PbS, HgS, CdS, $Bi_2S_3$, $As_2S_3$, $Sb_2S_3$, SnS) precipitate. Group IV sulphides with higher $K_{sp}$ (ZnS, MnS, NiS, CoS) stay in solution.

Group III ($NH_4OH$ + $NH_4Cl$):

$$NH_4OH \rightleftharpoons NH_4^+ + OH^- \qquad NH_4Cl \rightarrow NH_4^+ + Cl^-$$

$NH_4^+$ (common ion) suppresses the ionisation of $NH_4OH$, so $[OH^-]$ is low. This low $[OH^-]$ is enough to exceed the very low $K_{sp}$ of $Fe(OH)_3$, $Al(OH)_3$ and $Cr(OH)_3$, but not of $Zn(OH)_2$, $Mn(OH)_2$, $Ni(OH)_2$, $Co(OH)_2$ or $Mg(OH)_2$. These remain for later groups.

Group IV ($H_2S$ in ammoniacal medium): $OH^-$ from $NH_4OH$ removes $H^+$ ($H^+ + OH^- \rightarrow H_2O$). The ionisation of $H_2S$ shifts forward and $[S^{2-}]$ becomes high, so the ionic products of ZnS, MnS, NiS and CoS (higher $K_{sp}$) are exceeded and they precipitate.

Group V ($(NH_4)_2CO_3$ + $NH_4Cl$ + $NH_4OH$): $NH_4^+$ from $NH_4Cl$ suppresses the ionisation of $(NH_4)_2CO_3$, keeping $[CO_3^{2-}]$ low. This is enough to precipitate $BaCO_3$, $SrCO_3$ and $CaCO_3$ (low $K_{sp}$) but not $MgCO_3$. $Mg^{2+}$ remains in solution for group VI.

Summary

So, by adjusting the concentration of the precipitating ion through the common ion effect, the cations can be separated group by group on the basis of their solubility products.

(c) Greater than $1 \times 10^{-14}$

Ionisation of water ($H_2O \rightleftharpoons H^+ + OH^-$) is endothermic, so $K_w$ increases with temperature. At 393 K (120 °C), well above 298 K, $K_w$ is greater than $1 \times 10^{-14}$ (about $10^{-12}$).

(a) less than 7

The $10^{-8}$ M $H^+$ from HCl is added to the $\approx10^{-7}$ M $H^+$ from water. Solving $x(x+10^{-8}) = 10^{-14}$ gives total $[H^+] \approx 1.05\times10^{-7}$ M, so pH ≈ 6.98. The solution is slightly acidic, not pH 8.

i) Ionic product of water ($K_w$): the product of the molar concentrations of hydrogen ions and hydroxide ions in water (or in any aqueous solution) at a given temperature. At 25 °C, $K_w = [H^+][OH^-] = 1\times10^{-14}$ mol² L⁻².

ii) Deduction of $K_w = [H^+][OH^-]$

Water is a very weak electrolyte and ionises slightly:

$$H_2O \rightleftharpoons H^+ + OH^-$$

By the law of mass action,

$$K = \frac{[H^+][OH^-]}{[H_2O]}$$

Because ionisation is so small, the concentration of unionised water is practically constant ($[H_2O] = 1000/18 = 55.5$ M). So

$$K \times [H_2O] = [H^+][OH^-]$$ $$K_w = [H^+][OH^-]$$

where $K_w = K[H_2O]$ is a constant called the ionic product of water.

iii) $[OH^-]$ in 0.01 M HCl at 25 °C

HCl is a strong acid and ionises completely, so $[H^+] = 0.01 = 10^{-2}$ M.

$$[OH^-] = \frac{K_w}{[H^+]} = \frac{1\times10^{-14}}{1\times10^{-2}} = 1\times10^{-12} \text{ M}$$

$[OH^-] = 1\times10^{-12}$ mol/L.

iv) Effect of temperature: ionisation of water is endothermic. By Le Chatelier's principle, raising the temperature favours ionisation, so $[H^+]$ and $[OH^-]$ both increase and $K_w$ increases with temperature (about $10^{-14}$ at 25 °C, about $5.5\times10^{-14}$ at 50 °C, about $5\times10^{-13}$ at 100 °C). Water stays neutral ($[H^+] = [OH^-]$), but its pH falls below 7 at higher temperatures.

(a) $6.02\times10^{7}$

pH = 13 gives $[H^+] = 10^{-13}$ mol/L. 1 mL contains $10^{-13}\times10^{-3} = 10^{-16}$ mol, which is $10^{-16}\times6.02\times10^{23} = 6.02\times10^{7}$ ions.

Application of common ion effect and solubility product in wet-way detection of basic radicals

• Solubility product principle: a sparingly soluble salt precipitates only when its ionic product exceeds its $K_{sp}$.
• Common ion effect: adding a strong electrolyte with a common ion suppresses the ionisation of a weak electrolyte (or lowers the solubility of a sparingly soluble salt).

In salt analysis, basic radicals are separated into groups by controlling the concentration of the precipitating ion (Cl⁻, S²⁻, OH⁻, CO₃²⁻) with a common ion. Each group then precipitates only when its own $K_{sp}$ is exceeded:

Mass of AgCl precipitated

Given: $K_{sp}$(AgCl) = $1.8\times10^{-10}$, molar mass of AgCl = 143.5 g/mol. Assume 0.2 mol HCl is added to 1 L of saturated AgCl solution with negligible volume change.

Step 1: Solubility of AgCl in water

$$AgCl \rightleftharpoons Ag^+ + Cl^-, \qquad K_{sp} = s^2$$ $$s = \sqrt{1.8\times10^{-10}} = 1.342\times10^{-5} \text{ mol/L}$$

Step 2: Solubility in presence of 0.2 M HCl

HCl ionises completely, so $[Cl^-] \approx 0.2$ M (the common ion).

$$s' = [Ag^+] = \frac{K_{sp}}{[Cl^-]} = \frac{1.8\times10^{-10}}{0.2} = 9\times10^{-10} \text{ mol/L}$$

Step 3: AgCl precipitated

$$\text{Moles precipitated} = s - s' = 1.342\times10^{-5} - 9\times10^{-10} \approx 1.342\times10^{-5} \text{ mol}$$ $$\text{Mass} = 1.342\times10^{-5}\times143.5 = 1.93\times10^{-3} \text{ g}$$

About $1.93\times10^{-3}$ g (1.93 mg) of AgCl is precipitated per litre of saturated solution. Almost all the dissolved AgCl comes out because the common ion $Cl^-$ pushes $[Ag^+][Cl^-]$ above $K_{sp}$.

(b) $Na_2CO_3$

$Na_2CO_3$ is a salt of a strong base (NaOH) and a weak acid ($H_2CO_3$). It hydrolyses ($CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$) to give a basic solution, pH > 7. NaCl and KCl are neutral (pH 7), and $CuSO_4$ is acidic.

i) Lewis base: a substance (molecule or ion) that can donate a lone pair of electrons to form a coordinate bond. Examples: $NH_3$, $H_2O$, $OH^-$, $Cl^-$, $CN^-$.

$$H_3N{:} + BF_3 \rightarrow H_3N \rightarrow BF_3$$

ii) Basic buffer: a solution of a weak base and its salt with a strong acid that resists a change in pH when a small amount of acid or base is added or when it is diluted. It keeps pH above 7. Example: $NH_4OH + NH_4Cl$.

pH of the mixture of equal volumes of pH 3 and pH 5 solutions

Solution 1: pH = 3, so $[H^+]_1 = 1\times10^{-3}$ M. Solution 2: pH = 5, so $[H^+]_2 = 1\times10^{-5}$ M.

Let volume V of each be mixed, giving total volume 2V:

$$[H^+]_{\text{mix}} = \frac{[H^+]_1 V + [H^+]_2 V}{2V} = \frac{1\times10^{-3} + 1\times10^{-5}}{2}$$ $$[H^+]_{\text{mix}} = \frac{1.01\times10^{-3}}{2} = 5.05\times10^{-4} \text{ M}$$ $$pH = -\log(5.05\times10^{-4}) = 4 - \log 5.05 = 4 - 0.703 = 3.30$$

pH of the resulting solution ≈ 3.3.

(a) Applications of solubility product principle and common ion effect in qualitative salt analysis

The cations (basic radicals) are separated into groups by precipitation. A salt precipitates only when its ionic product exceeds $K_{sp}$. The concentration of the precipitating anion is controlled with a common ion, so that only the ions of one group precipitate at each stage.

(b) Brønsted–Lowry vs Lewis concept of bases

(c) Degree of ionisation, $[H^+]$ and pH of 0.01 M HCN

Given: C = 0.01 M, $K_a = 4.8\times10^{-10}$. HCN is a weak acid, so by Ostwald's dilution law:

$$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{4.8\times10^{-10}}{0.01}} = \sqrt{4.8\times10^{-8}} = 2.19\times10^{-4}$$

(i.e. 0.022% ionised)

$$[H^+] = C\alpha = 0.01 \times 2.19\times10^{-4} = 2.19\times10^{-6} \text{ M}$$ $$pH = -\log(2.19\times10^{-6}) = 6 - 0.34 = 5.66$$

α = $2.19\times10^{-4}$; $[H^+] = 2.19\times10^{-6}$ M; pH ≈ 5.66.

(a) Solubility product principle: a sparingly soluble electrolyte precipitates from its solution only when the ionic product (the product of the molar concentrations of its ions, each raised to the power of its coefficient) exceeds its solubility product ($K_{sp}$) at that temperature. If the ionic product is less than $K_{sp}$, the solution is unsaturated and no precipitate forms.

Application in group II analysis:

Group II cations ($Cu^{2+}$, $Pb^{2+}$, $Hg^{2+}$, $Cd^{2+}$, $Bi^{3+}$, $As^{3+}$, $Sb^{3+}$, $Sn^{2+}$) are precipitated as sulphides by passing $H_2S$ in the presence of dilute HCl.

$$H_2S \rightleftharpoons 2H^+ + S^{2-} \qquad HCl \rightarrow H^+ + Cl^-$$

• Common ion effect: $H^+$ from the strong acid HCl suppresses the ionisation of the weak acid $H_2S$, so $[S^{2-}]$ becomes very low.
• Solubility product: group II sulphides (CuS, PbS, HgS, CdS, etc.) have very low $K_{sp}$, so even this low $[S^{2-}]$ makes $[M^{2+}][S^{2-}] > K_{sp}$ and they precipitate. Group IV sulphides (ZnS, MnS, NiS, CoS) have much higher $K_{sp}$, so their ionic product stays below $K_{sp}$ and they are not precipitated. This separates group II from group IV.

(b) Brønsted–Lowry acid vs Lewis acid

(c) pH of the mixture of equal volumes of A (pH 3) and B (pH 5)

$[H^+]_A = 1\times10^{-3}$ M, $[H^+]_B = 1\times10^{-5}$ M. Mixing volume V of each gives total volume 2V:

$$[H^+] = \frac{(1\times10^{-3} + 1\times10^{-5})V}{2V} = \frac{1.01\times10^{-3}}{2} = 5.05\times10^{-4} \text{ M}$$ $$pH = -\log(5.05\times10^{-4}) = 4 - 0.703 = 3.30$$

pH of the resulting solution ≈ 3.3.

a) i) Weak electrolyte: a substance that ionises only partially in aqueous solution, so an equilibrium exists between the unionised molecules and the ions. It conducts electricity poorly. Examples: $CH_3COOH$, HCN, $NH_4OH$, $H_2CO_3$.

ii) Derivation of Ostwald's dilution law

Consider a weak binary electrolyte (e.g. a weak acid HA) of molar concentration $C$ and degree of ionisation $\alpha$:

$$HA \rightleftharpoons H^+ + A^-$$

Applying the law of mass action:

$$K_a = \frac{[H^+][A^-]}{[HA]} = \frac{C\alpha \cdot C\alpha}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}$$

For a weak electrolyte $\alpha \ll 1$, so $1-\alpha \approx 1$:

$$K_a = C\alpha^2 \quad\Rightarrow\quad \alpha = \sqrt{\frac{K_a}{C}}$$

If $V$ is the volume (in litres) holding 1 mole, $C = 1/V$, so $\alpha = \sqrt{K_aV}$. This is Ostwald's dilution law: the degree of ionisation of a weak electrolyte is directly proportional to the square root of dilution.

iii) Limitations of Ostwald's dilution law

1. It does not apply to strong electrolytes (HCl, NaCl, NaOH, etc.). They are almost completely ionised at all concentrations, so no ionisation equilibrium exists and the law of mass action cannot be applied.
2. It fails for weak electrolytes at high concentration, and the approximation $1-\alpha \approx 1$ fails when α is not small (very dilute solutions). It also ignores interionic attractions, so molar concentrations are used instead of activities.

b) Will AgCl precipitate?

Given: 20 mL of $1\times10^{-3}$ N NaCl + 80 mL of $1\times10^{-2}$ N $AgNO_3$. For these salts N = M (n-factor = 1). $K_{sp}$(AgCl) = $1.5\times10^{-10}$.

Total volume after mixing = 20 + 80 = 100 mL

$$[Cl^-] = \frac{1\times10^{-3}\times20}{100} = 2\times10^{-4} \text{ M}$$ $$[Ag^+] = \frac{1\times10^{-2}\times80}{100} = 8\times10^{-3} \text{ M}$$

Ionic product:

$$[Ag^+][Cl^-] = (8\times10^{-3})(2\times10^{-4}) = 1.6\times10^{-6}$$

Since $1.6\times10^{-6} > K_{sp} = 1.5\times10^{-10}$, the ionic product exceeds the solubility product.

Therefore AgCl will precipitate. (The atomic weight of Ag is not needed, since only molar concentrations are compared.)

(c) Acidic

$CuSO_4$ is the salt of a weak base, $Cu(OH)_2$, and a strong acid, $H_2SO_4$. $Cu^{2+}$ hydrolyses and releases $H^+$ ions, so the solution is acidic:

$$Cu^{2+} + 2H_2O \rightleftharpoons Cu(OH)_2 + 2H^+$$

According to the Bronsted-Lowry concept, an acid is a proton donor and a base is a proton acceptor. Water can do both, so it is amphiprotic.

Water as an acid (reaction with ammonia):

$$H_2O + NH_3 \rightleftharpoons OH^- + NH_4^+$$

Here $H_2O$ donates a proton. The conjugate pairs are $H_2O / OH^-$ (acid / conjugate base) and $NH_3 / NH_4^+$ (base / conjugate acid).

Water as a base (reaction with HCl):

$$H_2O + HCl \rightleftharpoons H_3O^+ + Cl^-$$

Here $H_2O$ accepts a proton. The conjugate pairs are $HCl / Cl^-$ (acid / conjugate base) and $H_2O / H_3O^+$ (base / conjugate acid).

Limitation: The concept cannot explain the acidic nature of substances that have no transferable proton, such as $BF_3$, $AlCl_3$ and $CO_2$, which react with bases. These are explained by the Lewis concept.

(c) $HCO_3^-$

A species that can both give and accept a proton is amphiprotic. $HCO_3^-$ acts as an acid ($HCO_3^- \rightarrow CO_3^{2-} + H^+$) and as a base ($HCO_3^- + H^+ \rightarrow H_2CO_3$). Note that $H_2PO_4^-$ (option b) behaves in the same way, giving $HPO_4^{2-}$ or $H_3PO_4$, so both (b) and (c) are amphiprotic. HCl is only an acid and $O_2$ is neither.

a) pH [1]

pH is the negative logarithm (to base 10) of the hydrogen ion concentration of a solution.

$$pH = -\log[H^+]$$

b) Application in daily life [1]

The pH of soil decides which crops can be grown, and farmers add lime to acidic soil. (Also: blood pH is kept at 7.35–7.45, and pH is controlled in water treatment and in food and medicine making.)

c) pOH and its relation to pH [2]

pOH is the negative logarithm of the hydroxide ion concentration.

$$pOH = -\log[OH^-]$$

Since the ionic product of water at 25 °C is $K_w = [H^+][OH^-] = 10^{-14}$, taking $-\log$ of both sides gives

$$pH + pOH = 14$$

d) pH of $5\times10^{-4}$ M $H_2SO_4$ [1]

Treating $H_2SO_4$ as fully dissociated into 2 $H^+$ per molecule:

$$[H^+] = 2 \times 5\times10^{-4} = 1\times10^{-3}\ M$$ $$pH = -\log(10^{-3}) = 3$$

Difference between solubility and solubility product [2]

Application in qualitative salt analysis [6]

Principle. If the ionic product of a salt is greater than its $K_{sp}$, the salt precipitates; if it is smaller, the salt stays in solution. By controlling the concentration of the precipitating ion (using the common-ion effect), cations can be separated into groups one after another.

• Group II (e.g. $Cu^{2+}$, $Pb^{2+}$, $Cd^{2+}$): $H_2S$ is passed in the presence of dilute HCl. The $H^+$ from HCl suppresses the ionisation of $H_2S$, so $[S^{2-}]$ is very low. Only the sulphides with very small $K_{sp}$ (CuS, PbS, CdS) have an ionic product above their $K_{sp}$ and precipitate. ZnS, MnS, NiS and CoS have larger $K_{sp}$ and stay in solution.
• Group III ($Fe^{3+}$, $Al^{3+}$, $Cr^{3+}$): $NH_4OH$ is added in the presence of $NH_4Cl$. The common ion $NH_4^+$ suppresses the ionisation of the weak base $NH_4OH$, so $[OH^-]$ is low. Only the hydroxides with very low $K_{sp}$ ($Fe(OH)_3$, $Al(OH)_3$, $Cr(OH)_3$) precipitate. $Mg(OH)_2$ (higher $K_{sp}$) does not precipitate at this stage.
• Group IV ($Zn^{2+}$, $Mn^{2+}$, $Ni^{2+}$, $Co^{2+}$): $H_2S$ is passed in the presence of $NH_4OH$. The $OH^-$ removes $H^+$, so $[S^{2-}]$ rises and the ionic product exceeds the larger $K_{sp}$ of ZnS, MnS, NiS and CoS, which precipitate.
• Group V ($Ca^{2+}$, $Sr^{2+}$, $Ba^{2+}$): $(NH_4)_2CO_3$ is added in the presence of $NH_4Cl$ and $NH_4OH$ to get the carbonates, because their $K_{sp}$ values are exceeded only when $[CO_3^{2-}]$ is high. $NH_4Cl$ prevents $Mg^{2+}$ from precipitating as carbonate.
• Group I ($Ag^+$, $Pb^{2+}$): dilute HCl precipitates the chlorides, because the ionic product $[Ag^+][Cl^-]$ exceeds the very small $K_{sp}$ of AgCl.
• Confirmatory tests: $BaCrO_4$ and $PbCrO_4$ are precipitated for confirmation, since their $K_{sp}$ are exceeded on adding chromate.

In every case, the cation with the smaller $K_{sp}$ precipitates first, so the solubility product principle explains why the groups separate in a fixed order without mixing.