Chapter 6: Equilibrium (Equilibrium)

(a) The same vapour now fills a larger volume, so the vapour pressure falls at first.

(b) The rate of evaporation is unchanged (it depends only on temperature and surface); the rate of condensation falls (fewer vapour molecules per unit volume strike the liquid).

(c) Evaporation exceeds condensation until the two rates become equal again. The final vapour pressure is the same as the original, since the equilibrium vapour pressure depends only on temperature.

(a) Falls (b) evaporation same, condensation slower (c) original vapour pressure is restored

$$K_c=\frac{[\mathrm{SO}{}_3{]}^2}{[\mathrm{SO}{}_2{]}^2[\mathrm{O}{}_2]}=\frac{(1.90{)}^2}{(0.60{)}^2\times 0.82}=\frac{3.61}{0.2952}$$

Kc = 12.23 M⁻¹

Volume % = mole % = pressure %: $p_{\mathrm{I}}=0.40\times 10^5=4.0\times 10^4$ Pa; $p_{\mathrm{I}{}_2}=6.0\times 10^4$ Pa.

$K_p=\frac{p_{\mathrm{I}}^2}{p_{\mathrm{I}{}_2}}=\frac{(4.0\times 10^4{)}^2}{6.0\times 10^4}$

Kp = 2.67 × 10⁴ Pa

Pure solids and liquids (including the solvent water) are left out.

(i) [NO]²[Cl₂]/[NOCl]² (ii) [NO₂]⁴[O₂] (iii) [CH₃COOH][C₂H₅OH]/[CH₃COOC₂H₅] (iv) 1/([Fe³⁺][OH⁻]³) (v) [IF₅]²/[F₂]⁵

$K_c=\frac{K_p}{(RT{)}^{\Delta n}}$; $\Delta n=+1$ in both.

(i) $K_c=\frac{1.8\times 10^{-2}}{0.0831\times 500}=4.33\times 10^{-4}$ (ii) $K_c=\frac{167}{0.0831\times 1073}=1.87$

(i) 4.33 × 10⁻⁴ (ii) 1.87 (NCERT key 1.90, with R = 0.0821 L atm)

$K_c^{\prime }=\frac{1}{K_c}=\frac{1}{6.3\times 10^{14}}$

1.59 × 10⁻¹⁵

The molar concentration of a pure solid or liquid = density ÷ molar mass, which is constant at a given temperature whatever its amount. Being constant, it is absorbed into the equilibrium constant (its activity is taken as 1). E.g. CaCO₃(s) ⇌ CaO(s) + CO₂(g): $K_p=p_{\mathrm{CO}{}_2}$.

Their concentration (activity) is constant, so it is included in K.

Kc is extremely small, so very little N₂O forms: $[\mathrm{N}{}_2]\approx 0.0482$ M, $[\mathrm{O}{}_2]\approx 0.0933$ M.

$[\mathrm{N}{}_2\mathrm{O}{]}^2=K_c[\mathrm{N}{}_2{]}^2[\mathrm{O}{}_2]=2.0\times 10^{-37}\times (0.0482{)}^2\times 0.0933=4.34\times 10^{-41}$

[N₂] = 0.0482 M, [O₂] = 0.0933 M, [N₂O] = 6.6 × 10⁻²¹ M

0.0518 mol NOBr uses 0.0518 mol NO and $\frac{0.0518}{2}=0.0259$ mol Br₂.

NO $=0.087-0.0518=0.0352$ mol; Br₂ $=0.0437-0.0259=0.0178$ mol.

NO 0.0352 mol; Br₂ 0.0178 mol

$\Delta n=2-3=-1$: $K_c=K_p(RT{)}^{-\Delta n}=2.0\times 10^{10}\times 0.0831\times 450$.

Kc = 7.48 × 10¹¹ M⁻¹

HI decomposed $=0.2-0.04=0.16$ atm, giving $p_{\mathrm{H}{}_2}=p_{\mathrm{I}{}_2}=0.08$ atm.

$K_p=\frac{0.08\times 0.08}{(0.04{)}^2}$

Kp = 4.0

$[\mathrm{N}{}_2]=0.0785$, $[\mathrm{H}{}_2]=0.096$, $[\mathrm{NH}{}_3]=0.4065$ M.

$Q_c=\frac{(0.4065{)}^2}{0.0785\times (0.096{)}^3}=2.38\times 10^3>K_c$.

Not at equilibrium; the net reaction goes backward (NH₃ decomposes)

Products are in the numerator; powers are the coefficients.

4NO(g) + 6H₂O(g) ⇌ 4NH₃(g) + 5O₂(g)

0.40 mol reacts: H₂O = CO = 0.60 mol; H₂ = CO₂ = 0.40 mol (in 10 L: 0.06 and 0.04 M).

$K_c=\frac{0.04\times 0.04}{0.06\times 0.06}$

K = 0.44

$[\mathrm{H}{}_2]=[\mathrm{I}{}_2]=x$: $\frac{(0.5{)}^2}{x^2}=54.8\Rightarrow x^2=4.56\times 10^{-3}$.

[H₂] = [I₂] = 0.068 mol L⁻¹

$\frac{x^2}{(0.78-2x{)}^2}=0.14\Rightarrow \frac{x}{0.78-2x}=0.374\Rightarrow x=0.167$.

[I₂] = [Cl₂] = 0.167 M; [ICl] = 0.446 M

$\frac{x^2}{4.0-x}=0.04\Rightarrow x^2+0.04x-0.16=0\Rightarrow x=\frac{-0.04+\sqrt{0.0016+0.64}}{2}=0.38$.

p(C₂H₆) = 4.0 − 0.38 = 3.62 atm

(i) $Q_c=\frac{[\mathrm{CH}{}_3\mathrm{COOC}{}_2\mathrm{H}{}_5][\mathrm{H}{}_2\mathrm{O}]}{[\mathrm{CH}{}_3\mathrm{COOH}][\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{OH}]}$

(ii) Ester = water = 0.171, acid = 0.829, ethanol = 0.009 mol (volume cancels): $K=\frac{(0.171{)}^2}{0.829\times 0.009}=3.92$.

(iii) Ester = water = 0.214, acid = 0.786, ethanol = 0.286: $Q=\frac{(0.214{)}^2}{0.786\times 0.286}=0.204<K$.

(ii) K = 3.92 (iii) Not yet; Q < K, the reaction will go further forward

$[\mathrm{PCl}{}_3]=[\mathrm{Cl}{}_2]=x$: $x^2=8.3\times 10^{-3}\times 0.05=4.15\times 10^{-4}$.

[PCl₃] = [Cl₂] = 0.020 mol L⁻¹

$Q_p=\frac{0.80}{1.4}=0.571>K_p$: the reaction goes backward.

$\frac{0.80-x}{1.4+x}=0.265\Rightarrow 0.80-x=0.371+0.265x\Rightarrow x=0.339$.

p(CO) = 1.739 atm; p(CO₂) = 0.461 atm

$Q_c=\frac{(0.5{)}^2}{3.0\times (2.0{)}^3}=\frac{0.25}{24}=0.0104<K_c$.

Not at equilibrium; it proceeds forward (more NH₃)

$\frac{x^2}{(3.3\times 10^{-3}-2x{)}^2}=32\Rightarrow \frac{x}{3.3\times 10^{-3}-2x}=5.66\Rightarrow x=1.52\times 10^{-3}$.

$[\mathrm{BrCl}]=3.3\times 10^{-3}-3.03\times 10^{-3}=2.7\times 10^{-4}$ mol L⁻¹.

About 3 × 10⁻⁴ mol L⁻¹

In 100 g: CO $=\frac{90.55}{28}=3.234$ mol, CO₂ $=\frac{9.45}{44}=0.215$ mol; $x_{\mathrm{CO}}=0.938$, $x_{\mathrm{CO}{}_2}=0.062$.

At 1 atm: $K_p=\frac{(0.938{)}^2}{0.062}=14.1$ atm. $K_c=\frac{K_p}{RT}=\frac{14.1}{0.0821\times 1127}$.

Kc ≈ 0.15 (NCERT key 0.149)

(a) $\Delta G^{\circ }=52.0-87.0-0=-35.0$ kJ mol⁻¹.

(b) $\ln K=\frac{35000}{8.314\times 298}=14.13\Rightarrow K=1.36\times 10^6$.

(a) −35.0 kJ mol⁻¹ (b) K = 1.36 × 10⁶

Lower pressure favours the side with more gas molecules.

(a) 1 → 2 gas moles: products increase. (b) 1 → 0: products decrease. (c) 4 → 4: no change.

(a) Increase (b) decrease (c) remain the same

Higher pressure favours fewer gas moles.

(i) 1 → 2: backward. (ii) 3 → 3: not affected. (iii) 1 → 2: backward. (iv) 3 → 1: forward. (v) 0 → 1: backward. (vi) 9 → 10: backward.

All except (ii) are affected: (iv) forward; (i), (iii), (v), (vi) backward

For 2HBr ⇌ H₂ + Br₂: $K=\frac{1}{1.6\times 10^5}=6.25\times 10^{-6}$.

$\frac{x^2}{(10-2x{)}^2}=6.25\times 10^{-6}\Rightarrow \frac{x}{10-2x}=2.5\times 10^{-3}\Rightarrow x=0.0249$ bar.

p(H₂) = p(Br₂) = 2.5 × 10⁻² bar; p(HBr) ≈ 10.0 bar

(a) $K_p=\frac{p_{\mathrm{CO}}{p}_{\mathrm{H}{}_2}^3}{p_{\mathrm{CH}{}_4}{p}_{\mathrm{H}{}_2\mathrm{O}}}$

(b) (i) $K_p$ unchanged; 2 → 4 gas moles, so the equilibrium shifts backward (less CO and H₂). (ii) Endothermic: $K_p$ increases and the equilibrium shifts forward. (iii) No change in $K_p$ or composition; equilibrium is only reached sooner.

(b) (i) Kp same, shifts backward (ii) Kp increases, shifts forward (iii) no change

By Le Chatelier's principle the system opposes each change.

(a) Forward (b) backward (c) backward (d) forward

(a) $K_c=\frac{[\mathrm{PCl}{}_3][\mathrm{Cl}{}_2]}{[\mathrm{PCl}{}_5]}$ (b) $\frac{1}{8.3\times 10^{-3}}=120.5$

(c) (i), (ii): Kc does not change (only the position of equilibrium shifts). (iii) The forward reaction is endothermic, so Kc increases.

(b) 120.5 (c) (i) no change (ii) no change (iii) Kc increases

$\frac{x^2}{(4.0-x{)}^2}=10.1\Rightarrow \frac{x}{4.0-x}=3.18\Rightarrow x=\frac{12.71}{4.18}=3.04$ bar.

p(H₂) = 3.04 bar (p(CO) = p(H₂O) = 0.96 bar, the value printed in the NCERT key)

(a) Almost all reactant; (b) almost all product; (c) Kc between 10⁻³ and 10³: both appreciable.

(c)

$[\mathrm{O}{}_3{]}^2=K_c[\mathrm{O}{}_2{]}^3=2.0\times 10^{-50}\times (1.6\times 10^{-2}{)}^3=8.19\times 10^{-56}$

[O₃] = 2.86 × 10⁻²⁸ M

$[\mathrm{CH}{}_4]=\frac{K_c[\mathrm{CO}][\mathrm{H}{}_2{]}^3}{[\mathrm{H}{}_2\mathrm{O}]}=\frac{3.90\times 0.30\times (0.10{)}^3}{0.02}$

[CH₄] = 5.85 × 10⁻² M

A conjugate acid–base pair is an acid and a base that differ by one proton (HA/A⁻).

Conjugate bases: HNO₂ → NO₂⁻; HClO₄ → ClO₄⁻. Conjugate acids: CN⁻ → HCN; F⁻ → HF; OH⁻ → H₂O (its conjugate base is O²⁻); CO₃²⁻ → HCO₃⁻; S²⁻ → HS⁻.

NO₂⁻, HCN, ClO₄⁻, HF, H₂O, HCO₃⁻, HS⁻

BF₃ (B has an empty orbital) and H⁺ (empty 1s) accept electron pairs. NH₄⁺ is also counted as an acid (it releases H⁺). H₂O, with lone pairs, is a Lewis base.

BF₃, H⁺ (and NH₄⁺)

Remove one H⁺ from each.

F⁻, HSO₄⁻ and CO₃²⁻

Add one H⁺ to each.

NH₃, NH₄⁺ and HCOOH

As in the table

(a) Base: donates a lone pair, OH⁻ + H⁺ → H₂O. (b) Base: F⁻ + BF₃ → BF₄⁻. (c) Acid: accepts a pair, H⁺ + :NH₃ → NH₄⁺. (d) Acid: BCl₃ + :NH₃ → Cl₃B←NH₃.

Lewis bases: OH⁻, F⁻; Lewis acids: H⁺, BCl₃

pH $=-\log (3.8\times 10^{-3})=3-0.58$

pH = 2.42

$[\mathrm{H}{}^{+}]=10^{-3.76}=10^{-4}\times 10^{0.24}$

1.74 × 10⁻⁴ M

$K_b=\frac{K_w}{K_a}$: F⁻: $\frac{10^{-14}}{6.8\times 10^{-4}}$; HCOO⁻: $\frac{10^{-14}}{1.8\times 10^{-4}}$; CN⁻: $\frac{10^{-14}}{4.8\times 10^{-9}}$.

F⁻ 1.5 × 10⁻¹¹; HCOO⁻ 5.6 × 10⁻¹¹; CN⁻ 2.08 × 10⁻⁶

$[\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{O}{}^{-}]=\sqrt{K_{a}c}=\sqrt{1.0\times 10^{-10}\times 0.05}=2.2\times 10^{-6}$ M; $\alpha =\frac{2.2\times 10^{-6}}{0.05}=4.5\times 10^{-5}$.

With 0.01 M phenolate (common ion): $[\mathrm{H}{}^{+}]=\frac{K_a[\mathrm{PhOH}]}{[\mathrm{PhO}{}^{-}]}=\frac{10^{-10}\times 0.05}{0.01}=5\times 10^{-10}$; $\alpha =\frac{5\times 10^{-10}}{0.05}=1.0\times 10^{-8}$.

[C₆H₅O⁻] = 2.2 × 10⁻⁶ M; α = 1.0 × 10⁻⁸ in the presence of phenolate

In water: $[\mathrm{HS}{}^{-}]=[\mathrm{H}{}^{+}]=\sqrt{9.1\times 10^{-8}\times 0.1}=9.54\times 10^{-5}$ M; $[\mathrm{S}{}^{2-}]=\frac{K_{a2}[\mathrm{HS}{}^{-}]}{[\mathrm{H}{}^{+}]}=K_{a2}=1.2\times 10^{-13}$ M.

In 0.1 M HCl ($[\mathrm{H}{}^{+}]=0.1$): $[\mathrm{HS}{}^{-}]=\frac{K_{a1}\times 0.1}{0.1}=9.1\times 10^{-8}$ M; $[\mathrm{S}{}^{2-}]=\frac{1.2\times 10^{-13}\times 9.1\times 10^{-8}}{0.1}=1.09\times 10^{-19}$ M.

[HS⁻] 9.54 × 10⁻⁵ M → 9.1 × 10⁻⁸ M in HCl; [S²⁻] 1.2 × 10⁻¹³ M → 1.09 × 10⁻¹⁹ M

$\alpha =\sqrt{\frac{1.74\times 10^{-5}}{0.05}}=0.0187$; $[\mathrm{CH}{}_3\mathrm{COO}{}^{-}]=[\mathrm{H}{}^{+}]=c\alpha =9.3\times 10^{-4}$ M; pH $=-\log (9.3\times 10^{-4})$.

α = 0.0187; [CH₃COO⁻] = 9.3 × 10⁻⁴ M; pH = 3.03

$[\mathrm{A}{}^{-}]=[\mathrm{H}{}^{+}]=10^{-4.15}=7.08\times 10^{-5}$ M; $K_a=\frac{(7.08\times 10^{-5}{)}^2}{0.01}=5.0\times 10^{-7}$; $\text{p}{K}_a=6.30$.

[A⁻] = 7.08 × 10⁻⁵ M; Ka = 5.0 × 10⁻⁷; pKa = 6.30

(a) $-\log 0.003=2.52$ (b) pOH = 2.30, pH = 11.70 (c) $-\log 0.002=2.70$ (d) pOH = 2.70, pH = 11.30

(a) 2.52 (b) 11.70 (c) 2.70 (d) 11.30

(a) M(TlOH) = 221.4: $[\mathrm{OH}{}^{-}]=\frac{2/221.4}{2}=4.52\times 10^{-3}$ M; pOH 2.35, pH 11.65.

(b) $[\mathrm{Ca}(\mathrm{OH}){}_2]=\frac{0.3/74}{0.5}=8.11\times 10^{-3}$ M, $[\mathrm{OH}{}^{-}]=1.62\times 10^{-2}$ M; pOH 1.79, pH 12.21.

(c) $[\mathrm{OH}{}^{-}]=\frac{0.3/40}{0.2}=0.0375$ M; pOH 1.43, pH 12.57.

(d) $[\mathrm{H}{}^{+}]=\frac{13.6\times 1}{1000}=0.0136$ M; pH 1.87.

(a) 11.65 (b) 12.21 (c) 12.57 (d) 1.87

$[\mathrm{H}{}^{+}]=c\alpha =0.0132$ M, pH = 1.88. $K_a=\frac{c{\alpha }^2}{1-\alpha }=\frac{0.1\times 0.0174}{0.868}=2.0\times 10^{-3}$; pKa = 2.70.

pH = 1.88; pKa = 2.70

pOH = 4.05, $[\mathrm{OH}{}^{-}]=8.91\times 10^{-5}$; $K_b=\frac{(8.91\times 10^{-5}{)}^2}{0.005}=1.59\times 10^{-6}$; pKb = 5.80.

Kb = 1.59 × 10⁻⁶; pKb = 5.80

$[\mathrm{OH}{}^{-}]=\sqrt{4.27\times 10^{-10}\times 0.001}=6.53\times 10^{-7}$ M; pOH 6.19, pH 7.81.

$\alpha =\frac{6.53\times 10^{-7}}{0.001}=6.53\times 10^{-4}$; $K_a(\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{NH}{}_3{}^{+})=\frac{10^{-14}}{4.27\times 10^{-10}}=2.34\times 10^{-5}$.

pH = 7.81; α = 6.53 × 10⁻⁴; Ka = 2.34 × 10⁻⁵

$K_a=10^{-4.74}=1.82\times 10^{-5}$; $\alpha =\sqrt{K_a/c}=0.019$.

With HCl, $[\mathrm{H}{}^{+}]\approx [\mathrm{HCl}]$ (common ion), and $\alpha =\frac{K_a}{[\mathrm{H}{}^{+}]}$: (a) $1.82\times 10^{-3}$ (b) $1.82\times 10^{-4}$. Ionisation is suppressed.

α = 0.019; (a) 1.82 × 10⁻³ (b) 1.82 × 10⁻⁴

α is not small here: $\frac{0.02{\alpha }^2}{1-\alpha }=5.4\times 10^{-4}\Rightarrow \alpha =0.15$ (the approximation $\sqrt{K_b/c}$ gives 0.16).

In 0.1 M NaOH: $\alpha =\frac{K_b}{[\mathrm{OH}{}^{-}]}=\frac{5.4\times 10^{-4}}{0.1}=0.0054$.

α ≈ 0.15 (0.16 by approximation); 0.54% ionised in 0.1 M NaOH

$[\mathrm{H}{}^{+}]=10^{-\text{pH}}$.

(a) 1.48 × 10⁻⁷ M (b) 0.063 M (c) 4.17 × 10⁻⁸ M (d) 3.98 × 10⁻⁷ M

$[\mathrm{H}{}^{+}]=10^{-\text{pH}}$.

Milk 1.58 × 10⁻⁷; coffee 1.0 × 10⁻⁵; tomato juice 6.31 × 10⁻⁵; lemon juice 6.31 × 10⁻³; egg white 1.58 × 10⁻⁸ M

$\frac{0.561}{56.1}=0.01$ mol in 0.2 L: $[\mathrm{K}{}^{+}]=[\mathrm{OH}{}^{-}]=0.05$ M; $[\mathrm{H}{}^{+}]=\frac{10^{-14}}{0.05}=2.0\times 10^{-13}$ M; pH = 12.70.

[K⁺] = [OH⁻] = 0.05 M; [H⁺] = 2.0 × 10⁻¹³ M; pH = 12.70

M(Sr(OH)₂) = 121.6: $[\mathrm{Sr}{}^{2+}]=\frac{19.23}{121.6}=0.1581$ M; $[\mathrm{OH}{}^{-}]=0.3162$ M; pOH 0.50, pH 13.50.

[Sr²⁺] = 0.1581 M; [OH⁻] = 0.3162 M; pH = 13.50

$\alpha =\sqrt{\frac{1.32\times 10^{-5}}{0.05}}=0.0163$; $[\mathrm{H}{}^{+}]=8.1\times 10^{-4}$, pH 3.09. In 0.01 M HCl: $\alpha =\frac{K_a}{0.01}=1.32\times 10^{-3}$.

α = 0.0163, pH = 3.09; α = 1.32 × 10⁻³ in 0.01 M HCl

$[\mathrm{H}{}^{+}]=10^{-2.34}=4.57\times 10^{-3}$; $K_a=\frac{(4.57\times 10^{-3}{)}^2}{0.1}=2.09\times 10^{-4}$; $\alpha =\frac{4.57\times 10^{-3}}{0.1}=0.0457$.

Ka = 2.09 × 10⁻⁴; α = 0.0457

$\mathrm{NO}{}_2{}^{-}+\mathrm{H}{}_2\mathrm{O}\rightleftharpoons \mathrm{HNO}{}_2+\mathrm{OH}{}^{-}$; $K_h=\frac{K_w}{K_a}=2.22\times 10^{-11}$.

$[\mathrm{OH}{}^{-}]=\sqrt{K_{h}c}=9.43\times 10^{-7}$; pOH 6.03, pH 7.97. $h=\sqrt{K_h/c}=2.36\times 10^{-5}$.

pH = 7.97; degree of hydrolysis 2.36 × 10⁻⁵

$\mathrm{C}{}_5\mathrm{H}{}_5\mathrm{NH}{}^{+}\rightleftharpoons \mathrm{C}{}_5\mathrm{H}{}_5\mathrm{N}+\mathrm{H}{}^{+}$: $[\mathrm{H}{}^{+}]=10^{-3.44}=3.63\times 10^{-4}$; $K_a=\frac{(3.63\times 10^{-4}{)}^2}{0.02}=6.6\times 10^{-6}$.

$K_b=\frac{K_w}{K_a}=\frac{10^{-14}}{6.6\times 10^{-6}}$.

Kb = 1.5 × 10⁻⁹

Strong acid + strong base: neutral. Weak acid + strong base (anion hydrolyses): basic. Strong acid + weak base (cation hydrolyses): acidic.

Neutral: NaCl, KBr; basic: NaCN, NaNO₂, KF; acidic: NH₄NO₃

Acid: $[\mathrm{H}{}^{+}]\approx \sqrt{1.35\times 10^{-3}\times 0.1}=0.0116$ M, pH ≈ 1.94.

Salt: $K_h=\frac{10^{-14}}{1.35\times 10^{-3}}=7.4\times 10^{-12}$; $[\mathrm{OH}{}^{-}]=\sqrt{7.4\times 10^{-12}\times 0.1}=8.6\times 10^{-7}$; pOH 6.06, pH 7.94.

Acid pH ≈ 1.94; salt pH ≈ 7.94

$[\mathrm{H}{}^{+}]=\sqrt{2.7\times 10^{-14}}=1.64\times 10^{-7}$ M; pH $=-\log (1.64\times 10^{-7})$. Water is still neutral ([H⁺] = [OH⁻]).

pH = 6.78

(a) OH⁻ = 10 × 0.2 × 2 = 4.0 mmol; H⁺ = 2.5 mmol; excess OH⁻ = 1.5 mmol in 35 mL = 0.0429 M; pOH 1.37, pH 12.63.

(b) H⁺ = 0.2 mmol, OH⁻ = 0.2 mmol: exactly neutralised, pH 7.

(c) H⁺ = 2 mmol, OH⁻ = 1 mmol: excess H⁺ 1 mmol in 20 mL = 0.05 M, pH 1.30.

(a) 12.63 (b) 7.00 (c) 1.30

Ag₂CrO₄: $4S^3=1.1\times 10^{-12}\Rightarrow S=6.5\times 10^{-5}$ M; [Ag⁺] = 1.3 × 10⁻⁴, [CrO₄²⁻] = 6.5 × 10⁻⁵ M.

BaCrO₄: $S^2=1.2\times 10^{-10}\Rightarrow S=1.1\times 10^{-5}$ M = [Ba²⁺] = [CrO₄²⁻].

Fe(OH)₃: $27S^4=1.0\times 10^{-38}\Rightarrow S=1.39\times 10^{-10}$ M; [Fe³⁺] = S, [OH⁻] = 4.17 × 10⁻¹⁰ M.

PbCl₂: $4S^3=1.6\times 10^{-5}\Rightarrow S=1.59\times 10^{-2}$ M; [Pb²⁺] = S, [Cl⁻] = 3.18 × 10⁻² M.

Hg₂I₂: $4S^3=4.5\times 10^{-29}\Rightarrow S=2.24\times 10^{-10}$ M = [Hg₂²⁺]; [I⁻] = 4.48 × 10⁻¹⁰ M.

Ag₂CrO₄ 6.5 × 10⁻⁵ M; BaCrO₄ 1.1 × 10⁻⁵ M; Fe(OH)₃ 1.39 × 10⁻¹⁰ M; PbCl₂ 1.59 × 10⁻² M; Hg₂I₂ 2.24 × 10⁻¹⁰ M

Ag₂CrO₄: $S={\left(\frac{1.1\times 10^{-12}}{4}\right)}^{1/3}=6.5\times 10^{-5}$ M; AgBr: $S=\sqrt{5.0\times 10^{-13}}=7.07\times 10^{-7}$ M.

Ratio $=\frac{6.5\times 10^{-5}}{7.07\times 10^{-7}}$: silver chromate is far more soluble although its Ksp is only about twice as large.

About 91.9

After mixing each concentration halves: [Cu²⁺] = [IO₃⁻] = 0.001 M. Ionic product $Q=[\mathrm{Cu}{}^{2+}][\mathrm{IO}{}_3{}^{-}{]}^2=10^{-3}\times 10^{-6}=1\times 10^{-9}<K_{sp}$.

No precipitate (Q < Ksp)

In water: $S=\sqrt{K_{sp}}$.

In the buffer, $[\mathrm{H}{}^{+}]=10^{-3.19}=6.46\times 10^{-4}$, so $\frac{[\mathrm{HA}]}{[\mathrm{A}{}^{-}]}=\frac{[\mathrm{H}{}^{+}]}{K_a}=10$. Total dissolved benzoate $S^{\prime }=[\mathrm{A}{}^{-}](1+10)$ and $[\mathrm{Ag}{}^{+}]=S^{\prime }$: $K_{sp}=S^{\prime }\cdot \frac{S^{\prime }}{11}\Rightarrow S^{\prime }=\sqrt{11K_{sp}}$.

$\frac{S^{\prime }}{S}=\sqrt{11}$

About 3.32 times

In the mixture $[\mathrm{Fe}{}^{2+}]=[\mathrm{S}{}^{2-}]=c^{\prime }$ with $c^{\prime 2}\le 6.3\times 10^{-18}\Rightarrow c^{\prime }=2.5\times 10^{-9}$ M. Mixing equal volumes halves each, so each solution may be at most $5.0\times 10^{-9}$ M before mixing.

2.5 × 10⁻⁹ M in the mixture (5.0 × 10⁻⁹ M for each solution before mixing)

$S=\sqrt{9.1\times 10^{-6}}=3.02\times 10^{-3}$ mol L⁻¹ $\times 136$ g mol⁻¹ $=0.410$ g L⁻¹. Volume for 1 g $=\frac{1}{0.410}$.

2.44 L

In 15 mL: $[\mathrm{S}{}^{2-}]=10^{-19}\times \frac{10}{15}=6.67\times 10^{-20}$ M; $[\mathrm{M}{}^{2+}]=0.04\times \frac{5}{15}=1.33\times 10^{-2}$ M. Ionic product $Q=8.9\times 10^{-22}$.

Ksp: FeS 6.3 × 10⁻¹⁸, MnS 2.5 × 10⁻¹³ (Q smaller: no precipitate); ZnS 1.6 × 10⁻²⁴, CdS 8.0 × 10⁻²⁷ (Q larger: precipitate).

ZnS and CdS precipitate (in ZnCl₂ and CdCl₂); the NCERT key names CdCl₂