Consider two Group IV metal ions X²⁺$X^{2+}$ and Y²⁺$Y^{2+}$. A solution containing 0.01 M X²⁺$0.01 \text{ M } X^{2+}$ and 0.01 M Y²⁺$0.01 \text{ M } Y^{2+}$ is saturated with H₂S$H_2S$. The pH at which the metal sulphide YS will form as a precipitate is ____ (Nearest integer) (Given : Kₛₚ(XS) = 1 × 10⁻²²$K_{sp}(XS) = 1 \times 10^{-22}$ at 25°C$25^{\circ}\text{C}$, Kₛₚ(YS) = 4 × 10⁻¹⁶$K_{sp}(YS) = 4 \times 10^{-16}$ at 25°C$25^{\circ}\text{C}$, [H₂S] = 0.1 M$[H_2S] = 0.1 \text{ M}$ in solution, Kₐ₁ × Kₐ₂(H₂S) = 1.0 × 10⁻²¹$K_{a1} \times K_{a2}(H_2S) = 1.0 \times 10^{-21}$, 2 = 0.30$\log 2 = 0.30$, 3 = 0.48$\log 3 = 0.48$, 5 = 0.70$\log 5 = 0.70$)

Core Logic

Initially at equilibrium (t = teq$t = t_{eq}$): 2 mole, 2 mole, 2 mole.

Calculate Keq$K_{eq}$:

(Since Δ ng = 0$\Delta n_g = 0$, volume V$V$ cancels out in the expression.)

Now 1 mole of each P₂$\mathrm{P}_{2}$ and Q₂$\mathrm{Q}_{2}$ is added. The reaction will move in the forward direction. t = t'eq$t = t'_{eq}$ : Moles of P₂ = 2 + 1 - x = 3 - x$\mathrm{P}_2 = 2 + 1 - x = 3 - x$ Moles of Q₂ = 2 + 1 - x = 3 - x$\mathrm{Q}_2 = 2 + 1 - x = 3 - x$ Moles of PQ = 2 + 2x$\mathrm{PQ} = 2 + 2x$

Step 1: Solve for x

Taking the square root on both sides:

2 + 2x = 3 - x$2 + 2x = 3 - x$

Step 2: Final Moles

At the new equilibrium: Moles of P₂ = 3 - (1)/(3) = (8)/(3) 2.67$\mathrm{P}_{2} = 3 - \frac{1}{3} = \frac{8}{3} \simeq 2.67$ Moles of Q₂ = 3 - (1)/(3) = (8)/(3) 2.67$\mathrm{Q}_{2} = 3 - \frac{1}{3} = \frac{8}{3} \simeq 2.67$ Moles of PQ = 2 + 2((1)/(3)) = (8)/(3) 2.67$\mathrm{PQ} = 2 + 2\left(\frac{1}{3}\right) = \frac{8}{3} \simeq 2.67$

Pattern Recognition

For reactions where Δ ng = 0$\Delta n_g = 0$, concentration can be directly substituted with moles in the Kc$K_c$ expression because the volume terms cancel entirely.

Chapter Mix

Class 11 Chemistry: Equilibrium

Related Formula

Step 1: Find Active Moles

Reaction: B + HCl arrow BH^+ + Cl^-$\mathrm{B} + \mathrm{HCl} \rightarrow \mathrm{BH}^+ + \mathrm{Cl}^-$ Moles of HCl = 0.02x$0.02x$ mmol Moles of Base B = 0.02y$0.02y$ mmol Since a buffer is formed, Base B is in excess. Moles of Salt formed (BH^+$\mathrm{BH}^+$) = 0.02x$0.02x$ Moles of Base remaining = 0.02y - 0.02x$0.02y - 0.02x$

Step 2: Apply Henderson-Hasselbalch Equation

Given pH = 9$\mathrm{pH} = 9$, so pOH = 14 - 9 = 5$\mathrm{pOH} = 14 - 9 = 5$.

We know 5 = 0.699$\log 5 = 0.699$, therefore (1/5) = -0.699$\log(1/5) = -0.699$.

Step 3: Solve for x and y

Total volume x + y = 100~mL$x + y = 100\mathrm{~mL}$.

Pattern Recognition

Buffer mixing implies partial neutralization. The ratio of acid volume to base volume defines the logarithmic salt-to-base ratio. Using inverse logs directly maps variables to total volume restrictions.

Chapter Mix

Class 11 Chemistry: Equilibrium

Related Formula

Step 1: Identify Parameters

pKₐ = 4$\mathrm{pK}_a = 4$

Concentration C = 10~mM = 10 × 10⁻³~M = 10⁻²~M$C = 10\mathrm{~mM} = 10 \times 10^{-3}\mathrm{~M} = 10^{-2}\mathrm{~M}$

Step 2: Execute Calculation

Pattern Recognition

For weak monoprotic acids where α ll 1$\alpha \ll 1$, the pH is always the arithmetic mean of the pKₐ$\mathrm{pK}_a$ and the negative logarithm of molarity.

Chapter Mix

Class 11 Chemistry: Equilibrium

Related Formula

Core Logic

Initial equilibrium: A leftharpoons B$A \rightleftharpoons B$ [A] = 0.5 M$[A] = 0.5\text{ M}$, [B] = 0.375 M$[B] = 0.375\text{ M}$

Now 0.1 mole$0.1\text{ mole}$ of A$A$ is added to the 1 L flask. New initial [A] = 0.5 + 0.1 = 0.6 M$[A] = 0.5 + 0.1 = 0.6\text{ M}$. Reaction shifts forward: A leftharpoons B$A \rightleftharpoons B$ Eq: 0.6 - x 0.375 + x$0.6 - x \quad 0.375 + x$

Step 1: Solve for x

1.75x = 0.075$1.75x = 0.075$

Step 2: Calculate New Concentrations

New concentration of A = 0.6 - 0.043 = 0.557 M$A = 0.6 - 0.043 = 0.557\text{ M}$ New concentration of B = 0.375 + 0.043 = 0.418 M$B = 0.375 + 0.043 = 0.418\text{ M}$

Pattern Recognition

Le Chatelier addition problems require maintaining Keq$K_{eq}$ constant. Set up ICE table after perturbation, equate to initial Keq$K_{eq}$.

Chapter Mix

Class 11 Chemistry: Equilibrium