Engineering

Chemistry

Solubility Equilibria

Question

Concentration of H₂SO₄ and Na₂SO₄ in a solution is 1 M and 1.8 × 10^–2 M, respectively. Molar solubility of PbSO₄ in the same solution is X × 10^–Y M (expressed in scientific notation). The value of Y is _________.

[Given: Solubility product of PbSO₄ (Ksp) = 1.6 × 10^–8. For H₂SO₄, K_a1 is very large and K_a2 = 1.2 × 10^–2]

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H₂SO₄ ⇌ HSO₄^– + H⁺

1M - -

- 1M 1M

Na₂SO₄ → 2Na⁺ + SO₄^2–

1.8 × 10^–2 M - -

- 3.6 × 10^–2 M 1.8 × 10^–2 M

HSO₄^2– ⇌ H+ + SO₄^2– ; $K_{a_{2}}$ = 1.2 × 10^–2 M

1M 1M 1.8× 10^–2 M

Since Q_C > K_C it will move in backward direction.

1 + x 1 – x 1.8 × 10^–2 – x

$K_{a_{2}} = 1.2 \times 10^{- 2} = \frac{(1 - x) (1.8 \times 10^{- 2} - x)}{(1 + x)}$

Since x is very small (1 + x) ≃ 1 and (1 – x) ≃ 1

x = (1.8×10^–2 – 1.2 × 10^–2) M

[SO₄^–2 ] = ( 1.8 × 10^–2 – 0.6×10^–2) M

= 1.2 × 10^–2 M

PbSO₄ → Pb²⁺ + SO₄^2–

s - 1.2 ×10^–2 M

- s (s + 1.2 × 10^–2)

K_sp = s (s + 1.2 × 10^–2) = 1.6 × 10^–8

(PbSO₄)

Here, (s + 1.2 × 10^–2) 1.2 × 10^–2 (since 's' is very small)

s(1.2 × 10^–2) = 1.6 × 10^–8

⟹ s = $\frac{1.6}{1.2}$ 10^–6 M = X × 10^–y M

⟹ Y = 6

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