Concentration of H 2 SO 4 and Na 2 SO 4 in a solution is 1 M and 1 . 8 × 10 - 2 M , respectively. Molar solubility of PbSO 4 in the same solution is X × 10 - Y M (expressed in scientific notation). The value of Y is [Given: Solubility product of PbSO 4 K sp = 1 . 6 × 10 - 8 . For H 2 SO 4 ,   K a 1 is very large and K a 2 = 1 . 2 × 10 - 2 ] If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places.
A. A
H 2 SO 4 → H + + HSO 4 2 - 1 M 1 M ( K a 1 is very large) HSO 4 - ⇌ H + + SO 4 2 -    K a 2 = 1 . 2 × 10 - 2 SO 4 2 - coming from Na 2 SO 4 = 1 . 8 × 10 - 2 SO 4 2 - H + HSO 4 - = 1 . 8 × 10 - 2 × 1 1 > K a 2 ∴ Rather than dissociation of HSO 4 - into H + and SO 4 2 - ions, association between already present H + and SO 4 2 - will take place. Assuming ' x ' mol / L of SO 4 2 - and H + combines to form HSO 4 - ∴    SO 4 2 - = 1 . 8 × 10 - 2 - x H + = 1 - x ≈ 1 HSO 4 - = 1 + x ≈ 1 (assuming x < < 1 ) 1 . 8 × 10 - 2 - x 1 1 = 1 . 2 × 10 - 2 ⇒ x = 0 . 6 × 10 - 2 SO 4 2 - = 1 . 2 × 10 - 2 M PbSO 4 s ⇌ Pb 2 + aq + SO 4 2 - aq If solubility of PbSO 4 = sM ∴    Pb 2 + = s SO 4 2 - = s + 1 . 2 × 10 - 2 ≈ 1 . 2 × 10 - 2 (assuming s ≪ 1 . 2 × 10 - 2 ) ∴    s × 1 . 2 × 10 - 2 = 1 . 6 × 10 - 8 s = 1 . 6 1 . 2 × 10 - 6 = 1 . 33 × 10 - 6 On comparing with X × 10 - Y Y = 6
Related: Chemistry — Ionic Equilibrium · All PYQ Banks