Question

Sodium chromate solution is gradually added to a mixture containing 0.05 M Pb²⁺ ions and 0.10 M Ba²⁺ ions. The concentration of the ion precipitating first when the second ion begins to form a precipitate is –

[Note K_sp of BaCrO₄ = 2.4 × 10^–10 and K_sp of PbCrO₄ = 1.8 × 10^–14]

• A. 7.5 × 10^–6
• B. 2.5 × 10^–5
• C. 8.2 × 10^–3
• D. 5.0 × 10^–4

Answer 1

Answer: (A)

7.5 × 10^–6

Answer 2

2^nd ion precipitated = Ba²⁺

$\left[ \mathrm { CrO } _ { 4 } ^ { - 2 } \right] = \frac { \mathrm { K } _ { \mathrm { sp } } \left( \mathrm { BaCrO } _ { 4 } \right) } { \left[ \mathrm { Ba } ^ { 2 + } \right] } = \frac { 2.4 \times 10 ^ { - 10 } } { 10 ^ { - 1 } }$ = 2.4 × 10^–9

At this concentration of $\left[ \mathrm { CrO } _ { 4 } ^ { - 2 } \right]$ ions,

$\left[ \mathrm { Pb } ^ { 2 + } \right] = \frac { 1.8 \times 10 ^ { - 14 } } { 2.4 \times 10 ^ { - 9 } }$ = 0.75 × 10^–5 = 7.5 × 10^–6