Question

K_sp of SrF₂ (s) in water is 3.2 × 10^–11. The solubility of SrF₂ (s) in 0.1 (M) NaCl solution is –

• A. 3.2 × 10^–9 (M)
• B. 2 × 10^–4 (M)
• C. 4 × 10^–4 (M)
• D. slightly higher than 2 × 10^–4 (M)

Answer 1

Answer: (D)

slightly higher than 2 × 10^–4 (M)

Answer 2

SrF₂ (s) $\rightleftarrows$ Sr²⁺ + 2F^– s 2s

(where s is the solubility)

\ 4s³ = 32 × 10^–12 or s = 2 × 10^–4 (M)

But practically the solubility of SrF₂ (s) in NaCl solution is slightly greater than 2 × 10^–4 because NaCl increases ionic strength of the solution.