Solubility product of radium sulphate is 4 × 10-11. What will be the solubility of Ra2+ in 0.10 M Na2SO4?

Question

Solubility product of radium sulphate is 4 × 10-11. What will be the solubility of Ra2+ in 0.10 M Na2SO4? (A) 4×10-10 M (B) 2×10-5 M (C) 4×10-5 M (D) 2×10-10 M

Tardigrade Admin · Accepted Answer

RaSO4 leftharpoons Ra2+ +SO2-4 Ksp=[Ra2+][SO2-4] Concentration of SO2-4 from Na2SO4=0.10 M Ra2+=(4×10-11/0.10) =4×10-10 M

Q. Solubility product of radium sulphate is $4 \times 1 0^{- 11}$ .
What will be the solubility of $R a^{2 +}$ in $0.10 M N a_{2} S O_{4}$ ?

$4 \times 1 0^{- 10} M$

$2 \times 1 0^{- 5} M$

$4 \times 1 0^{- 5} M$

$2 \times 1 0^{- 10} M$

$R a S O_{4} ⇌ R a^{2 +} + S O^{2 -} 4$
$K_{s p} = [R a^{2 +}] [S O_{4}^{2 -}]$
Concentration of $S O_{4}^{2 -}$ from $N a_{2} S O_{4} = 0.10 M$
$R a^{2 +} = \frac{4 \times 1 0 ^{- 11}}{0.10}$
$= 4 \times 1 0^{- 10} M$

Q. Solubility product of radium sulphate is 4×10−11. What will be the solubility of Ra2+ in 0.10MNa2​SO4​?

4×10−10M

2×10−5M

4×10−5M

2×10−10M