The solubility product of barium sulphate is 1.5 × 10-9 at 18°C. Its solubility in water at 18°C is

Question

The solubility product of barium sulphate is 1.5 × 10-9 at 18°C. Its solubility in water at 18°C is (A) 1.5 × 10-9 mol L-1 (B) 1.5 × 10-5 mol L-1 (C) 3.9 × 10-9 mol L-1 (D) 3.9 × 10-5 mol L-1.

Tardigrade Admin · Accepted Answer

BaSO4 → Ba+2+SO2-4 ∴ Solubility =(Ksp)1/2 =(1.5 × 10-9)1/2=3.87×10-5 mol L-1

Q. The solubility product of barium sulphate is $1.5 \times 1 0^{- 9}$ at 18 $^{\circ}$ C. Its solubility in water at 18 $^{\circ}$ C is

$1.5 \times 1 0^{- 9} m o l L^{- 1}$

$1.5 \times 1 0^{- 5} m o l L^{- 1}$

$3.9 \times 1 0^{- 9} m o l L^{- 1}$

$3.9 \times 1 0^{- 5} m o l L^{- 1}$ .

$B a S O_{4} \to B a^{+ 2} + S O_{4}^{2 -}$
$∴$ Solubility $= (K_{s p})^{1/2}$
$= (1.5 \times 1 0^{- 9})^{1/2} = 3.87 \times 1 0^{- 5} m o l L^{- 1}$

Q. The solubility product of barium sulphate is 1.5×10−9 at 18∘C. Its solubility in water at 18∘C is

1.5×10−9molL−1

1.5×10−5molL−1

3.9×10−9molL−1

3.9×10−5molL−1.