A saturated solution of BaSO 4 at 25° C is 4 × 10-5 M. The solubility of BaSO 4 in 0.1 M Na 2 SO 4 at this temperature will be

Question

A saturated solution of BaSO 4 at 25° C is 4 × 10-5 M. The solubility of BaSO 4 in 0.1 M Na 2 SO 4 at this temperature will be (A) 1.6 × 10-9 M (B) 1.6 × 10-8 M (C) 4 × 10-6 M (D) 4 × 10-4 M

Tardigrade Admin · Accepted Answer

BaSO4 leftharpoons undersetsBa2+ + undersets + 0.1SO42- Na2SO4 arrow underset0.22Na+ + underset0.1SO42- Ksp = [Ba+2][SO4-2] ∴ 4 × 10-5 = s × (s + 0.1) or 4 × 10-5 ≈ s× 0.1 ∴ s = 4 × 10-4 as s < < < 0.1 ∴ s + 0.1 ≈ 0.1

Q. A saturated solution of $B a S O_{4}$ at $2 5^{\circ} C$ is $4 \times 1 0^{- 5} M$ . The solubility of $B a S O_{4}$ in $0.1 M N a_{2} S O_{4}$ at this temperature will be

$1.6 \times 1 0^{- 9} M$

$1.6 \times 1 0^{- 8} M$

$4 \times 1 0^{- 6} M$

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

Q. A saturated solution of BaSO4​ at 25∘C is 4×10−5M. The solubility of BaSO4​ in 0.1MNa2​SO4​ at this temperature will be

1.6×10−9M

1.6×10−8M

4×10−6M

4×10−4M

BaSO4​⇌sBa2+​+s+0.1SO42−​​ Na2​SO4​→0.22Na+​+0.1SO42−​​ Ksp​=[Ba+2][SO4−2​] ∴4×10−5=s×(s+0.1) or 4×10−5≈s×0.1 ∴s=4×10−4 as s<<<0.1 ∴s+0.1≈0.1

Q. A saturated solution of $B a S O_{4}$ at $2 5^{\circ} C$ is $4 \times 1 0^{- 5} M$ . The solubility of $B a S O_{4}$ in $0.1 M N a_{2} S O_{4}$ at this temperature will be

$1.6 \times 1 0^{- 9} M$

$1.6 \times 1 0^{- 8} M$

$4 \times 1 0^{- 6} M$

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