What minimum concentration of SO42 - is required to precipitate BaSO4 from the solution containing 1.0× 10- 4 moles of Ba2 + ? (Given Ks p for BaSO4 is 4× 10- 10 )

Question

What minimum concentration of SO42 - is required to precipitate BaSO4 from the solution containing 1.0× 10- 4 moles of Ba2 + ? (Given Ks p for BaSO4 is 4× 10- 10 ) (A) 2× 10- 3M (B) 2× 10- 7M (C) 4× 10- 6M (D) 4× 10- 10M

Tardigrade Admin · Accepted Answer

BaSO4 â Ba2 ++SO42 - Ks p=[.Ba2 +].× [.SO42 -]. Ks p=S2⇒ S=√Ks p ; 4× 10- 10=[.1× 10- 4].× [.SO42 -]. [.SO42 -].=(4 × 1 0- 10/1 × 1 0- 4)=4× 10- 6 .

Q. What minimum concentration of $S O_{4}^{2 -}$ is required to precipitate $B a S O_{4}$ from the solution containing $1.0 \times 1 0^{- 4} m o l eso f B a^{2 +}$ ? (Given $K_{s p}$ for $B a S O_{4}$ is $4 \times 1 0^{- 10}$ )

$2 \times 1 0^{- 3} M$

$2 \times 1 0^{- 7} M$

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

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

$B a S O_{4}$ ⇌ $B a^{2 +} + S O_{4}^{2 -}$

$K_{s p} = [B a^{2 +}] \times [S O_{4}^{2 -}]$

$K_{s p} = S^{2} \Rightarrow S = K_{s p}$ ;

$4 \times 1 0^{- 10} = [1 \times 1 0^{- 4}] \times [S O_{4}^{2 -}]$

$[S O_{4}^{2 -}] = \frac{4 \times 1 0 ^{- 10}}{1 \times 1 0 ^{- 4}} = 4 \times 1 0^{- 6}$ .

Q. What minimum concentration of SO42−​ is required to precipitate BaSO4​ from the solution containing 1.0×10−4molesofBa2+ ? (Given Ksp​ for BaSO4​ is 4×10−10 )

2×10−3M

2×10−7M

4×10−6M

4×10−10M

BaSO4​ ⇌ Ba2++SO42−​ Ksp​=[Ba2+]×[SO42−​] Ksp​=S2⇒S=Ksp​​ ; 4×10−10=[1×10−4]×[SO42−​] [SO42−​]=1×10−44×10−10​=4×10−6 .

Q. What minimum concentration of $S O_{4}^{2 -}$ is required to precipitate $B a S O_{4}$ from the solution containing $1.0 \times 1 0^{- 4} m o l eso f B a^{2 +}$ ? (Given $K_{s p}$ for $B a S O_{4}$ is $4 \times 1 0^{- 10}$ )

$2 \times 1 0^{- 3} M$

$2 \times 1 0^{- 7} M$

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

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