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Question: The solubility product of \(BaSO_{4}\) at \(25{^\circ}C\) is \(1.0 \times 10^{- 9}\). What would be ...

The solubility product of BaSO4BaSO_{4} at 25∘C25{^\circ}C is 1.0×10−91.0 \times 10^{- 9}. What would be the concentration of H2SO4H_{2}SO_{4} necessary to precipitate BaSO4BaSO_{4} from a solution of 0.01MBa2+0.01MBa^{2 +}ions.

A

10−910^{- 9}

B

10−810^{- 8}

C

10−710^{- 7}

D

10−610^{- 6}

Answer

10−710^{- 7}

Explanation

Solution

BaSO4BaSO_{4}⇌Ba++(S)0.01+SO4−−(S)5\underset{0.01}{\underset{(S)}{Ba^{+ +}}} + \underset{5}{\underset{(S)}{SO_{4}^{- -}}}

Ksp=S2=S×S=0.01×SK_{sp} = S^{2} = S \times S = 0.01 \times S

S(SO42−)=KspS(Ba++)=1×10−90.01=10−7S_{(SO_{4}^{2 -})} = \frac{K_{sp}}{S_{(Ba^{+ +})}} = \frac{1 \times 10^{- 9}}{0.01} = 10^{- 7}mole/litre.