Question
A sparingly soluble salt gets precipitated only when the product of the concentration of its ions in the solution $\left( Q _{ sp }\right)$ becomes greater than its solubility product. If the solubility of $BaSO _4$ in water is $8 \times 10^{-4} mol dm ^{-3}$, calculate its solubility in $0.01 mol dm ^{-3}$ of $H _2 SO _4$.

Answer


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Given solubility of $BaSO _4$ is $8 \times 10^{-4} mol dm ^{-3}$.
Therefore, for $BaSO _4$ in water, $K _{ sp }=\left[ Ba ^{2+}\right]\left[ SO _4^{2-}\right]=(S)(S)=S^2$
$
\therefore K_{s p}=\left(8 \times 10^{-4}\right)^2=64 \times 10^{-8} \ldots(1)
$
The expression for solubility product ( $K _{ sp }$ ) in the presence of sulphuric acid will be as follows:
$
K^{\prime} sp=(S)(S+0.01)\ldots(2)
$
Since the value of $K _{ sp }$ will not change in the presence of sulphuric acid.
Therefore, from (1) and (2), we have,
$
\begin{aligned}
& (S)(S+0.01)=64 \times 10^{-8} \\
& S^2+0.01 S=64 \times 10^{-8} \\
& S^2+0.01 S-64 \times 10^{-8}=0
\end{aligned}
$
$\begin{aligned} & \Rightarrow S=\frac{-0.01 \pm \sqrt{(0.01)^2+\left(4 \times 64 \times 10^{-8}\right)}}{2} \\ & \Rightarrow \frac{-0.01 \pm \sqrt{10^{-4}+\left(256 \times 10^{-8}\right)}}{2} \\ & \Rightarrow \frac{-0.01 \pm \sqrt{1.0256 \times 10^{-4}}}{2} \\ & \Rightarrow \frac{-0.01 \pm 1.0127 \times 10^{-2}}{2} \\ & \Rightarrow \frac{-0.01-1.0127 \times 10^{-2}}{2}, \frac{-0.01+1.0127 \times 10^{-2}}{2} \\ & \Rightarrow-1.006 \times 10^{-2}, 6.35 \times 10^{-5}\end{aligned}$
Concentration can never be negative. Hence required concentration is $6.35 \times 10^{-5} mol dm ^{-3}$

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