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A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution $\left(\mathrm{Q}_{\mathrm{sp}}\right)$ becomes greater than its solubility product. If the solubility of $\mathrm{BaSO}_4$ in water is $8 \times 10^{-4} \mathrm{~mol} \mathrm{~dm}{ }^{-3}$. Calculate its solubility in $0.01 \mathrm{~mol} \mathrm{~dm}^{-3}$ of $\mathrm{H}_2 \mathrm{SO}_4$.
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$\mathrm{BaSO}_4(\mathrm{~s}) \rightleftharpoons \mathrm{Ba}^{2+}(\mathrm{aq})+\mathrm{SO}_4^{2-}(\mathrm{aq})$
$\mathrm{K}_{\mathrm{sp}}$ for $\mathrm{BaSO}_4$ in water $=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{-2}\right]=[\mathrm{s}][\mathrm{s}]=\mathrm{s}^2$
But $\mathrm{s}=8 \times 10^{-4} \mathrm{~mol} \mathrm{~dm}^{-3}$
$\therefore \quad \mathrm{K}_{\mathrm{sp}}=\left(8 \times 10^{-4}\right)^2=64 \times 10^{-8}\quad\quad...(i)$
The expression for $K_{s p}$ in the presence of sulphuric acid will be as follows :
$\mathrm{K}_{\mathrm{sp}}=(\mathrm{s})(\mathrm{s}+0.01)\quad\quad...(ii)$
Since value of $K_{s p}$ will not change in the presence of sulphuric acid, therefore from (i) and (ii)
$\begin{aligned}
&\text { (s) }(\mathrm{s}+0.01)=64 \times 10^{-8} \\
&\Rightarrow \mathrm{s}^2+0.01 \mathrm{~s}=64 \times 10^{-8} \\
&\Rightarrow \mathrm{s}^2+0.01 \mathrm{~s}-64 \times 10^{-8}=0
\end{aligned}$
$\begin{aligned} \Rightarrow \quad \mathrm{s} &=\frac{-0.01 \pm \sqrt{(0.01)^2+\left(4 \times 64 \times 10^{-8}\right)}}{2} \\ &=\frac{-0.01 \pm \sqrt{10^{-4}+\left(256 \times 10^{-8}\right)}}{2} \\ &=\frac{-0.01 \pm \sqrt{10^{-4}+\left(1+256 \times 10^{-2}\right)}}{2} \\ &=\frac{-0.01 \pm 10^{-2} \sqrt{1+0.256}}{2} \\ &=\frac{-0.01 \pm 10^{-2} \sqrt{1.256}}{2} \\ &=\frac{-10^{-2}+\left(1.12 \times 10^{-2}\right)}{2} \end{aligned}$
$=\frac{-(-1+1.12) \times 10^{-2}}{2}=\frac{0.12}{2} \times 10^{-2}$
$=6 \times 10^{-4} \mathrm{~mol} \mathrm{~dm}^{-3}$
$\mathrm{K}_{\mathrm{sp}}$ for $\mathrm{BaSO}_4$ in water $=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{-2}\right]=[\mathrm{s}][\mathrm{s}]=\mathrm{s}^2$
But $\mathrm{s}=8 \times 10^{-4} \mathrm{~mol} \mathrm{~dm}^{-3}$
$\therefore \quad \mathrm{K}_{\mathrm{sp}}=\left(8 \times 10^{-4}\right)^2=64 \times 10^{-8}\quad\quad...(i)$
The expression for $K_{s p}$ in the presence of sulphuric acid will be as follows :
$\mathrm{K}_{\mathrm{sp}}=(\mathrm{s})(\mathrm{s}+0.01)\quad\quad...(ii)$
Since value of $K_{s p}$ will not change in the presence of sulphuric acid, therefore from (i) and (ii)
$\begin{aligned}
&\text { (s) }(\mathrm{s}+0.01)=64 \times 10^{-8} \\
&\Rightarrow \mathrm{s}^2+0.01 \mathrm{~s}=64 \times 10^{-8} \\
&\Rightarrow \mathrm{s}^2+0.01 \mathrm{~s}-64 \times 10^{-8}=0
\end{aligned}$
$\begin{aligned} \Rightarrow \quad \mathrm{s} &=\frac{-0.01 \pm \sqrt{(0.01)^2+\left(4 \times 64 \times 10^{-8}\right)}}{2} \\ &=\frac{-0.01 \pm \sqrt{10^{-4}+\left(256 \times 10^{-8}\right)}}{2} \\ &=\frac{-0.01 \pm \sqrt{10^{-4}+\left(1+256 \times 10^{-2}\right)}}{2} \\ &=\frac{-0.01 \pm 10^{-2} \sqrt{1+0.256}}{2} \\ &=\frac{-0.01 \pm 10^{-2} \sqrt{1.256}}{2} \\ &=\frac{-10^{-2}+\left(1.12 \times 10^{-2}\right)}{2} \end{aligned}$
$=\frac{-(-1+1.12) \times 10^{-2}}{2}=\frac{0.12}{2} \times 10^{-2}$
$=6 \times 10^{-4} \mathrm{~mol} \mathrm{~dm}^{-3}$
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