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A saturated solution of $\mathrm{BaSO}_{4}$ at $25^{\circ} \mathrm{C}$ is $4 \times 10^{-5} \mathrm{M}$. The solubility of $\mathrm{BaSO}_{4}$ in $0.1 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}$ at this temperature will be
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The correct answer is:
$4 \times 10^{-4} \mathrm{M}$
$\begin{array}{r}
\mathrm{BaSO}_{4} \rightleftharpoons \mathrm{Ba}^{2+}+\mathrm{SO}_{4}^{2-} \\
\mathrm{s} \quad \mathrm{s}+0.1 \\
\mathrm{Na}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{Na}^{+}+\mathrm{SO}_{4}^{2-} \\
0.2 \quad 0.1
\end{array}$
$\begin{array}{l}
\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ba}^{+2}\right]\left[\mathrm{SO}_{4}^{-2}\right] \\
\therefore 4 \times 10^{-5}=\mathrm{s} \times(\mathrm{s}+0.1)
\end{array}$
or $4 \times 10^{-5} \approx \mathrm{s} \times 0.1 \quad \therefore \mathrm{s}=4 \times 10^{-4}$ as $\mathrm{s}<< < 0.1 \quad \therefore \mathrm{s}+0.1 \approx 0.1$
\mathrm{BaSO}_{4} \rightleftharpoons \mathrm{Ba}^{2+}+\mathrm{SO}_{4}^{2-} \\
\mathrm{s} \quad \mathrm{s}+0.1 \\
\mathrm{Na}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{Na}^{+}+\mathrm{SO}_{4}^{2-} \\
0.2 \quad 0.1
\end{array}$
$\begin{array}{l}
\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ba}^{+2}\right]\left[\mathrm{SO}_{4}^{-2}\right] \\
\therefore 4 \times 10^{-5}=\mathrm{s} \times(\mathrm{s}+0.1)
\end{array}$
or $4 \times 10^{-5} \approx \mathrm{s} \times 0.1 \quad \therefore \mathrm{s}=4 \times 10^{-4}$ as $\mathrm{s}<< < 0.1 \quad \therefore \mathrm{s}+0.1 \approx 0.1$
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