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The solubility of \(\mathrm{AgBr}\) with solubility product \(5.0 \times 10^{-13}\) at \(298 \mathrm{~K}\) in \(0.1 \mathrm{M} \mathrm{NaBr}\) solution would be
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Verified Answer
The correct answer is:
\(5 \times 10^{-12} \mathrm{M}\)
Let, the solubility of \(\mathrm{AgBr}\) be \(S \mathrm{~mol} / \mathrm{L}\).
\(\mathrm{AgBr} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Br}^{-}\)
Hence, \(\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Br}^{-}\right]=5 \times 10^{-13}\)
Given that, \(\left[\mathrm{Br}^{-}\right]=0.1\) (from \(\mathrm{NaBr}\))
So, \(\left[\mathrm{Ag}^{+}\right]=\left(5 \times 10^{-13}\right) / 0.1=5 \times 10^{-12} \mathrm{M}\)
It means solubility in \(\mathrm{NaBr}\) is \(5 \times 10^{-12}\).
Hence, option (b) is correct.
\(\mathrm{AgBr} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Br}^{-}\)
Hence, \(\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Br}^{-}\right]=5 \times 10^{-13}\)
Given that, \(\left[\mathrm{Br}^{-}\right]=0.1\) (from \(\mathrm{NaBr}\))
So, \(\left[\mathrm{Ag}^{+}\right]=\left(5 \times 10^{-13}\right) / 0.1=5 \times 10^{-12} \mathrm{M}\)
It means solubility in \(\mathrm{NaBr}\) is \(5 \times 10^{-12}\).
Hence, option (b) is correct.
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