The solubility of \(\mathrm{AgBr}\) with solubility product \(5.0 \times 10^{-13}\) at \(298 \mathrm{~K}\)…
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
\(7 \times 10^{-6} \mathrm{M}\)
\(5 \times 10^{-12} \mathrm{M}\)
\(5 \times 10^{-14} \mathrm{M}\)
\(5 \times 10^{-6} \mathrm{M}\)
Solution
Let, the solubility of \(\mathrm{AgBr}\) be \(S \mathrm{~mol} / \mathrm{L}\).
\(\mathrm{AgBr} ightleftharpoons \mathrm{Ag}^{+}+\mathrm{Br}^{-}\)
Hence, \(\left[\mathrm{Ag}^{+}ight]\left[\mathrm{Br}^{-}ight]=5 \times 10^{-13}\)
Given that, \(\left[\mathrm{Br}^{-}ight]=0.1\) (from \(\mathrm{NaBr}\))
So, \(\left[\mathrm{Ag}^{+}ight]=\left(5 \times 10^{-13}ight) / 0.1=5 \times 10^{-12} \mathrm{M}\)
It means solubility in \(\mathrm{NaBr}\) is \(5 \times 10^{-12}\).
Hence, option (b) is correct.