AP EAMCET · Chemistry · Ionic Equilibrium
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
- A \(7 \times 10^{-6} \mathrm{M}\)
- B \(5 \times 10^{-12} \mathrm{M}\)
- C \(5 \times 10^{-14} \mathrm{M}\)
- D \(5 \times 10^{-6} \mathrm{M}\)
Answer & Solution
Correct Answer
(B) \(5 \times 10^{-12} \mathrm{M}\)
Step-by-step Solution
Detailed explanation
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,…
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