MHT CET · Chemistry · Ionic Equilibrium
Solubility of \(\mathrm{AgCl}\) is \(7.2 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}\). What is it's solubility product?
- A \(3.6 \times 10^{-13}\)
- B \(7.2 \times 10^{-14}\)
- C \(2.59 \times 10^{-14}\)
- D \(5.18 \times 10^{-14}\)
Answer & Solution
Correct Answer
(D) \(5.18 \times 10^{-14}\)
Step-by-step Solution
Detailed explanation
Solubility product of \(\operatorname{AgCl}\left(\mathrm{K}_{\mathrm{sp}}\right)=\mathrm{s}^2\)
Solubility of \(\mathrm{AgCl}(\mathrm{s})=7.2 \times 10^{-7} \mathrm{~mol} / \mathrm{L}\)
\(
\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=\left(7.2 \times 10^{-7}\right)^2 \\
& =51.8 \times 10^{-14} \\
& \mathrm{~K}_{\mathrm{sp}}=5.18 \times 10^{-13}
\end{aligned}
\)
Solubility of \(\mathrm{AgCl}(\mathrm{s})=7.2 \times 10^{-7} \mathrm{~mol} / \mathrm{L}\)
\(
\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=\left(7.2 \times 10^{-7}\right)^2 \\
& =51.8 \times 10^{-14} \\
& \mathrm{~K}_{\mathrm{sp}}=5.18 \times 10^{-13}
\end{aligned}
\)
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