MHT CET · Chemistry · Ionic Equilibrium
The solubility product of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) is \(32 \times 10^{-12}\). What is the concentration of \(\mathrm{CrO}_{4}^{2-}\) ions in that solution?
- A \(2 \times 10^{-4}\)
- B \(16 \times 10^{-4}\)
- C \(8 \times 10^{-4}\)
- D \(12 \times 10^{-4}\)
Answer & Solution
Correct Answer
(A) \(2 \times 10^{-4}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Ag}_{2} \mathrm{CrO}_{4} \longrightarrow 2 \mathrm{Ag}^{+}+\mathrm{CrO}_{4}^{2-} \) \( 2 s \)
\( K_{\mathrm{sp}} =\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{CrO}_{4}^{2-}\right] \)
\( =(2 s)^{2}(s) \)
\( K_{\mathrm{sp}} =4 s^{3} \)
\( s =\left(\frac{K_{\mathrm{sp}}}{4}\right)^{1 / 3} \)
\( =\left(\frac{32 \times 10^{-12}}{4}\right)^{1 / 3} \)
\( =2 \times 10^{-4} \mathrm{M}\)
\( K_{\mathrm{sp}} =\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{CrO}_{4}^{2-}\right] \)
\( =(2 s)^{2}(s) \)
\( K_{\mathrm{sp}} =4 s^{3} \)
\( s =\left(\frac{K_{\mathrm{sp}}}{4}\right)^{1 / 3} \)
\( =\left(\frac{32 \times 10^{-12}}{4}\right)^{1 / 3} \)
\( =2 \times 10^{-4} \mathrm{M}\)
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