MHT CET · Chemistry · Ionic Equilibrium
Calculate the solubility in \(\mathrm{mol} \mathrm{dm}^{-3}\) of sparingly soluble salt BA if its solubility product \(4.9 \times 10^{-13}\) at same temperature.
- A \(7.0 \times 10^{-7}\)
- B \(7.5 \times 10^{-7}\)
- C \(8.0 \times 10^{-7}\)
- D \(4.9 \times 10^{-7}\)
Answer & Solution
Correct Answer
(A) \(7.0 \times 10^{-7}\)
Step-by-step Solution
Detailed explanation
For BA,
\(\mathrm{BA}_{(\mathrm{s})} \rightleftharpoons \mathrm{B}_{(\mathrm{aq})}^{+}+\mathrm{A}_{(\mathrm{aq})}^{-}\)
Here, \(x=1, y=1\)
\(\therefore \mathrm{K}_{\mathrm{sp}}=x^x \mathrm{y}^y \mathrm{~S}^{x+y}=(1)^1(1)^1 \mathrm{~S}^{1+1}=\mathrm{S}^2\)
\(\therefore \mathrm{~S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}=\sqrt{4.9 \times 10^{-13}}\) \(=7.0 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}\)
\(\mathrm{BA}_{(\mathrm{s})} \rightleftharpoons \mathrm{B}_{(\mathrm{aq})}^{+}+\mathrm{A}_{(\mathrm{aq})}^{-}\)
Here, \(x=1, y=1\)
\(\therefore \mathrm{K}_{\mathrm{sp}}=x^x \mathrm{y}^y \mathrm{~S}^{x+y}=(1)^1(1)^1 \mathrm{~S}^{1+1}=\mathrm{S}^2\)
\(\therefore \mathrm{~S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}=\sqrt{4.9 \times 10^{-13}}\) \(=7.0 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}\)
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