MHT CET · Chemistry · Ionic Equilibrium
Find solubility in terms of \(\mathrm{mol} \mathrm{L}^{-1}\) if solubility product of silver bromide is \(6.4 \times 10^{-13}\).
- A \(4.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)
- B \(8.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\)
- C \(7.5 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)
- D \(6.4 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\)
Answer & Solution
Correct Answer
(B) \(8.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{AgBr}_{(\mathrm{s})} \rightleftharpoons \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{Br}_{(\mathrm{sq})}^{-} \)
\( \therefore x=1, y=1 \)
\( \mathrm{~K}_{\mathrm{sp}}=x^x y^y \mathrm{~S}^{x+y}=(1)^1(1)^1 \mathrm{~S}^{1+1}=\mathrm{S}^2 \)
\( \therefore \mathrm{S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}=\sqrt{6.4 \times 10^{-13}}=\sqrt{64 \times 10^{-14}} \)
\( =8 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\)
\( \therefore x=1, y=1 \)
\( \mathrm{~K}_{\mathrm{sp}}=x^x y^y \mathrm{~S}^{x+y}=(1)^1(1)^1 \mathrm{~S}^{1+1}=\mathrm{S}^2 \)
\( \therefore \mathrm{S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}=\sqrt{6.4 \times 10^{-13}}=\sqrt{64 \times 10^{-14}} \)
\( =8 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\)
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