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MHT CET · Chemistry · Ionic Equilibrium

Find solubility in terms of \(\mathrm{mol} \mathrm{L}^{-1}\) if solubility product of silver bromide is \(6.4 \times 10^{-13}\).

  1. A \(4.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)
  2. B \(8.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\)
  3. C \(7.5 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)
  4. D \(6.4 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(8.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\)

Step-by-step Solution

Detailed explanation

\(\mathrm{AgBr}_{(\mathrm{s})} \rightleftharpoons \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{Br}_{(\mathrm{sq})}^{-} \)
\( \therefore x=1, y=1 \)
\( \mathrm{~K}_{\mathrm{sp}}=x^x y^y \mathrm{~S}^{x+y}=(1)^1(1)^1 \mathrm{~S}^{1+1}=\mathrm{S}^2 \)
\( \therefore \mathrm{S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}=\sqrt{6.4 \times 10^{-13}}=\sqrt{64 \times 10^{-14}} \)
\( =8 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\)