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Question: Answered & Verified by Expert
The solubility product constants of \(\mathrm{Ag}_2 \mathrm{CrO}_4\) and \(\mathrm{AgBr}\) are \(1.1 \times 10^{-12}\) and \(5.0 \times 10^{-13}\) respectively. Calculate the ratio of the molarities of their saturated solutions.
ChemistryEquilibrium
Solution:
1252 Upvotes Verified Answer


\(\mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~s})^2(\mathrm{~s})=4 \mathrm{~s}^3, \cdot \cdot \mathrm{s}^3=\mathrm{K}_{\mathrm{sp}} / 4\)
\(\therefore \quad \mathrm{s}=6.5 \times 10^{-5}\)
\(\mathrm{AgBr} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Br}^{-}\)
\(\mathrm{K}_{\mathrm{sp}}=\mathrm{s}^2=5 \times 10^{-13}\)
\(\therefore \quad s=\sqrt{5 \times 10^{-13}}=7.07 \times 10^{-7}\)
\(\therefore\) ratio of \(\left[\mathrm{Ag}_2 \mathrm{CrO}_4\right]:[\mathrm{AgBr}]\)
\(=6.5 \times 10^{-5}: 7.07 \times 10^{-7}\)
\(=91.9\)
Silver chromate is more soluble than silver bromide.

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