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An excess of $\mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{~s})$ is added to a $5 \times 10^{-3} \mathrm{M}$ $\mathrm{K}_{2} \mathrm{CrO}_{4}$ solution. The concentration of $\mathrm{Ag}^{+}$in the solution is closest to
$\left[\right.$ Solubility product for $\left.\mathrm{Ag}_{2} \mathrm{CrO}_{4}=1.1 \times 10^{-12}\right]$
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$\left[\right.$ Solubility product for $\left.\mathrm{Ag}_{2} \mathrm{CrO}_{4}=1.1 \times 10^{-12}\right]$
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Verified Answer
The correct answer is:
$1.5 \times 10^{-5} \mathrm{M}$
$1.1 \times 10^{-12}=\left[\mathrm{Ag}^{+}\right]^{2}\left[5 \times 10^{-13}\right]$
$\therefore\left[\mathrm{Ag}^{+}\right]=1.5 \times 10^{-5}$
$\therefore\left[\mathrm{Ag}^{+}\right]=1.5 \times 10^{-5}$
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