An excess of Ag_2 CrO_4(~s) is added to a 5 × 10^-3 M K_2 CrO_4 solution. The concentration of Ag^+ in the solution is closest to [ . Solubility product for .Ag_2 CrO_4=1.1 × 10^-12 ]

Correct answer: (B) 1.5 × 10^-5 M. 1.1 × 10^-12= [Ag^+ ]^2 [5 × 10^-13 ] ∴ [Ag^+ ]=1.5 × 10^-5

Q24 / 90 · KVPY 2017 (19 Nov SB/SX) · Chemistry

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 $\left.\mathrm{Ag}_{2} \mathrm{CrO}_{4}=1.1 \times 10^{-12}\right]$

$1.1 \times 10^{-12}=\left[\mathrm{Ag}^{+}\right]^{2}\left[5 \times 10^{-13}\right]$ $\therefore\left[\mathrm{Ag}^{+}\right]=1.5 \times 10^{-5}$

Hence, the correct answer is option: 'B'.