- Home
- Standard 11
- Chemistry
6-2.Equilibrium-II (Ionic Equilibrium)
hard
An excess of $Ag _{2} CrO _{4}(s)$ is added to a $5 \times 10^{-3}\, M$ $K _{2} CrO _{4}$ solution. The concentration of $Ag ^{+}$in the solution is closest to
[ Solubility product for $Ag _{2} CrO _{4}=1.1 \times 10^{-12}$ ]
A$2.2 \times 10^{-10} \,M$
B$15 \times 10^{-5} \,M$
C$10 \times 10^{-6} \,M$
D$5.0 \times 10^{-3}\, M$
(KVPY-2017)
Solution
(b)
Given,
solubility product of $Ag _{2} CrO _{4}=11 \times 10^{12}$
Concentration of $CrO _{4}^{2-}$ ions, $\left[ CrO _{4}\right]^{2-}$ $=5 \times 10^{-3}\, M$
For reaction,
$Ag _{2} CrO _{4} \rightleftharpoons 2 Ag ^{+}+ CrO _{4}^{2-}$
$K_{ sp } =\left[ Ag ^{2+}\right]^{2}\left[ CrO _{4}^{2-}\right]$
$11 \times 10^{-12}=\left[ Ag ^{2+}\right]^{2}\left[5 \times 10^{-13}\right]$
$\therefore \quad\left[ Ag ^{+}\right] =148 \times 10^{-5} \,M \approx 15 \times 10^{-5} \,M$
Given,
solubility product of $Ag _{2} CrO _{4}=11 \times 10^{12}$
Concentration of $CrO _{4}^{2-}$ ions, $\left[ CrO _{4}\right]^{2-}$ $=5 \times 10^{-3}\, M$
For reaction,
$Ag _{2} CrO _{4} \rightleftharpoons 2 Ag ^{+}+ CrO _{4}^{2-}$
$K_{ sp } =\left[ Ag ^{2+}\right]^{2}\left[ CrO _{4}^{2-}\right]$
$11 \times 10^{-12}=\left[ Ag ^{2+}\right]^{2}\left[5 \times 10^{-13}\right]$
$\therefore \quad\left[ Ag ^{+}\right] =148 \times 10^{-5} \,M \approx 15 \times 10^{-5} \,M$
Standard 11
Chemistry
