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6-2.Equilibrium-II (Ionic Equilibrium)
hard
The $K _{\text {s }}$ of $Ag _2 CrO _4$ is $1.1 \times 10^{-12}$ at $298 K$. The solubility (in $mol / L$ ) of $Ag _2 CrO$, in a $0.1 M AgNO _2$, solution is
A
$(A)$ $1.1 \times 10^{-11}$
B
$(B)$ $1.1 \times 10^{-10}$
C
$(C)$ $1.1 \times 10^{-12}$
D
$(D)$ $1.1 \times 10^{-9}$
(IIT-2013)
Solution
$\begin{array}{c} Ag _2 CrO _4 \rightleftharpoons 2 Ag ^{+}+ CrO _4^{2-} \\ s \\ 0.1+2 s \quad s \\ \approx 0.1 \\ 1.1 \times 10^{-12}=(0.1)^2 s \\ s =1.1 \times 10^{-10}\end{array}$
Standard 11
Chemistry
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