Question

The solubility of $A g C l$ in $0.2\, M\, NaCl$ solution is $\left(K_{s p}\right.$ of $\left.A g C l=1.20 \times 10^{-10}\right)$

• A. $ 6.0\times 10^{-10}\,M $
• B. $ 0.2\,M $
• C. $ 1.2\times 10^{-10}\, M $
• D. $ 0.2\times 10^{-10}\, M $

Answer 1

Answer: (A)

Given, concentration of $N a C l=0.2\, M$
$K_{sp}(AgCl)=1.20 \times 10^{-10}$
Let the solubility of $AgCl$ in $NaCl =x$
$ AgCl \to Ag^{+} +Cl^{-}$

$\therefore \left[A g^{+}\right]=x$ and $\left[C l^{-}\right]=(x+0.2)$
$\therefore K_{s p}(A g C l)=\left[A g^{+}\right]\left[C l^{-}\right]$
$=x(x+0.2)$
$=x^{2}+0.2\, x$
$\therefore K_{s p}=0.2 x\left(x^{2} < < 1\right)$
or $1.2 \times 10^{-10}=0.2 x$
$\therefore x=6 \times 10^{-10}$