Question

At $25$, the $K_{sp}$ value of $AgCl$ is $1.8\times {10}^{-10}$. If ${10}^{-5}$ moles of ${Ag}^{+}$ are added to solution then $K_{sp}$ will be:

• A. $1.8\times {10}^{-15}$
• B. $1.8\times {10}^{-10}$
• C. $5.04\times {10}^{-5}$
• D. $1.8\times {10}^{+10}$

Answer 1

The correct option is C $5.04\times {10}^{-5}$

$AgCl\rightleftharpoons {Ag}^{+}+{Cl}^{-}initialssafter{s}^{\prime }{s}^{\prime }adding{Ag}^{+}{Ag}^{+}=C={10}^{-5}{K^{\prime }}_{sp}=s^{\prime }(s^{\prime }+C)$

Let us assume $C>>s^{\prime }$

${K^{\prime }}_{sp}=s^{\prime }\times C{s}^{\prime }=\frac{1.8\times {10}^{-10}}{{10}^{-5}}=1.8\times {10}^{-5}{K^{\prime }}_{sp}=s^{\prime }(s^{\prime }+C)=1.8\times {10}^{-5}(1.8\times {10}^{-5}+{10}^{-5})=5.04\times {10}^{-5}$