Question

$200$ml of $0.005$M $AgN{O}_3$ reacts with $300$ mL of $0.01$ M KCl. If $K_{sp}$ of $AgCl$ is $1.8\times {10}^{-10}$. Then maximum conc. of $A{g}^{+}$ in mixture is:

• A. $2\times {10}^{-8}$
• B. $4.5\times {10}^{-8}$
• C. $4.8\times {10}^{-5}$
• D. $1.34\times {10}^{-5}$

Answer 1

The correct option is B $4.5\times {10}^{-8}$

No. of moles of $A{g}^{+}$ ions $=200\times \frac{0.005}{1000}={10}^{-3}moles$

No. of moles of $C{l}^{-}$ ions $=300\times \frac{0.01}{100}=3\times {10}^{-3}moles$

$A{g}_{\left(aq\right)}^{+}+C{l}_{\left(aq\right)}^{\ominus }\to AgCl\left(s\right)$

$C{l}^{-}$ ions are in excess, therefore, it will consume all $A{g}^{+}$ ions and form $AgCl$

Thus, Amount of $C{l}^{-}$ ions left $=3\times {10}^{-3}-{10}^{-3}=2\times {10}^{-3}moles$

$[C{l}^{-}{]}_{left}=\frac{2\times {10}^{-3}}{300+200}\times 1000=4\times {10}^{-3}M=0.004M$

Dissociation of $AgCl$ to its ions will produce an equal amount of each ion ( say S)

Then $\left[A{g}^{+}\right]=S$ and $\left[a^{-}\right]=S+0.004\simeq 0.004$

$K_{cp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]=1.8\times {10}^{-10}$

$\Rightarrow \left[A{g}^{+}\right]=\frac{1.8\times {10}^{-10}}{\left[C{l}^{-}\right]}=\frac{1.8\times {10}^{-10}}{0.004}$

Maximum conc of $A{g}^{+}$ in the mixture is $4.5\times {10}^{-8}M$

Hence, the correct option is $B$