Question

When equal volumes of the following solutions are mixed precipitation of $AgCl(K_{sp}=1.8\times {10}^{-10})$ will occur only with:

• A. ${10}^{-4}MA{g}^{+}$ and ${10}^{-4}MC{l}^{-}$
• B. ${10}^{-5}MA{g}^{+}$ and ${10}^{-5}MC{l}^{-}$
• C. ${10}^{-6}MA{g}^{+}$ and ${10}^{-6}MC{l}^{-}$
• D. ${10}^{-10}MA{g}^{+}$ and ${10}^{-10}MC{l}^{-}$

Answer 1

The correct option is A ${10}^{-4}MA{g}^{+}$ and ${10}^{-4}MC{l}^{-}$

Given, $AgCl(K_{sp}=1.8\times {10}^{-10})$
For precipitation the ionic product should exceed solubility product $\left(K_{sp}\right)$.

(a) ${10}^{-4}MA{g}^{+}$ and ${10}^{-4}MC{l}^{-}$
$\left[A{g}^{+}\right]\left[C{l}^{-}\right]=({10}^{-4}\times {10}^{-4})={10}^{-8}{M}^2$
$\left[A{g}^{+}\right]\left[C{l}^{-}\right]={10}^{-8}>K_{sp}^0=1.8\times {10}^{-10})$
Hence, precipitation will take place.

b) ${10}^{-5}MA{g}^{+}$ and ${10}^{-5}MC{l}^{-}$
$\left[A{g}^{+}\right]\left[C{l}^{-}\right]=({10}^{-5}\times {10}^{-5})={10}^{-8}{M}^2$
$\left[A{g}^{+}\right]\left[C{l}^{-}\right]={10}^{-10}<K_{sp}^0=1.8\times {10}^{-10})$
Hence, no precipitation will take place.

c) ${10}^{-6}MA{g}^{+}$ and ${10}^{-6}MC{l}^{-}$
$\left[A{g}^{+}\right]\left[C{l}^{-}\right]=({10}^{-6}\times {10}^{-6})={10}^{-12}{M}^2$
$\left[A{g}^{+}\right]\left[C{l}^{-}\right]={10}^{-12}<K_{sp}^0=1.8\times {10}^{-10})$
Hence, no precipitation will take place.

d) ${10}^{-10}MA{g}^{+}$ and ${10}^{-10}MC{l}^{-}$
$\left[A{g}^{+}\right]\left[C{l}^{-}\right]=({10}^{-10}\times {10}^{-10})={10}^{-20}{M}^2$
$\left[A{g}^{+}\right]\left[C{l}^{-}\right]={10}^{-20}<K_{sp}^0=1.8\times {10}^{-10})$
Hence, no precipitation will take place.