Question

When equal volumes of the following solutions are mixed, precipitation of AgCl $(K_{sp}=1.8\times {10}^{-10})$ will not occur only with:

• A. ${10}^{-4}$ M $\left[A{g}^{+}\right]$ and ${10}^{-4}M\left[HCl\right]$
• B. ${10}^{-3}M\left[A{g}^{+}\right]$ and ${10}^{-3}M\left[C{l}^{-}\right]$
• C. ${10}^{-6}M\left[A{g}^{+}\right]$ and ${10}^{-6}M\left[C{l}^{-}\right]$
• D. ${10}^{-2}M\left[A{g}^{+}\right]$ and ${10}^{-2}M\left[C{l}^{-}\right]$

Answer 1

Answer: (C)

${10}^{-6}M\left[A{g}^{+}\right]$ and ${10}^{-6}M\left[C{l}^{-}\right]$

Given : Ksp = 1.8 x 10${}^{-}$${}^1$${}^0$

Solution : condition for precitipation is,

ionic product > solubility product

ionic product = [$A{g}^{+}$][$C{l}^{-}$]

for option A

10${}^{-}$${}^4$ M [$A{g}^{+}$] and 10${}^{-}$${}^4$ M [$C{l}^{-}$] present since equal volumes are mixed so

$\frac{M1V1}{V1+V2}$

so ,

$\frac{{10}^{-}{.}^4}{2}\times \frac{{10}^{-}{.}^4}{2}=\frac{{10}^{-}{.}^8}{4}$

for option B.

$\frac{{10}^{-}3}{2}\times \frac{{10}^{-}3}{2}=\frac{{10}^{-}6}{4}$

for option C.

$\frac{{10}^{-}6}{2}\times \frac{{10}^{-}6}{2}=\frac{{10}^{-}12}{4}$

for option D.

$\frac{{10}^{-}2}{2}\times \frac{{10}^{-}2}{2}=\frac{{10}^{-}4}{4}$

only in option C ionic product < solubility product and only here precitipation will not occur

Hence , option C is correct