Question

The solubility product of a saturated solution of $A{g}_{2}Cr{O}_4$ in water at 298 K will be................., if the EMF of the cell:

$Ag|A{g}^{\oplus }\left(satA{g}_{2}Cr{O}_{4}sol\right)||Ag\left(0.1M\right)|Ag$ is $0.164V$ at 298 K.

• A. $2.287\times {10}^{-12}{M}^3$
• B. $2.659\times {10}^{-12}{M}^3$
• C. $1.965\times {10}^{-12}{M}^3$
• D. $Noneofthese$

Answer 1

Answer: (A)

$2.287\times {10}^{-12}{M}^3$

Cell circuit is given as below:

$Ag|A{g}^{\oplus }$ (saturated) $A{g}_{2}Cr{O}_4||A{g}^{\oplus }\left(0.1M\right)|Ag$

$E_{cell}=0.164$ at 298 K.

For the given cell (concentration cell)

$E_{cell}=\frac{0.0591}{n}log\frac{c_2}{c1}$

$c_1=[A{g}^{\oplus }$ in $A{g}_{2}Cr{O}_4$ (left side)

$c_2=\left(A{g}^{\oplus }\right)$ (right side)

N = 1 (number of exchange electron)

So, $0.164=\frac{0.0591}{1}log\frac{0.1}{c_1}$

Or $log\frac{0.1}{c_1}=\frac{0.164}{0.0591}$ = 2.774

On solving

$c_1=\left[A{g}^{\oplus }\right]inA{g}_{2}Cr{O}_4=1.66\times {10}^{-4}M$

We get $A{g}_{2}Cr{O}_4\rightleftharpoons 2A{g}^{\oplus }+Cr{O}_4^{2-}$

$\left[Cr{O}_4^{2-}\right]=\frac{\left[A{g}^{\oplus }\right]}{2}$

So, $[Cr{O}_4^{2-}=\frac{1.66\times {10}^{-4}}{2}M=0.83\times {10}^{-4}M$

$\therefore K_{sp}ofA{g}_{2}Cr{O}_4=\left[A{g}^{\oplus }{]}^2\right[Cr{O}_4^{2-}]$

$=(1.66\times {10}^{-4})(0.83\times {10}^{-4})$

$=2.287\times {10}^{-12}{M}^3$