Chemical and Ionic Equilibrium 2 Question 57
57. (a) Find the solubility product of a saturated solution of $\mathrm{Ag}{2} \mathrm{CrO}{4}$ in water at $298 \mathrm{~K}$ if the emf of the cell $\mathrm{Ag} \mid \mathrm{Ag}^{+}$(saturated. $\mathrm{Ag}{2} \mathrm{CrO}{4}$ solution.) $| \mathrm{Ag}^{+}(0.1 \mathrm{M}) \mid \mathrm{Ag}$ is $0.164 \mathrm{~V}$ at $298 \mathrm{~K}$.
(1998, 6M)
(b) What will be the resultant $\mathrm{pH}$ when $200 \mathrm{~mL}$ of an aqueous solution of $\mathrm{HCl}(\mathrm{pH}=2.0)$ is mixed with $300 \mathrm{~mL}$ of an aqueous solution of $\mathrm{NaOH}(\mathrm{pH}=12.0)$ ?
$(1998,6 M)$
Show Answer
Solution:
- (a) $E=0.164=-0.059 \log \frac{\left[\mathrm{Ag}^{+}\right]_{\text {anode }}}{0.10}$
$$ \left[\mathrm{Ag}^{+}\right]_{\text {anode }}=1.66 \times 10^{-4} \mathrm{M} $$
$$ \begin{aligned} {\left[\mathrm{CrO}{4}^{2-}\right] } & =\frac{\left[\mathrm{Ag}^{+}\right]}{2}=8.3 \times 10^{-5} \mathrm{M} \ K{\text {sp }} & =\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{CrO}_{4}^{2-}\right] \ & =\left(1.66 \times 10^{-4}\right)^{2}\left(8.3 \times 10^{-5}\right) \ & =2.3 \times 10^{-12} \end{aligned} $$
(b) $\mathrm{pH}$ of $\mathrm{HCl}=2$
$$ \therefore \quad\left[\mathrm{H}^{+}\right]=10^{-2} \mathrm{M} $$
Moles of $\mathrm{H}^{+}$ions in $200 \mathrm{~mL}$ of $10^{-2} \mathrm{M} \mathrm{HCl}$ solution
$$ =\frac{10^{-2}}{1000} \times 200=2 \times 10^{-3} $$
Similarly, $\mathrm{pH}$ of $\mathrm{NaOH}=12$
$$ \begin{aligned} & \therefore \quad\left[\mathrm{H}^{+}\right]=10^{-12} \mathrm{M} \ & \text { or }\left[\mathrm{OH}^{-}\right]=10^{-2} \mathrm{M} \ & {\left[\because\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-14} \mathrm{~m}\right]} \end{aligned} $$
Moles of $\mathrm{OH}^{-}$ion in $300 \mathrm{~mL}$ of $10^{-2} \mathrm{M} \mathrm{NaOH}$ solution
$$ =\frac{10^{-2}}{1000} \times 300=3 \times 10^{-3} $$
Total volume of solution after mixing $=500 \mathrm{~mL}$
Moles of $\mathrm{OH}^{-}$ion left in $500 \mathrm{~mL}$ of solution
$$ =\left(3 \times 10^{-3}\right)-\left(2 \times 10^{-3}\right)=10^{-3} $$
Molar concentration of $\mathrm{OH}^{-}$ions in the resulting
$$ \begin{aligned} \text { solution } & =\frac{10^{-3}}{500} \times 1000=2 \times 10^{-3} \mathrm{M} \ \mathrm{pOH} & =-\log \left(2 \times 10^{-3}\right) \ & =-\log 2+3 \log 10 \ & =-0.3 \simeq 103=2.699 \ \therefore \quad \mathrm{pH} & =14-2.699=11.301 \end{aligned} $$
