Chemical and Ionic Equilibrium - Result Question 113
####57. (a) Find the solubility product of a saturated solution of $Ag _2 CrO _4$ in water at $298 K$ if the emf of the cell $Ag \mid Ag^{+}$(saturated. $Ag _2 CrO _4$ solution.) $| Ag^{+}(0.1 M) \mid Ag$ is $0.164 V$ at $298 K$.
(1998, 6M)
(b) What will be the resultant $pH$ when $200 mL$ of an aqueous solution of $HCl(pH=2.0)$ is mixed with $300 mL$ of an aqueous solution of $NaOH(pH=12.0)$ ?
$(1998,6 M)$
Show Answer
Solution:
- (a) $E=0.164=-0.059 \log \frac{\left[Ag^{+}\right] _{\text {anode }}}{0.10}$
$$ \left[Ag^{+}\right] _{\text {anode }}=1.66 \times 10^{-4} M $$
$$ \begin{aligned} {\left[CrO _4^{2-}\right] } & =\frac{\left[Ag^{+}\right]}{2}=8.3 \times 10^{-5} M \ K _{\text {sp }} & =\left[Ag^{+}\right]^{2}\left[CrO _4^{2-}\right] \ & =\left(1.66 \times 10^{-4}\right)^{2}\left(8.3 \times 10^{-5}\right) \ & =2.3 \times 10^{-12} \end{aligned} $$
(b) $pH$ of $HCl=2$
$$ \therefore \quad\left[H^{+}\right]=10^{-2} M $$
Moles of $H^{+}$ions in $200 mL$ of $10^{-2} M HCl$ solution
$$ =\frac{10^{-2}}{1000} \times 200=2 \times 10^{-3} $$
Similarly, $pH$ of $NaOH=12$
$$ \begin{aligned} & \therefore \quad\left[H^{+}\right]=10^{-12} M \ & \text { or }\left[OH^{-}\right]=10^{-2} M \ & {\left[\because\left[H^{+}\right]\left[OH^{-}\right]=10^{-14} m\right]} \end{aligned} $$
Moles of $OH^{-}$ion in $300 mL$ of $10^{-2} M NaOH$ solution
$$ =\frac{10^{-2}}{1000} \times 300=3 \times 10^{-3} $$
Total volume of solution after mixing $=500 mL$
Moles of $OH^{-}$ion left in $500 mL$ of solution
$$ =\left(3 \times 10^{-3}\right)-\left(2 \times 10^{-3}\right)=10^{-3} $$
Molar concentration of $OH^{-}$ions in the resulting
$$ \begin{aligned} \text { solution } & =\frac{10^{-3}}{500} \times 1000=2 \times 10^{-3} M \ pOH & =-\log \left(2 \times 10^{-3}\right) \ & =-\log 2+3 \log 10 \ & =-0.3 \simeq 103=2.699 \ \therefore \quad pH & =14-2.699=11.301 \end{aligned} $$
