Chemical and Ionic Equilibrium - Result Question 56
57. (a) Find the solubility product of a saturated solution of $\mathrm{Ag}_2 \mathrm{CrO}_4$ in water at 298 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 V at 298 K .
(1998, 6M)
(b) What will be the resultant pH when 200 mL of an aqueous solution of $\mathrm{HCl}(\mathrm{pH}=2.0)$ is mixed with 300 mL of an aqueous solution of $\mathrm{NaOH}(\mathrm{pH}=12.0)$ ?
(1998,6 M)
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Solution:
(a)
$ \begin{aligned} 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} \\ {\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) pH of $\mathrm{HCl}=2$
$ \therefore \quad\left[\mathrm{H}^{+}\right]=10^{-2} \mathrm{M} $
Moles of $\mathrm{H}^{+}$ ions in 200 mL of $10^{-2} \mathrm{M} \mathrm{HCl}$ solution
$ =\frac{10^{-2}}{1000} \times 200=2 \times 10^{-3} $
Similarly, pH of $\mathrm{NaOH}=12$
$\therefore \quad\left[\mathrm{H}^{+}\right]=10^{-12} \mathrm{M}$
or $\left[\mathrm{OH}^{-}\right]=10^{-2} \mathrm{M}\quad$ $ \left[\because\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-14} \mathrm{~m}\right] $
Moles of $\mathrm{OH}^{-}$ion in $300$ 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 $500$ 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} $
BETA
