$50 \mathrm{ml}$ of $0.1 \mathrm{M}$ HA is titrated with $0.1 \mathrm{M}$ strong base (NaOH). The minimum…

$50 \mathrm{ml}$ of $0.1 \mathrm{M}$ HA is titrated with $0.1 \mathrm{M}$ strong base (NaOH). The minimum value of equilibriwn constant (k) so that when $49.5 \mathrm{ml}$ of titrant has been added, the reaction between $H A ~\&~ O H$ is essentially complete and the $p H$ changes by $2.00$ units on addition of two more drops $(0.1 \mathrm{~mL})$ of titrant is
  1. $2 \times 10^{10}$
  2. $2 \times 10^{9}$
  3. $5 \times 10^{8}$
  4. $2 \times 10^{7}$

Solution

The $\mathrm{pH} 0.05 \mathrm{~mL}$ beyond the equivalence point can be calculated as follows
$\left[\mathrm{OH}^{-}ight]=\frac{0.05 \times 0.1}{100.05}=5 \times 10^{-5} \mathrm{M}$
$\mathrm{POH}=4.3 \quad \mathbf{p H}=9.7$ before
If $\Delta \mathrm{pH}=2$ units, the $\mathrm{pH} 0.05 \mathrm{~mL}$ before the equivalence point must be $7.7 .$ At this point if the
reaction is complete, we have only $0.005$ m,mol of HA unreacted,
Hence
$\mathbf{p H}=\mathbf{p k}_{\mathrm{a}}+\log \frac{\left[\mathrm{A}^{-}ight]}{[\mathrm{HA}]} \quad 7.7=\mathrm{pk}_{\mathrm{a}}+\log \frac{4.995}{0.005}$
$p k_{a}=4.7 k_{a}=10^{-4.7}=2 \times 10^{-5} \quad k=\frac{k_{a}}{k_{w}}=\frac{2 \times 10^{-5}}{10^{-14}}=2 \times 10^{9}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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