What volume of $0.80 \mathrm{M}$ NaOH must be added to $50 \mathrm{~mL} \mathrm{M}$ solution of…

What volume of $0.80 \mathrm{M}$ NaOH must be added to $50 \mathrm{~mL} \mathrm{M}$ solution of $\mathrm{HA}$, a weak acid $\left(\mathrm{K}_{\mathrm{b}}ight.$
for $A^{-}=10^{-10}$ and $\left.K_{w}=10^{-14}ight)$ so as to adjust the $\mathrm{pH}$ of the mixture at $4.6$ ?
  1. $62 \mathrm{~mL}$
  2. $60 \mathrm{~mL}$
  3. $50 \mathrm{~mL}$
  4. $15 \mathrm{~mL}$

Solution

Let, v $\mathrm{mL}$ of $0.80 \mathrm{M}$ NaOH be added then
No. of $m$ mole of salt $($ NaA $)$ formed $=0.8 v$
No. of $\mathrm{m}$ mole of acid remaining unreacted $=50-0.8 \mathrm{v}$
$\mathrm{K}_{\mathrm{a}}$ for $\mathrm{HA}=10^{-14} / 10^{-10}=10^{-4}$
$\therefore \mathrm{pK}_{\mathrm{a}}=4$
$4.6=4+\log \frac{0.8 v}{50-0.8 v}$
$\therefore \frac{0.8 v}{50-0.8 v}=4, \quad$ So $v=50 \mathrm{~mL}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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