Chemical and Ionic Equilibrium - Result Question 56
78. $20$ mL of $0.2$ M sodium hydroxide is added to $50$ mL of $0.2$ M acetic acid solution to give $70$ mL of the solution. What is the pH of this solution?
Calculate the additional volume of $0.2$ M NaOH required to make the pH of the solution $4.74$ . (Ionisation constant of $\mathrm{CH}_3 \mathrm{COOH}=1.8 \times 10^{-5}$ ).
(1982,3 M)
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Solution:
- mmol of $\mathrm{NaOH}=20 \times 0.2=4$
mmol of acetic acid $=50 \times 0.2=10$
After neutralisation, buffer solution is formed which contain $6$ $\mathrm{mmol} $ $\mathrm{CH}_3 \mathrm{COOH}$ and $4 $ $\mathrm{mmol} $ $\mathrm{CH}_3 \mathrm{COONa}$.
$\begin{aligned} \mathrm{pH} & =\mathrm{p} K_a+\log \frac{\left[\mathrm{CH}_3 \mathrm{COONa}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \\ & =-\log \left(1.8 \times 10^{-5}\right)+\log \frac{4}{6}=4.56 \end{aligned}$
Now, let $x$ mmol of NaOH is further added so that pH of the resulting buffer solution is $4.74$ .
Now, the buffer solution contains $(4+x)$ $ \mathrm{mmol} $ $\mathrm{CH}_3 \mathrm{COONa}$ and $(6-x) \mathrm{mmol}$ of $\mathrm{CH}_3 \mathrm{COOH}$.
$4.74 =-\log \left(1.8 \times 10^{-5}\right)+\log \frac{4+x}{6-x} $
$\Rightarrow \frac{4+x}{6-x} =1 $
$\Rightarrow x =1.0$ $ \mathrm{mmol}=0.2 \times V $
$\Rightarrow V =5.0 $ $\mathrm{mmol}$ $ \mathrm{NaOH} .$
BETA
