Chemical and Ionic Equilibrium 2 Question 80
78. $20 \mathrm{~mL}$ of $0.2 \mathrm{M}$ sodium hydroxide is added to $50 \mathrm{~mL}$ of $0.2 \mathrm{M}$ acetic acid solution to give $70 \mathrm{~mL}$ of the solution. What is the $\mathrm{pH}$ of this solution?
Calculate the additional volume of $0.2 \mathrm{M} \mathrm{NaOH}$ required to make the $\mathrm{pH}$ 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:
- $\mathrm{mmol}$ 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 $x \mathrm{mmol}$ of $\mathrm{NaOH}$ is further added so that $\mathrm{pH}$ 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 $\mathrm{CH}{3} \mathrm{COOH}$.
$$ \begin{array}{rlrl} & & 4.74 & =-\log \left(1.8 \times 10^{-5}\right)+\log \frac{4+x}{6-x} \ \Rightarrow & & \frac{4+x}{6-x} & =1 \ \Rightarrow & & x & =1.0 \mathrm{mmol}=0.2 \times V \ \Rightarrow & V & =5.0 \mathrm{mmol} \mathrm{NaOH} . \end{array} $$
