Chemical and Ionic Equilibrium - Result Question 136
####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 $CH _3 COOH=1.8 \times 10^{-5}$ ).
$(1982,3 M)$
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Solution:
- $mmol$ of $NaOH=20 \times 0.2=4$
mmol of acetic acid $=50 \times 0.2=10$
After neutralisation, buffer solution is formed which contain 6 $mmol CH _3 COOH$ and $4 mmol CH _3 COONa$.
$$ \begin{aligned} pH & =p K _a+\log \frac{\left[CH _3 COONa\right]}{\left[CH _3 COOH\right]} \ & =-\log \left(1.8 \times 10^{-5}\right)+\log \frac{4}{6}=4.56 \end{aligned} $$
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 $(4+x) mmol CH _3 COONa$ and $(6-x) mmol$ of $CH _3 COOH$.
$$ \begin{array}{rlrl} & & 4.74 & =-\log \left(1.8 \times 10^{-5}\right)+\log \frac{4+x}{6-x} \ \Rightarrow & & \frac{4+x}{6-x} & =1 \ \Rightarrow & & x & =1.0 mmol=0.2 \times V \ \Rightarrow & V & =5.0 mmol NaOH . \end{array} $$
