Question

$20\, ml$. 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_3COOH = 1.8 \times 10^{ - 5 } )$.

Answer 1

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 $6 \,mmol$ $CH_3COOH$ and $4 \,mmol$ $CH_3COONa. $
$pH = pK_a + log \frac{ [ CH_3COONa ]}{ CH_3COOH ]}$
$= - log ( 1.8 \times 10^{ - 5 }) + log \frac{ 4 }{ 6 } = 4 . 56 $
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_3COONa $
and $(6 - x) \,mmol$ of $CH_3COOH.$
$4.74 = - log ( 1.9 \times 10^{ - 5 }) + log \frac{ 4 + x}{ 6 - x} $
$\Rightarrow \frac{ 4 + x}{ 6 - x} = 1 \Rightarrow x = 1.0\, mmol = 0.2 \times\, V$
$\Rightarrow V = 5.0\, mmol \,NaOH$.