Question

$20\, mL$ of $M / 10 \, CH _3 COOH$ solution is titrated with $M/10 \, NaOH$ solution. After addition of $16 \, mL$ solution of $NaOH$. What is the $pH$ of the solution $\left( p K_a=4.47\right)$

• A. $5.05$
• B. $4.15$
• C. $4.75$
• D. $5.35$

Answer 1

Answer: (D)

$mEq$ of $CH _3 COOH =20 \times \frac{1}{10}=2$
$mEq$ of $NaOH =16 \times \frac{1}{10}=1.6$
So $1.6\, m\,Eq$ of $NaOH$ reacts with $1.6 \,m\,Eq$ of $CH _3 COOH$ to form $1.6\, m\, Eq$ of salt $CH _3 COONa$ and $(2-1.6=0.4 \, mEq )$ of $W_A\left( CH _3 COOH \right)$ is left and an acidic buffer is formed. So to calculate $pH$, buffer equation is used
$ pH = p K_a+\log \left(\frac{\text { Salt }}{\text { Acid }}\right)=4.74+\log \left(\frac{1.6}{0.4}\right) $
$=4.74+2 \log 2 $
$=4.74+2 \times 0.3010=5.35$