Question

$50\, mL$ of $0.02\, M \,NaOH$ solution is mixed $50\, mL$ of $0.06 \,M$ acetic acid solution, the $pH$ of resulting solution is $\left( p K_{a}\right.$ of acetic acid is $4.76, \log 5=0.70$ )

• A. 5.06
• B. 4.06
• C. 5.46
• D. 4.46

Answer 1

Answer: (D)

4.46

Answer 2

The correct answer is option (D) : 4.46
Given, Volume of $NaOH$ solution $=50 \,mL$

Molarity of $NaOH$ solution $=0.02 \,M$

Me of $NaOH =50 \times 0.02=1 \,meq$

Volume of $CH _{3} COOH$ solution $=50 \,mL$

Molarity of $CH _{3} COOH =0.06\, M$

Me of $CH _{3} COOH =50 \times 0.06=3\, meq$
$1$ me of $NaOH$ combines with $3$ me of

$CH _{3} COOH$ and forms $1$ me of $CH _{3} COONa$.

$\therefore 2$ me of $CH _{3} COOH$ is left.

So, it forms an acidic buffer.

Now, for an acidic buffer,

$pH = p K_{a}+\log \left[\frac{\text { Salt }}{\text { Acid }}\right]$
$pH = p K_{a}+\log \frac{\left[ CH _{3} COONa \right]}{\left[ CH _{3} COOH \right]}$
$\therefore pH =4.76+\log \left[\frac{ l }{2}\right]$
$pH =4.76+\log \left(5 \times 10^{-1}\right)$
$pH =4.76+\log 5-\log 10$
$=4.76+0.70- 1$
$=4.46$