Question

$20\, mL$ of $0.2 \,M$ sodium hydroxide solution is added to $40\, mL$ of $0.2\, M$ acetic acid solution. What is the $pH$ of the solution?
$\left( p K_{a}\right.$ of $\left.CH _{3} COOH =4.8\right)$

• A. 9.2
• B. 4.8
• C. 8.4
• D. 2.9

Answer 1

Answer: (B)

Given,

$p K_{a}$ of $CH _{3} COOH =4.8$

Number of millimoles of $NaOH$ in $20\, mL$ of $0.2\, M$

solution $=20 \times 0.2=4 $ mmol

Number of millimoles of $CH _{3} COOH$ in $40\, mL$ of $0.2\, M$ solution $=40 \times 0.2=8$ mmol

$20$ mmol of $0.2 \,M ,NaOH$ will neutralise $20$ mmol

of $0.2\, M$ acetic acid and produce $4$ mmol of sodium acetate as follows:

$CH _{3} COOH + NaOH \longrightarrow CH _{3} COONa + H _{2} O$

That is equivalent to $4$ mmol of sodium acetate.

Now, $ pH = p K_{a}+\log \left[\frac{ Salt }{ Acid }\right]$

$pH = p K_{a}+\log \left[\frac{\left[ CH _{3} COONa \right]}{ CH _{3} COOH }\right]$

$pH =4.8+\log \frac{[4]}{[4]}$

$pH =4.8$