Question

$20mL$ of $0.1N$ acetic acid is mixed with $10mL$ of $0.1N$ solution of $NaOH$. The $pH$ of the resulting solution is :

[Given : $p{K}_a$ of acetic acid is $4.74$]

• A. $3.74$
• B. $4.74$
• C. $5.74$
• D. $6.74$

Answer 1

Answer: (B)

$4.74$

$C{H}_{3}COOH+NaOH⟶C{H}_{3}COONa+H_{2}O$

Initial moles of acetic acid $=0.02\times 0.1=0.002moles$

Initial moles of NaOH = $0.01\times 0.1=0.001moles$

Moles after reaction :

Moles of acetic acid $=0.002-0.001=0.001moles$

Moles of $C{H}_{3}COONa=0.001moles$

From Henderson's equation,
$pH=p{K}_a+\log \frac{\left[C{H}_{3}COONa\right]}{\left[C{H}_{3}COOH\right]}$

$=4.74+\log \frac{1}{1}=4.74$