Question

$20\, mL$ of $0.1\, M$ acetic acid is taken in a beaker. To this, $10\, mL$ of $0.1\, M\, NaOH$ solution is mixed. The $pH$ of the resulting solution is______.
$\left[ pK _{ a }\right.$ of acetic acid $\left.=4.74\right]$

Answer 1

Moles of acetic acid $=0.1 \times 20 / 1000$
$=0.002\, mol$
Moles of $NaOH =0.1 \times 10 / 1000=0.001\, mol$

$pH = pK _{ a }+\log _{10} \frac{[\text { salt }]}{[\text { acid }]}$ (Henderson's equation)
$\therefore pH =4.74+\log _{10} \frac{(0.001)}{(0.001)}$
$=4.74+\log _{10} 1=4.74+0=4.74$