Question

A 50.0 mL solution of 0.150 M acetic acid (CH3
COOH) is titrated with 0.150 M NaOH.
What is the pH after 20.0 mL of base has been added? Ka
CH3
COOH = 1.8 x 10-5

Answer 1

$\begin{aligned} CH_3COOH + NaOH \to CH_3COONa + H_2O \end{aligned}$

$\begin{aligned} \textsf{No. of moles of }CH_3COOH &= Conc. × volume\\ &= 0.150M×\frac{50}{1000}\\ &= 0.0075 moles \end{aligned}$

$\begin{aligned} \textsf{No. of moles of }NaOH &= Conc. × volume\\ &= 0.150M×\frac{20}{1000}\\ &= 0.003moles \end{aligned}$

After Neutralization, moles of excess acid = 0.0075 - 0.003 = 0.0045moles

$pH = pKa + log\dfrac{[A^-]}{[HA]}$

$\begin{aligned} = -\log{1.8×10^{-5}} + \log\dfrac{0.003}{0.0045} \end{aligned}$

$= 4.745 + 0.176$

$= 4.921$