Question

A solution contains 0.525 M NaCH_3COO and 0.465 M CH_3COOH(aq). If 15.0 mL of 0.2 M NaOH is added to 50.0 mL of the solution, calculate the final pH. (k_a for acetic acid is 1.75 Times 10^-5)

Answer 1

For acetic acid, $Ka = 1.75 \times 10^{-5}$
$pKa = -log Ka = - log 1.75 \times 10^{-5} = 4.76$
Initially, number of millimoles of sodium acetate $= 50.0 \times 0.525 = 26.25$
Number of millimoles of acetic acid $= 50.0 \times 0.465 = 23.25$
15.0 mL of 0.2 M NaOH is added.
Number of millimoles of NaOH $= 15.0 \times 0.2 = 3.0$
3.0 mmol NaOH will neutralize 3.0 mmol acetic acid to form 3.0 mmol sodium acetate.
Total number of mmol of sodium acetate
Total number of mmol of acetic acid

$pH = pKa + log \dfrac {[CH_3COONa]}{[CH_3COOH]} \\ pH = 4.76 + log \dfrac {29.25}{20.25} \\ pH = 4.92$