Answer to Question #286700 in General Chemistry for June

Question

A 50.0 mL solution of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH.

What is the pH after 20.0 mL of base has been added? Ka CH3COOH = 1.8 x 10^-5

Answer 1

$HA+H_2O\rightleftharpoons\>A^-+H_3O^+$

Initial moles of HA (CH$_3$ COOH)

$=0.150×\frac{50}{1000}=0.0075\>moles$

Moles of NAOH added

$=0.150×\frac{20}{1000}=0.003\>moles$

Moles of HA remaining $=0.0075-0.003$

$=0.0045\>moles$

Ka$=\frac{[H^+_{aq}][A^-_{aq}]}{[HA_{aq}]}$

$1.8×10^{-5}=\frac{[H^+_{aq}]×0.003}{0.0045}$

$[H^+_{aq}]=2.7×10^{-5}$

PH$=-log\>2.7×10^{-5}$

$=4.5686$