Untuk titrasi 50 ml larutan CH3COOH 0,1 M dg larutan NaOH 0,1 M, perkiraan pH pada titik ekivalen....

(=)

na x Ma x Va = nb x Mb x Vb
1 x 0,1 x 50 = 1 x 0.1 x Vb
Vb = 50 mL

(=)

mol CH3COOH = 0,1 x 50 = 5 mmol

mol NaOH = 0,1 x 50 = 5 mmol

(=) CH3COOH + NaOH -> CH3COONa + H2O
m_5 mmol____5 mmol ______-________-_
r__5 mmol____5 mmol ___5 mmol___5 mmol
s____-_________-_______5 mmol __5 mmol


M campuran = nCH3COONa/Vtotal = 5/100 = 0.05 M

pH = - log [H+] = - log 0.05 = 2-log 5