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7.) 0,41 gram NaCH3COO ditambahkan dalam 100 mL CH3COOH 0,4 M.
Ka = 10^-5
Ar Na = 23; O = 16; H = 1; C = 12

perubahan pH ... ?

*Mr NaCH3COO
Mr = Ar Na + (2.Ar C) + (3.Ar H) + (2.Ar O)
Mr = 23 + (2 . 12) + (3 . 1) + (2 . 16)
Mr = 23 + 24 + 3 + 32
Mr = 82

¤ pH larutan CH3COOH sebelum ditambah NaCH3COO
[H^+] = √( Ka × Ma )
[H^+] = √( 10^-5 × 0,4 )
[H^+] = √( 10^-6 × 4 )
[H^+] = 2 × 10^-3

pH = - log [H^+]
pH = - log 2 × 10^-3
pH = 3 - log 2

¤ pH larutan CH3COOH setelah ditambah NaCH3COO

*mol CH3COOH = M × V
mol CH3COOH = 0,4 × 100/1000
mol CH3COOH = 0,4 × 0,1
mol CH3COOH = 0,04 mol

*mol NaCH3COO = gram/Mr
mol NaCH3COO = 0,41/82
mol NaCH3COO = 0,005 mol

*[H^+] = Ka × mol asam lemah/mol basa konjugasi
[H^+] = 10^-5 × 0,04/0,005
[H^+] = 10^-5 × 8

pH = - log [H^+]
pH = - log 10^-5 × 8
pH = 5 - log 8

¤ pH larutan berubah dari 3 - log 2 menjadi 5 - log 8.