sebanyak 100 ml larutah CH3COOH 0,2 M dicampur dengan 100 ml larutan NaOH 0,1 M. Tentukan ph larutan sebelum dan sesudah reaksi (ka CH3COOH=1×10 pangkat -5)

Mol ch3cooh: m×v
: 0.2×100
:20 mmol
mol naOh:m×v
: 100×0.1
:10
Ch3cooh+naoh=ch3coona+h20
M: 20 10
R: 10 10 10
Stb: 10 0 10

[OH-]= ka× mol[A]
mol[G]
= 1×10^-5 × 10
10
= 1×10^-5
Ph = - log [H+]
= - log [ 1×10^-5]
= 5 - log 1
= 5
Jadi, ph= 5

pH sebelum reaksi

CH3COOH
[H+] = √Ka x √Ma = √10^-5 x √2x10^-1
= 10^-3 √2
pH = -log 10^-3 √2 = 3 -log √2 = 3 -log ½

NaOH
[H+] = a x Ma = 1 x 10^-1 = 10^-1
pH = -log 10^-1 = 1

pH sesudah reaksi

nCH3COOH = 100 x 0,2 = 20 mmol
nNaOH = 100 x 0,1 = 10 mmol

CH3COOH + NaOH --> CH3COONa + H2O
m....20..................10
r.......-10...............-10.…..............10
s......10 mmol.......-....................10 mmol

CH3COONa --> CH3COO- + Na+
10 mmol..................10mmol

[H+] = Ka . mmol as / mmol bs kjgs.
[H+] = 10^-5 . 10 / 10
[H+] = 10^-5

pH = -log 10^-5 = 5