sebanyak 50 ML larutanKOH dicampurkan dengan 450ml larutan asam asetat (Ka=2 kali 10 pangkat min5)kedua laruta itu memiliki konsentrasi yang sama.ph larutan yan

anggap saja konsentrasi CH3COOH dan KOH = 1 M
mol CH3COOH= M x V= 1 x 50 = 50 mmol
mol KOH = M x V = 1 x 45 = 45 mmol
CH3COOH + KOH ---> CH3COOK + H2O
mula2. 50. 45. 0. 0
reaksi. 45. 45. 45. 45.
sisa. 5. 0. 45. 45
karena tersisa mol asam lemah dan garam, maka pakai rumus buffer:
[H+] = ka. mol asam/mol garam
= 2x10^-5 x (5mmol/45mmol)
= 2,2 x 10^-6
pH = -log [H+]
= -log 2,2 x 10^-6
= 6-log 2,2 = 5,66

Smg bermanfaat.

50 mL KOH + 450 mL CH3COOH
Ka = 2 × 10^-5
Konsentrasi KOH = Konsentrasi CH3COOH = M

pH ... ?

n KOH = M × V
n KOH = M × 50 mL
n KOH = ( 50 M ) mmol

n CH3COOH = M × V
n CH3COOH = M × 450 mL
n CH3COOH = ( 450 M ) mmol

#Reaksi
... CH3COOH + KOH => CH3COOK + H2O
M : .. 450 M .... 50 M
R : ..... 50 M .... 50 M
________________---
S : .. 400 M ....... --- ........... 50 M ...... 50 M

[H^+] = Ka × Asam Lemah/Basa Konjugasi
[H^+] = 2 × 10^-5 × 400 M/50 M
[H^+] = 2 × 10^-5 × 8
[H^+] = 16 × 10^-5
[H^+] = 1,6 × 10^-4

pH = - log [H^+]
pH = - log 1,6 × 10^-4
pH = 4 - log 1,6