tolong bantu dijawab beserta caranya kk

CH3COOH 0,1 M + CH3COONa 0,1 M
pH = 4
Ka CH3COOH = 1,8 × 10^-5

V CH3COOH : V CH3COONa ... ?

*pH = 4
- log [H^+] = 4
[H^+] = 10^-4

*n CH3COOH = M × V
n CH3COOH = 0,1 × x
n CH3COOH = ( 0,1 x ) mmol

*n CH3COONa = M × V
n CH3COONa = 0,1 × y
n CH3COONa = ( 0,1 y ) mmol

*[H^+] = Ka × n Asam lemah/n B. konjugasi
10^-4 = 1,8 × 10^-5 × ( 0,1 x ) / ( 0,1 y )
...... 1 = 1,8 × 10^-1 × ( 0,1 x ) / ( 0,1 y )
0,1 y = 0,18 × ( 0,1 x )
......... 0,1 y = 0,018 x
0,1 / 0,018 = x / y
... 100 / 18 = x / y
...... 50 / 9 = x / y
......... x / y = 50 / 9

Jadi, perbandingan V CH3COOH : V CH3COONa = 50 : 9.