sejumlah NaOH 1M diperlukan untuk membuat 200 mL larutan buffer dgn larutan CH3COOH 1M ka=10^-5 .agar diperoleh pH=5-log 2,maka volume NaOH tersebut adalh

pH = 5 - log2
[H+] = 2×10^-5

[H+] = ka× mol sisa asam/ mol basa
mol sisa asam / mol basa = [H+]/ka
= 2×10^-5 ÷ (10^-5) = 2
( mol sisa asam = mol asam - mol basa)
( na = mol asam , nb = mol basa)

( na - nb ) / nb = 2
na - nb = 2nb
na = 3nb
MaVa = 3 MbVb
Va/Vb = 3 Mb/Ma
Va/Vb = 3 (1)/(1)
Va/Vb = 3 ( Va + Vb = 200ml )
( Va = 200-Vb )

(200-Vb) = 3Vb
200 = 4Vb
Vb = 50ml

NaOH 1 M + 200 mL CH3COOH 1 M
Ka = 10^-5
pH = 5 - log 2

V NaOH ... ?

*pH = 5 - log 2
[H^+] = 2 × 10^-5

*n NaOH = M × V
n NaOH = 1 × V
n NaOH = ( V ) mmol

*n CH3COOH = M × V
n CH3COOH = 1 × 200
n CH3COOH = 200 mmol

#Reaksi
... NaOH + CH3COOH => CH3COONa + H2O
M : . V ............. 200
R : .. V ............... V
________________---
S : .. --- ........ (200 - V) .............. V ................ V

[H^+] = Ka × Asam lemah/B. konjugasi
2 × 10^-5 = 10^-5 × (200 - V) / V
............ 2 = (200 - V ) / V
........ 2 V = 200 - V
. 2 V + V = 200
........ 3 V = 200
........... V = 200 / 3
........... V = 66,67 mL

Jadi, Volume NaOH tersebut adalah 66,67 mL.