Question

Untuk mengubah 110 mL larutan CH3COOH 0,1 M yang
pHnya 3 agar menjadi 6 diperlukan larutan NaOH 0,1 sebanyak.....

Tolong jawab beserta caranya, mohon
bantuannya ya terima kasih banyak

Answer 1

CH3COOH merupakan asam lemah 
pH = -log[H+] 
[H+] = 10^-pH = 10^-3 = 0,001 M 
Tetapan ionisasi asam 
[H+] = vKa.M 
Ka = [H+]^2 /M = (10^-3)^2 / 0,1 = 10^-5 
Supaya pH menjadi 6, maka: 
pH = 6 → [H+] = 10^-6 M 
[H+] = vKa.M 
M = [H+]^2 /Ka = (10^-6)^2 / 10^-5 = 10^-7 M 

Reaksi asam-basa 
CH3COOH + NaOH → CH3COONa + H2O 
Mol CH3COOH bereaksi = mol CH3COOH mula – mol CH3COOH akhir 
= (110/1000)x0,1 – (110/1000)x10^-7 
= 0,0109 mol 
Mol NaOH dibutuhkan = mol CH3COOH bereaksi = 0,0109 mol 
VNaOH = mol/M = 0,0109/0,1 = 0,109 liter = 109 ml