Untuk mengubah 110 ml larutan CH3COOH 0,1 M yang pHnya 3 agar menjadi 6 diperlukan larutan NaOH 0,1 M sebanyak ... A.10 ml B. 55 ml C. 100 ml D. 210 ml E. 1.100 ml
Amaldoft
Diket dan dit: *n CH3COOH = 110 mL x 0,1 M = 11 mmol *n NaOH = 0,1 M x v mL = 0,1V *pH awal = 3 (asam-basa) *pH akhir = 6 --> [H+] = 10^-6 (buffer) *Volume NaOH ?
CH3COOH + NaOH --> CH3COONa + H2O m 11 0,1V b -0,1V -0,1V +0,1V s 11-0,1V - 0,1V
[H+] = Ka x a/g ---------> [H+] yang dipakai miliki pH akhir 10^-6 = 10^-5 (11-0,1V)0,1V 0,01V = 11 - 0,1V 0,11V = 11 V = 100 mL (C)
*n CH3COOH = 110 mL x 0,1 M = 11 mmol
*n NaOH = 0,1 M x v mL = 0,1V
*pH awal = 3 (asam-basa)
*pH akhir = 6 --> [H+] = 10^-6 (buffer)
*Volume NaOH ?
CH3COOH + NaOH --> CH3COONa + H2O
m 11 0,1V
b -0,1V -0,1V +0,1V
s 11-0,1V - 0,1V
[H+] = Ka x a/g ---------> [H+] yang dipakai miliki pH akhir
10^-6 = 10^-5 (11-0,1V)0,1V
0,01V = 11 - 0,1V
0,11V = 11
V = 100 mL (C)