Campuran satu liter larutan CH3COOH 2M dengan larutan garam CH3COOK 1M mempunyai ph=5-log 2.Jika ka= CH3COOH =10^-5,maka tentukan perbandingan volume CH3COOH

Campuran tsb menghasilkan larutan penyangga asam dg
pH = 5-log 2
<=> [H+] = 2 x 10^-5 M
[H+] = Ka (mmol CH3COOH/mmol CH3COO-)
2 x 10^-5 = 10^-5 (2Va/1Vg)
2Vg = 2Va ; maka Vg = Va
Vol CH3COOH : Vol CH3COOK = 1 : 1

volume larutan 1 L

asumsikan
VCH3COOH = v1
VCH3COOK = 1 - v1

mol CH3COOH = v1 x 2M = 2v1 mol
mol CH3COOK = (1 -v1 ) x 1 M = 1 - v1 mol

pH = 5 - log2
[H+] = 2×10^-5M

[H+] = Ka x mol asam/mol garam

2×10^-5 = 10^-5 x ( 2v1 / 1 - v1 )
2 = 2v1 / 1 - v1
2 - 2v1 = 2v1
2 = 4v1
v1 = 0,5L

volume CH3COOH = v1 = 0,5L
volume CH3COOK = 1 - v1 = 1-0,5 = 0,5L

VCH3COOH : VCH3COOK
0,5 : 0,5
1 : 1