Question

Kedalam 10 mL larutan CH3COOH 0,1 M ditambahkan 6 mL larutan NaOH 0,1 M. bila diketahui Ka = 2 X 10^-5 dan log 3= 0,5 ; maka pH larutan yang terjadi.......
mohon uraian nya yaa....

Answer 1

Mol CH3COOH = M x V = 10 x 0,1 = 1 mol
Mol NaOH = M x V = 6 x 0.1 = 0,6 mol

CH3COOH + NaOH -> CH3COONa + H2O
Mula2 = 1 mol 0.6 mol - -
Reaksi = 0,6 mol 0,6 mol 0,6 mol 0,6 mol
_________________________________________ -
Sisa = 0.4. _. 0,6 mol. 0,6 mol

[H+] = ka. Mol asam / mol basa
= 2 x 10-5 . 0,4 / 0, 6
= 1,33 x 10^-5

pH = 5 - log 1,33
= 5 - 0,12
= 4,88

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