Question

Sebanyak x mL NaOH 0,3 M ditambahkan 2x mLCH3COOH 0,3 M. jika Ka = 10^-5 maka pH larutannya adalah

Answer 1

PH= -log [H]
    = -log [0,3]
     = 0,5

Answer 2

N NaOH = x * 0.3 = 0.3x
n CH3COOH = 2x * 0.3 = 0.6x
n CH3COOH = 1/1 . 0.3x = 0.3x

[H+] = Ka * na/nbk = 10^-5 * 0.3x/0.3x
= 10^-5 * 1
= 10^-5
pH = - log [H+]
= - log (10^-5)
= 5