Jika 100 ml larutan NaOH 0,1 M dicampurkan dengan 100 ml larutan CH3COOH 0,1 M, maka hitunglah PH larutan yang dihasilkan  ! Dik : Ka CH3COOH = ( 10 pangkat -5

n NaOH = M × V = 0,1 M × 100 mL = 10 mmol
n CH3COOH = M × V = 0,1 M × 100 mL = 10 mmol

NaOH + CH3COOH -> CH3COONa + H2O
10 mmol 10 mmol
-10 mmol -10 mmol + 10mmol +10mmol
0 0 +10 mmol +10mmol

zat pereaksi habis bereaksi dan menghasilkan garam CH3COONa yg bersifat basa. jadi digunakan rumus hidrolisis garam.

M CH3COONa= 10 mmol / (100 + 100) mL = 0,05 M.

[OH-] = akar [(Kw/ka) × M CH3COONa]
= akar [(10pgkt-14/10pgkt-5) × 5 × 10pgkt-2]
= akar (10pgkt-9 × 5 × 10pgkt-2]
= akar (5 × 10pgkt-11)
= 10pgkt-5,5 akar5

pOH = - log [OH-]
= - log 10pgkt-5,5 akar5
= 5,5 - log akar5

pH = 14 - pOH
= 14 - 5,5 - log akar5
= 8,5 - 0,35
= 8,85.

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