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mol NaOH = 20 x 0,1 = 2 mmol
Persamaan reaksi :
HCl + NaOH → NaCl + H₂O
a 5 2
r 2 2
__________________________
s 3 – 2 2
tersisa 3 mmol HCl (asam kuat)
mol HCl = 3 mmol
M = mol/Volume total
= 3/(50 + 20)
= 3/70
= 0,0428 M
[H⁺] = M x Valensi asam
= 0,0428 x 1
= 4,28 x 10⁻²
pH = -log [H⁺]
= -log [4,28 x 10⁻²]
= 2 - log 4,28
= 1,368
Jadi, pH campuran adalah 1,368