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M= 0,02
[OH-] = MxV = 0,02 x 1 = 0,02
pOH = 2 - log2
pH = 12 + log 2
Volume = 500 ml = 0,5 L
Mr.NaOH = 40
Dit : pH = ......?
Jwb : n.NaOH = masa/Mr
= 0,4/40
= 40.10⁻²/40
= 10⁻²
= 0,01 mol
[NaOH] = n/V
= 0,01 mol/0,5 L
= 0,02 M
NaOH -----> Na⁺ + OH⁻
0,02 M 0,02 M
pOH = - log [OH⁻]
= - log 2.10⁻²
= 2 - log 2
pH = 14 - ( 2 - log 2 )
= 12 + log 2
= 12 + 0,301
= 12,301