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n NaOH = 0,1 liter x 0,5 M = 0,05 mol
H₂SO₄ + 2NaOH ---> Na₂SO₄ + 2H₂O
m 0,03 0,05 - -
r -0,025 -0,05 0,025 0,05
s 0,005 - 0,025 0,05
A) Molaritas larutan setelah bereaksi
M [H⁺] Sisa = mol sisa
Volume total
= 0,005
0,25
= 0,02 M
B)pH campuran = ....?
[H⁺] = a x m
= 2 x 0,02
= 0,04
= 4 x 10⁻²
pH = -log [H⁺]
= - log 4 x 10⁻²
= 2 - log 4
= 2 - 0,6
= 1,4