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Volume = 500 cm³ = 500 ml = 0,5 L
Ka = 10⁻³
Mr Ba(CH₃COOH)₂ = 255
Dit : pH larutan = ···· ?
Jwb :
mol Ba(CH₃COOH)₂ = gram = 2,55 = 0,01
Mr 255
Molaritas = mol = 0,01 = 0,02 = 2 . 10⁻²
volume larutan 0,5
(H⁺) =
=
=
= 2⁻¹/² . 10⁻²·⁵
pH = - log 2⁻¹/² . 10⁻²·⁵
= 2,5 - 1/2 log 2
= 2,5 - 1/2 (0,301)
= 2,5 - 0,1505
= 2,3495