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mol CH3COOH = 20 x 0.001 = 0.2 mol
ka CH3COOH = 1.7 x 10^-5
NaOH + CH3COOH ------> NaCH3COO + H2O
M 0.2 0.2
R 0.2 0.2 0.2 0.2
__________________________________________________
S - - 0.2 0.2
karena asam maupun basanya habis maka pakai rumus hidrolisis sebagian atau karena sisa garamnya saja (asam lemah dan basa kuat)
OH- = √((Kw/Ka) xmol garam)
=√( (10^-14 / 1.7 x 10^-5) x 0.2)
= √0.588 x 10^-9 x 0.2
= √0.1176 x 10^-9
= √0.01176 x 10^-8
= 10^-4 x √0.01176
= 0.11 x 10^-4
POH = -log OH-
= -log (0.11 x 10^-4)
= -log 0.11 - log 10^-4
= 4 - log 0.11
= 4 - (-0.95)
= 4 + 0.95
= 4.95
PH = 14 - POH
= 14 - 4.95
= 9.05
jadi PH nya 9.05
semoga benar dan membantu