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= 0,002 mol
mol NaOH = 0,1 M x 0,015 L
= 0,0015 mol
CH3COOH + NaOH -> NaCH3COO + H2O
0,002 0,0015 - -
0,0015 0,0015 0,0015 0,0015
0,0005 - 0,0015 0,0015
(H+) = Ka . (asam lemah)
(garam)
= 10^-5 . 5.10^-4
1,5.10^-3
= 5.10^-9
1,5.10^-3
= 3,3.10^-6
pH= 6 - log 3,3
=6 - 0,5
= 5,5
mol NaOH=15.0,1=1,5 mmol
ka=10^-5
dit: pH=..?
CH3COOH + NaOH-----> CH3COONa+H2O
M: 2 1,5
T: 1,5 1,5 1,5 1,5
-------------------------------------------------------------------- -
S: 0,5 1,5 1,5
[H+]= ka .[CH3COOH]/[CH3COONa]
= 10^-5 * 0,5/1,5
= 3,3.10^-6
pH= -log[H+]
= -log[3,3.10^-6]
= 6-log3,3