What is the pH of a 0.402 M aqueous solution of NaCH3COO? Ka (CH3COOH) = 1.8x10-5
CH3COONa ----------> CH3COO- + Na+
0.402M 0.402M
CH3COO- + H2O --------> CH3COOH + OH-
I 0.402M 0 0
C -x +x +x
E 0.402-x +x +x
kb = Kw/Ka
= 1*10-14/1.8*10-5 = 5.6*10-10
Kb = [CH3COOH][OH-]/[CH3COOH]
5.6*10-10 = x*x/0.402-x
5.6*10-10 *(0.402-x) = x2
x = 1.5*10-5
[OH-] = x=1.5*10-5 M
POH = -log[OH-]
= -log1.5*10-5
= 4.8239
PH = 14-POH
= 14-4.8239 = 9.1761
What is the pH of a 0.402 M aqueous solution of NaCH3COO? Ka (CH3COOH) = 1.8x10-5
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