Ka of CH3OH = 3.2*10^-16
a)
for simplicity lets write weak acid CH3OH as HA
HA -----> H+ + A-
0.02 0 0
0.02-x x x
Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
since ka is small, x will be small and it can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((3.2*10^-16)*0.02) = 2.53*10^-9
pH = -log [H+] = -log (2.53*10^-9) = 8.6
Answer: 8.60
you can go with option f
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