Ans 9
assume a monoprotic weak acid
Dissociation reaction with ICE TABLE
HA(aq)+H2O(l)⇌A−(aq)+H3O+(aq)
I 3
C - x +x +x
E 3-x x x
Equilibrium constant expression of the reaction
Ka=[A−]⋅[H3O+]/[HA]
Now, given that
pKa = 1.01
pKa = -log Ka
Ka = 10^-1.01 = 0.0977
0.0977 = x2 / (3-x)
0.2931 - 0.0977 x - x2 = 0
x = 0.4947
[H3O+] = 0.4947
pH = - log[H3O+] = - log (0.4947)
pH = 0.3
Option e is the correct answer
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