What is the pH of a 0.179 M monoprotic acid whose
Ka is 5.887 × 10−3?
pH =
Dissociation equilibrium of monoprotic acid(HA) is
HA(aq) + H2O(l) -------> A-(aq) + H3O+(aq)
Ka = [A-][H3O+] / [HA] = 5.887 ×10-3
Initial concentration
[HA] = 0.179
[A-] = 0
[H3O+] = 0
change in concentration
[HA] = - x
[A-] = + x
[HA] = + x
equilibrium concentration
[HA] = 0.179 - x
[A-] = x
[HA] = x
so,
x2/ ( 0.179 - x) = 5.887 ×10-3
solving for x
x = 0.02965
[H3O+] = 0.02965M
pH = -log[H3O+]
pH = - log( 0.02965M)
pH = 1.53
What is the pH of a 0.179 M monoprotic acid whose Ka is 5.887 × 10−3?...
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