let HA represent monoprotic acid. HA +H2O -------->H3O+ + A-
pH= -log[H3O+] , 2.38= -log [H3O+] , [H3O+] = 10(-2.38) = 0.0042 = [A-]
[HA] at equilibrium = 0.0178-0.0042 = 0.0136
Ka= [H3O+] [A-]/ [HA] =0.0042*0.0042/ 0.0136 = 0.001297
pKa= 2.88
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