NaF will dissociate 100%, and therefore the F- will have extra 0.25 concentration... HF --- > H+ + F- [HF] = 0.3-0.3* [H+] = 0.3* [F-] = 0.3* + 0.25 Ka = (0.3*)*(0.3* + 0.25)/(0.3-0.3*) 7.2*10^-4 = *(0.3* + 0.25)/(1-) = 0.0028 [H+] = 0.3*0.0028 = 0.00084 M pH = -log(0.00084) = 3.07 Solution NaF will dissociate 100%, and therefore the F- will have extra 0.25 concentration... HF --- > H+ + F- [HF] = 0.3-0.3* [H+] = 0.3* [F-] = 0.3* + 0.25 Ka = (0.3*)*(0.3* + 0.25)/(0.3-0.3*) 7.2*10^-4 = *(0.3* + 0.25)/(1-) = 0.0028 [H+] = 0.3*0.0028 = 0.00084 M pH = -log(0.00084) = 3.07.