Question

44.70 ml of 0.100 M NaOH are required to completely neutralize 50.00 ml of a weak monoprotic acid, HA. When 34.43 ml of NaOH are added to another 50.00 ml of the acid, the pH reading was 4.80. Calculate Ka for the acid using the partial neutralization method.

Answer 1

anyone else get the same answer?

0.0447 L total neut'n with NaOH
0.03443 L partial neut'n
0.1 M NaOH
0.00447 moles T Total
0.04720169 T (M)
0.003443 moles P Partial
0.040779344 P (M)
4.8 pH
1.58489E-05 [H] = [A-]
3.91117E-08 Ka

Answer 2

Please ignore my previous answer in which I assumed that [A-] = [H+]. I think [A-] should be calculated using the partial titration data, and [H+] is known from the pH. My latest attempt is below:
0.0447 L total neut'n with NaOH
0.03443 L partial neut'n
0.1 M NaOH
0.00447 moles T Total
0.04720169 T (M)
0.003443 moles P Partial
0.040779344 P (M)
4.8 pH
1.58489E-05 [H]

0.040779344 [A-] = P
0.000100634 Ka