Ph of a an equal volume mixture?

Question

Given the following:
0.200M NH3(aq) with pKb = 4.76
0.200M HC2H3O2(aq) with pKa = 4.76
0.200M NaOH(aq)
0.200M HCl(aq)
A mixture is made using 50.0 mL of the HC2H3O2 and 50.0 mL of the NaOH. What is its pH?

Answer 1

Well, you've now got 100ml of the base CH3COONa, with a concentration of 0.100M and a pKb of 9.24.

So you can now work out its pOH and then its pH.

Answer 2

Use the Henderson-Hasselblach equation that's pH= pKa + log (conc. of base/conc. of acid). pKa = -log(Ka) and could equivalent 4.89 in this question. The conc. of acid and base could be 0.280, which delivers that pH = pKa for this occasion. So the pH is 4.89.