If 25 mL of 020M NaOH is added to 20 mL of 025M boric acid w

If 25 mL of 0.20M NaOH is added to 20 mL of 0.25M boric acid, what is the pH of the resultant solution?

Solution

Ka of boric acid = 5.4×10^–10

pKa = -log Ka = 9.27

mmoles of NaOH = 25 x 0.20 = 5

mmoles of boric acid = 20 x 0.25 = 5 .

here mmoles of NaOH = mmoles of base.

this is equivalence point. here salt only remains.

salt concentration = 5 / (25 + 20) = 0.111 M

pH = 7 + 1/2 (pKa + log C)

     = 7 + 1/2 (9.27 + log 0.111)

pH = 11.16