Question

Weak Acid tritrated with a Base and BUFFERS

In a titration of a 100.0mL 1.00M HA weak acid solution with 1.00M NaOH, what is the pH of the solution after the addition of 126 mL of NaOH?   K_a = 1.80 x 10^-5 for HA. (3 significant figures)

Answer 1

100.0mL 1.00M HA weak acid; 1.00M NaOH volume = 126 mL

Due to large volume of base all the acid has been neutralised. With this large excess of a strong base , the salt does not play any role in the pH of the final solution.

Here we added 126 mL of NaOH = 0.126 L * 1.00 M= 0.126 mol NaOH
This has reacted with 100mL =0.100 L* 1.0 M = 0.100 mol acid.

Moles of NaOH = 0.126-0.100 = 0.026 mol NaOH

Thus we have 0.026 mol NaOH remaining unreacted in a final volume of 100+126= 226 mL
Molarity of NaOH solution: 0.026/0.226 = 0.115 M NaOH

In 0.115M NaOH solution , [OH-] = 0.115 M

pOH = -log [OH-] = 0.94

pH+ pOH = 14

pH = 14-pOH

=14-0.94

= 13.06 0