Question

What volume (in milliliters) of 0.240 M NaOH should be added to a 0.140 L solution of 0.0250 M glycine hydrochloride (pKa1 = 2.350, pKa2 = 9.778) to adjust the pH to 2.85?

Answer 1

ph = pka + log (A-/HA)
2.85 = pKa1 + log(A-/HA)
2.5- 2.350 = log(A-/HA)
0.5= log (A-/HA)
3.162 = A- / HA
A- + HA = Total concentration glycine hydrochloride = 0.024M

Solve the system of equation to get

3.162 = A- / HA

A- = 3.162HA

A- + HA=0.025

3.162HA+ HA=0.025

so
A- = 0.019
HA = 0.006

In order to obtain 0.006 moles of HA, you need to deprotonate (0.025 - 0.006) = 0.019 moles
of NaOH are required:

0.019 Moles * 0.240 moles / L = 0.00456 amount of NaOH needed (Liters) (Liters)