Chemistry Question

Question

What volume, in milliliters, of 0.220 M NaOH should be added to a 0.140 L solution of 0.018 M glycine hydrochloride (PKa1 = 2.350, pKa2 = 9.778) to adjust the pH to 2.51?

Answer 1

This is glycine hydrochloride

H₃N⁺CH2COOH Cl^-

It has 2 acidic hydrogens so we can write it as H₂A (ignoring the Cl^- spectator). Since the desired pH (2.51) is close to pKa1, that is the acidic hydrogen we will focus on. This happens to be the carboxyl hydrogen (COOH).

H₂A + OH^- ==> HA^- + H2O

Using the Henderson Hasselbalch eq

pH = pKa + log [HA-] / [H2A]

2.51 = 2.35 + log [HA-] / [H2A]

0.16 = log [HA-] / [H2A]

[HA-] / [H2A] = 1.45

x / 0.018 - x =1.45

x = 0.0261 - 1.45x

x = .0107 M = [HA-]

0.0107 mol/L x 0.140L = 0.001498 mols HA-

This is also moles of NaOH needed

0.001498 mol x 1L/0.220 mol. = 0.00681 L = 6.81 ml NaOH

Be sure to check all the math.