Question

Caleulate the PH after the addition of 5and 4o mL f o.1M to 20.oo mL of o.I M H2A kaielxit, Kaeixto). NaoH

Answer 1

Given that Ka1 = 1 x 10^-4 ,

Ka2 = 1 x 10^-8

After 5 mL:

moles H₂A = 0.1 * 0.020

= 0.002 moles

moles OH^- = 0.1 * 0.005

= 0.0005 moles

H₂A + OH^- --------> HA^- + H₂O

moles acid remaining = 0.002 - 0.0005

= 0.0015 moles

pH = pKa + log A^-/HA

pH = -log(1x10^-4) + log(0.0015/0.002)

pH = 3.88

After 40 mL, this is near the equivalence point, because the volume at the equivalence point is:

H₂A + 2NaOH ---------> Na₂A + 2H₂O

moles A/moles B = 1/2 --> 2 moles A = moles B

Vb = 2 * 0.002 / 0.1

= 0.040 L or 40 mL.

At the equivalence point the pH can be calculated as:

pH = pKa1 + pKa2 / 2 pKa2 = -log(1x10^-8) = 8

pH = 4 + 8 / 2

pH = 6