Question

What is the pH of a solution prepared by mixing 800 mL of 0.075M potassium dihydrogen phosphate with 650 mL of 0.130 M potassium hydrogen phosphate? Then calculate the pH buffer after the additon of 3.00 mL of 4.00 M hydrochloric acid.

Answer 1

The pKa values are 2.15, 6.82 and 12.38.

For the first case, let's calculate first the moles:

moles of KH₂PO₄ = 0.075 * 0.8 = 0.06 moles

moles of K₂HPO₄ = 0.130 * 0.65 = 0.085 moles

use the HH equation:

pH = pKa + log Base/Acid

Now from H2PO4 to HPO4 is the second pKa so:

pH = 6.82 + log (0.085 / 0.06)

pH = 6.97

after the mixing with HCl:

moles of HCl = 4 * 0.003 = 0.012 moles

This acid will raise the concentration of the acid (In this case the H2PO4) and lower the concentration of base. This will cause that the pH get's lower so:

pH = 6.82 + log (0.06-0.012/0.085+0.012)

pH = 6.51