Question

You add 100 mL of 60 mM NaH₂PO4 and 100 mL of 40 mM Na₂HPO4 to 300 mL water. What is the final pH of the solution?

Answer 1

Answer – We are given the 100 mL of 60 mM NaH₂PO4 and 100 mL of 40 mM Na₂HPO4 , volume of water = 300 mL

First we need to calculate the moles of each

Moles of NaH₂PO₄ = 0.100 L * 0.060 M

                                 = 0.006 moles

Moles of Na₂HPO₄ = 0.100 L * 0.040 M

                                 = 0.004 moles

Total volume = 100 +100 +300 = 500 mL

New molarity

[H₂PO₄^-] = 0.006 moles / 0.500 L = 0.012 M

[HPO₄^2-] = 0.004 moles / 0.500 L = 0.008 M

We know the pKa₂ for the H₃PO₄ = 7.20

So using the Henderson Hasselbalch equation-

pH = pKa + log [HPO₄^2-] / [H₂PO₄^-]

      = 7.20 + log 0.008 /0.0120

      = 7.02

So, the final pH of the solution is 7.02