Buffer solution pH calculation

Question

What is the pH of the buffer formed by mixing 50 mL of 0.2 M NaH2PO4 with 50 mL of 0.12 M HCl. ( Ka1(H3PO4)= 7.11x10-3 Ka2(H3PO4)= 6.32x10-8 Ka3(H3PO4)= 4.5x10-13 )

Answer 1

You start with NaH₂PO₄. A buffer is a weak acid or weak base plus the salt of that weak acid or weak base.

NaH₂PO₄ can act as an acid or a base (amphoteric), but in the presence of HCl (a strong acid), the H₂PO₄^- acts as a base. The reaction would be H⁺ + H₂PO₄^- ==> H₃PO₄

So here you have made the weak acid, and then the salt of this weak acid is the NaH₂PO₄ or simply H₂PO₄^-.

This makes the buffer.

moles H₂PO₄^- = 50 ml x 1 L/1000 ml x 0.2 mol/L = 0.01 mols H₂PO₄^- initial amount

moles HCl added = 50 ml x 1 L/1000 ml x 0.12 mol/L = 0.006 mols HCl added

moles H₂PO₄^- used up = 0.006 mols

moles H₂PO₄^- remaining = 0.01 - 0.006 = 0.004 mols final

moles H₂PO₄^- formed = 0.006 mols final

So, final concentrations of buffer ingredients in the 100 mls (0.1 L) are as follows:

[H₃PO₄] = 0.006 mols / 0.1 L = 0.06 M

[H₂PO₄^-] = 0.004 mols / 0.1 L = 0.04 M

pH = pKa + log [salt]/[acid]

pKa = -log Ka and since the acid is H₃PO₄, the pKa = -log 7.11x10^-3. pKa = 2.15

pH = 2.15 + log (0.04/0.06) = 2.15 + log 0.667

pH = 2.15 - 0.18

pH = 1.97