Question

You have available 1.00 L volumes of the three following solutions: 1.00 M HCl, 1.00 M NaOH and 1.00 M NaH2PO4. You wish to prepare 500.0 mL of a buffer solution that maintains a pH of 2.50. Indicate which two of these three solutions you would use to produce this buffer, and calculate the volume of each of these two solution you would need to mix to get the desired volume and pH value for the buffer.

Answer 1

We would use HCl and NaH₂PO₄

Ka1 of H₃PO₄ = 6.9 x 10^-3

pKa1 = 2.16

pH = pKa1 + log([H₂PO₄^-] / [H₃PO₄])

2.5 = 2.16 + log([H₂PO₄^-] / [H₃PO₄])

log([H₂PO₄^-] / [H₃PO₄]) = 0.34

[H₂PO₄^-] = 2.18 [H₃PO₄] … equation (1)

Let B denote base H₂PO₄^- and A denote acid H₃PO₄

H₂PO₄^- + HCl -> H₃PO₄ + Cl-

(B – A) + 0 -> A + Cl-

[H₃PO₄] = Moles of HCl added / Total volume

[H₂PO₄^-] = (Moles of NaH₂PO₄ added – Moles of HCl added) / Total Volume

(Moles of NaH₂PO₄ added – Moles of HCl added) = 2.18 * Moles of HCl added (using equation 1)

Moles of NaH₂PO₄ added = 3.18 * Moles of HCl added

Total volume = 500 mL = 0.5 L

Volume of NaH₂PO₄ added + Volume of Moles of HCl added = 0.5 L

Volume of NaH₂PO₄ added * 1 M = 3.18 * Volume of Moles of HCl added * 1 M

Solving we get,

Volume of NaH₂PO₄ added = 0.38 L

Volume of Moles of HCl added = 0.12 L