Question

What volumes of 0.5 M Sodium phosphate monobasic (NaH2PO4), 0.5 M Sodium phosphate dibasic ( Na2HPO4) and water would you have to mix to make 500mL of a 0.250M Phosphate buffer at pH 7.6?

Answer 1

Answer

Volume of 0.5M NaH2PO4 required = 72.36ml

Volume of 0.5M Na2HPO4 required = 177.64ml

Volume of water required = 250ml

Explanation

Henderson - Hasselbalch equation is

pH = pKa + log([A^-]/[HA])

A^- = HPO4^2-

HA = H2PO4^-

pKa of H2PO4^- = 7.21

7.60 = 7.21 + log([HPO4^2-]/[H2PO4^-])

log([HPO4^2-]/[H2PO4^-]) = 0.39

[HPO4^2-]/[H2PO4^-】= 2.455

[HPO4^2-] = 2.455 × [H2PO4^-]

moles of HPO4^2- = 2.455 × moles of H2PO4^-

moles of buffer = (0.250mol/1000ml) × 500ml = 0.125mol

moles of HPO4^2- + moles of H2PO4^- = 0.125mol

2.455 ×moles of H2PO4^- + moles of H2PO4^- = 0.125mol

3.455×moles of H2PO4^- = 0.125mol

moles of H2PO4^- = 0.03618mol

moles of HPO4^2- = 0.125mol - 0.03618mol = 0.08882mol

Volume of 0.5M NaH2PO4 required = (1000ml/0.5mol) × 0.03618mol = 72.36ml

Volume of 0.5M Na2HPO4 required = (1000ml/0.5mol) × 0.08882mol = 177.64ml

Volume of water required= 500ml - ( 72.36ml + 177.64ml) = 250ml