Question

As a technician in a large pharmaceutical research firm, you need to produce 350. mL of 1.00 mol L−1 potassium phosphate buffer solution of pH = 6.91. The pKa of H2PO4− is 7.21.

You have the following supplies: 2.00 L of 1.00 mol L−1 KH2PO4 stock solution, 1.50 L of 1.00 mol L−1 K2HPO4 stock solution, and a carboy of pure distilled H2O.

How much 1.00 mol L−1 KH2PO4 will you need to make this solution?

Answer 1

Answer

Volume of 1.00 molL^-1 KH2PO4 required = 233.15ml

Explanation

Henderson- Hasselbalch equation is

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

6.91 = 7.21 + log([HPO4^2-]/[H2PO4^-])

log([HPO4^2-]/[H2PO4^-]) = - 0.3

[HPO4^2-] /[H2PO4^-] = 0.5012

[HPO4^2-] = [H2PO4^-] × 0.5012

moles of HPO4^2- = moles of H2PO4^- × 0.5012

moles of buffer required = (1.00mol /1000mol)×350ml = 0.350mol

moles of HPO4^2-  + moles of H2PO4^- = 1.00mol

(0.5012mol × H2PO4^- ) + moles of H2PO4^- = 0.350mol

1.5012× moles of H2PO4^- = 0.350mol

moles of H2PO4^- = 0.23315 mol

moles of HPO4^2- = 0.350mol - 0.23315mol = 0.11685mol

Volume of 1.00mol L^-1 KH2PO4 required = (1000ml/1.00mol)× 0.23315mol = 233.15ml

Volume of 1.00molL^-1 KHPO4 solution required = (1000ml/1.00mol)× 0.11685mol = 116.85ml