Question

What volume (to the nearest 0.1 mL) of 4.70-M HCl must be added to 0.450 L of 0.200-M K2HPO4 to prepare a pH = 7.00 buffer?

Answer 1

11.7ml

Explanation

Henderson - Hasselbalch equation is

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

pKa of H2PO4- = 7.20

pH = 7.00

Therefore,

7.00 = 7.20 + log([HPO4^2-]/[H2PO4^-])

log([HPO4^2-]/[H2PO4^-] = - 0.20

[ HPO4^2-]/[H2PO4^-] = 0.6309

[ HPO4^2- ] = [H2PO4^- ] 0.6309

Therefore,

No of mole of HPO4^2- = No of mole of H2PO4^- × 0.6309

Total no of mole = 0.0900

mole of H2PO4^- ×( 0.6309 × mole of H2PO4-) = 0.0.0900

1.6309 × mole of H2PO4^- = 0.0900

mole of H2PO4^- = 0.0552

mole of HPO4^- = 0.0900- 0.0552 = 0.0348

HCl + HPO4^- ---------> H2PO4^- + Cl-

Therefore,

to produce 0.0552 mole of H2PO4^- 0.0552mole of HCl required

Volume of HCl to be added =( 1000ml/4.70mol)× 0.0552mol = 11.7ml