Question

To have a buffer with a pH of 2.29, what volume of 0.0650 M NaOH must be added to 100 mL of 0.264 M H,PO? K. (H3PO4) = 7.5 x 10-3 K(H3PO4) = 6.2 x 10-8 K(H2PO4) = 3.6 x 10-13 Volume = ml

Answer 1

Required pH = 2.29 which near to pKa1(2.12) , so required acid/conjucate base species are H3PO4/H2PO4^-

Applying Henderson - Hasselbalch equation

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

2.29 = 2.12 + log ( [H2PO4^-]/[H3PO4])

log([H2PO4^-]/[H3PO4]) = 0.17

[H2PO4^-]/ [H3PO4] = 1.479

moles of H2PO4^-/moles of H3PO4 = 1.479

moles of H2PO4^- = 1.479 × moles of H3PO4

moles of buffer = (0.264mol/1000ml)×100ml = 0.0264mol

moles of H2PO4^- + moles of H3PO4 = 0.0264mol

1.479×moles of H3PO4 + moles of H3PO4 = 0.0264mol

2.479 × moles of H3PO4 = 0.0264mol

moles of H3PO4 = 0.01065mol

moles of H2PO4^- = 0.0264mol - 0.01065mol = 0.01575mol

H3PO4 + OH^- -------> H2PO4^- + H2O

to get 0.01575moles of H2PO4^- 0.01575 moles of NaOH must be added

Volume 0.0650M NaOH containing 0.01575moles of NaOH = (1000ml/0.0650mol)× 0.01575mol = 242.3ml

Therefore,

242.3ml of 0.0650M NaOH must be added