Question

How many milliliters of 1.07 M NaOH must be added to 175 mL of 0.13 M NaH₂PO₄ to make a buffer solution with a pH of 7.20?

Answer 1

Second dissociation H3PO4 is

H2PO4^- <-------> HPO4^2-  + H⁺

^{pKa2 =7.20}

pH = pKa2 + log([HPO4^2-]/[H2PO4^-])

log([HPO4^2-]/[H2PO4^-]) = 7.20 - 7.20 = 0

[HPO4^2-] /[H2PO4^-] = 1

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

Total buffer concentration = 0.13M

[H2PO4^- ] =0.065M

[HPO4^2-] = 0.065M

For 175ml

No of moles of H2PO4^- reaquired = (0.065mol/1000ml)×175ml = 0.011375mol

No of moles of HPO4^2- required = 0.011375mol

No of moles of NaOH required = 0.011375mol

Volume of 1.07M NaOH should be added = (1000ml/1.07mol)×0.011375mol = 10.63ml