Question

What is the pH of a H3PO4 buffer solution when 10 mL of H3PO4 is mixed 1 mL of NaOH, assuming that their concentration (H3PO4 and NaOH) are similar? The Ka of H3PO4 is 7.5x10-3

Answer 1

Let concentration of H3PO4 and NaOH be 1 M each

mol of H3PO4 = M * V

= 1 M * 10 mL

= 10 mmol

mol of NaOH added = M*V

= 1 M * 1 mL

= 1 mmol

1 mmol of both will react to form 1 mmol of H2PO4-

and 9 mmol of H3PO4 remains.

Ka = 7.5*10^-3

pKa = - log (Ka)

= - log(7.5*10^-3)

= 2.125

use:

pH = pKa + log {[conjugate base]/[acid]}

pH = pKa + log {[H2PO4-]/[H3PO4]}

= 2.125+ log {1/9}

= 1.171

Answer: 1.17