Question

A buffer solution contains 0.384 M hydrofluoric acid and 0.460 M sodium fluoride.

If 0.0183 moles of nitric acid are added to 125 mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume change does not change upon adding nitric acid)

pH = ????

Answer 1

mol of HNO3 added = 0.0183 mol

F- will react with H+ to form HF

Before Reaction:

mol of F- = 0.46 M *0.125 L

mol of F- = 0.0575 mol

mol of HF = 0.384 M *0.125 L

mol of HF = 0.048 mol

after reaction,

mol of F- = mol present initially - mol added

mol of F- = (0.0575 - 0.0183) mol

mol of F- = 0.0392 mol

mol of HF = mol present initially + mol added

mol of HF = (0.048 + 0.0183) mol

mol of HF = 0.0663 mol

Ka = 6.6*10^-4

pKa = - log (Ka)

= - log(6.6*10^-4)

= 3.18

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

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

= 3.18+ log {3.92*10^-2/6.63*10^-2}

= 2.952

Answer: 2.95