Question

Calculate pH after the addition of 0.10 mole of NaOH to a 500 ml buffer made up of 0.500 M HF (Ka = 6.8 x 10^-4) & 0.500 M NaF? Assume no change in volume of solution upon addition of base.

Answer 1

mol of NaOH added = 0.1 mol

HF will react with OH- to form F-

Before Reaction:

mol of F- = 0.5 M *0.5 L

mol of F- = 0.25 mol

mol of HF = 0.5 M *0.5 L

mol of HF = 0.25 mol

after reaction,

mol of F- = mol present initially + mol added

mol of F- = (0.25 + 0.1) mol

mol of F- = 0.35 mol

mol of HF = mol present initially - mol added

mol of HF = (0.25 - 0.1) mol

mol of HF = 0.15 mol

Ka = 6.8*10^-4

pKa = - log (Ka)

= - log(6.8*10^-4)

= 3.167

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

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

= 3.167+ log {0.35/0.15}

= 3.54

Answer: 3.54