Question

A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in LiF. Calculate the pH of the solution after the addition of 0.150 moles of solid LiOH. Assume no volume change upon the addition of base. The Ka for HF is 3.5x10^-4.

Please show how to work out this problem.

Answer 1

we know that

pKa = -log Ka

so

pKa = -log 3.5 x 10-4

pKa = 3.456

now

we know that

for buffers

pH = pKa + log [ salt / acid]

pKa for HF = 3.456

so

pH = pKa + log [ LiF / HF]

now

we know that

moles = molarity x volume (L)

so

moles of HF = 0.25 x 1 = 0.25

moles of LiF = 0.25 x 1 = 0.25

moles of LiOH added = 0.15

now

the reaction is

HF + LiOH ---> LiF + H20

we can see that

moles of HF reacted = moles of LiOH added = 0.15

moles of LiF formed = moles of LioH added = 0.15

so

finally

moles of HF = 0.25 - 0.15 = 0.1

moles of LiF = 0.25 + 0.15 = 0.4

now

pH = pKa + log [ LiF / HF]

as final volumes are same

ratio of concentrations = ratio of moles

so

pH = 3.456 + log [ 0.4 / 0.1 ]

pH = 4.058

so

pH of the solution is 4.058