Question

A.) Calculate the pH of 0.100 L of a buffer solution that is 0.29M in HF (Ka = 3.5 x 10-4 ) and 0.55M in NaF.

B.) What is the pH after adding 0.004mol of HNO3 to the buffer described in Part A?

C.) What is the pH after adding 0.002mol of KOH to the buffer described in Part A?

Answer 1

Answer 2

A)

for an acidic buffer

pH = pKa + log [ conjugate base / acid ]

pH = -log Ka + log [ NaF / HF]

pH = -log 3.5 x 10-4 + log [ 0.55 / 0.29 ]

pH = 3.734

so the pH is 3.734

B )

initial moles of HF = molarity x volume

initial moles of HF = 0.29 x 0.1

moles of HF = 0.029

moles of NaF = 0.55 x 0.1 = 0.055

moles of HN03 added = 0.004

as HN03 is acid it reacts with the conjugate base NaF

the reaction is given by

HN03 + NaF ---->   HF + NaN03

from the above reaction

moles of NaF reacted = moles of HN03 = 0.004

moles of HF formed = moles of NaF reacted = 0.004

new moles of HF = 0.029 + 0.004 = 0.033

new moles of NaF = 0.055 - 0.004 = 0.051

now

pH = pKa + log [ NaF / HF ]

as final volume is constant

ratio of conc = ratio of moles

so

pH = -log 3.5 x 10-4 + log [ 0.051 / 0.033]

pH = 3.645

C)

moles of HF = 0.029

moles of NaF = 0.055

moles of KOH added = 0.002

as KOH is base it reacts with the acid HF

the reaction is given by

OH- + HF ---->   F-   + H20

from the above reaction

moles of HF reacted = moles of KOH = 0.002

moles of F- formed = moles of HF reacted = 0.002

new moles of HF = 0.029 - 0.002 = 0.027

new moles of F- = 0.055 + 0.002 = 0.057

now

pH = pKa + log [ NaF / HF ]

as final volume is constant

ratio of conc = ratio of moles

so

pH = -log 3.5 x 10-4 + log [ 0.057 / 0.027]

pH = 3.78