Calculate the pH of 0100 L of a buffer solution that is 020

Calculate the pH of 0.100 L of a buffer solution that is 0.20 M in HF (Ka = 3.5 x 10^-4 ) and 0.47 M in NaF.

Express your answer using three significant figures.

3.83

Part B

What is the pH after adding 0.003 mol of HNO3 to the buffer described in Part A?

Express your answer using three significant figures.

Part C

What is the pH after adding 0.003 mol of KOH to the buffer described in Part A?

Express your answer using three significant figures.

Solution

Solution:-

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 [ 0.47 / 0.20 ]

pH = 3.83


Therefore the pH is 3.83


B )


initial moles of HF = molarity x volume

initial moles of HF = 0.20 x 0.1

moles of HF = 0.02

moles of NaF = 0.47 x 0.1 = 0.047


moles of HN03 added = 0.003

here HN03 is acid and it reacts with the conjugate base NaF

The reaction is


HN03 + NaF ---->   HF + NaN03

from the above reaction

moles of NaF reacted = moles of HN03 = 0.003

moles of HF formed = moles of NaF reacted = 0.003


new moles of HF = 0.020 + 0.003 = 0.023

new moles of NaF = 0.047 - 0.003 = 0.044

Now,
pH = pKa + log [ NaF / HF ]


as final volume is constant

the ratio of conc = ratio of moles

so


pH = -log 3.5 x 10^-4 + log [ 0.044 / 0.023]

pH = 3.738

C)


moles of HF = 0.020

moles of NaF = 0.047


moles of KOH added = 0.003

Now KOH is base it reacts with the acid HF

the reaction is


OH- + HF ---->   F-   + H20

from the above reaction

moles of HF reacted = moles of KOH = 0.003

moles of F- formed = moles of HF reacted = 0.003


new moles of HF = 0.020 - 0.003 = 0.017

new moles of F- = 0.047 + 0.003 = 0.05

now


pH = pKa + log [ NaF / HF ]


as final volume is constant

the ratio of conc = ratio of moles

so


pH = -log 3.5 x 10^-4 + log [ 0.05 / 0.017]

pH = 3.92