Question

1.0 L of a buffer is prepared to produce a solution that is 0.400 M in...

1.0 L of a buffer is prepared to produce a solution that is 0.400 M in HF and 0.250 M in LiF. What is the pH after the addition of 0.044 mol.of base? Assume no volume change. pKa for HF = 3.46

a) 3.25

b) 3.12

c) 3.37

d) 3.54

e) 3.79

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Homework Answers

Answer #1

The initial pH of solution will be due to the buffer action of HF and LiF which can be calculated by Hendersen Hassalbalch equation

pH = pKa + + log [salt] / [acid]

pKa of HF = 3.46

[Salt] = [LiF] = 0.250

[Acid] = [HF] = 0.4

pH = 3.46 + log [0.250 / 0.4] = 3.26

Now if we will add base to the given solution, then it will react with acid to form salt and will decrease the concentration of acid

Moles of salt formed = Moles of base added = 0.044

NaOH + HF --> NaF + H2O

Moles of acid decreased = Moles of base added = 0.044

pH = pKa + log [salt + base] / [acid - base]

pH = 3.46 + log [0.25 + 0.044] / [0.4-0.044] = 3.46 + log 0.294 / 0.356 = 3.46 + log 0.826 = 3.376

Answer: (c)

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