Question

Determine the pH of an HF solution of each of the following concentrations.

Part A: .280 M

PartB : 4.5*10^-2 M

Part C: 2*10^-2

Part D

In which cases can you not make the simplifying assumption that x is small?

In which cases can you not make the simplifying assumption that x is small?

only in (a)

only in (b)

in (a) and (b)

in (b) and (c)

Answer 1

PART a
              HF <---> H+ + F-

initial      0.28 M     0M   0M

change         x         x    x

equilibrium 0.28-x      x    x

Ka = [H+][F-]/[HF]

Ka of HF = 6.6*10^-4

6.6*10^-4 = (x^2/(0.28-x))

x = 0.0132 M

[H+] = 0.0132 M

pH = -log[H+]

    = -log(0.0132)

   = 1.88

part B

              HF <---> H+ + F-

initial      0.045 M     0M   0M

change         x         x    x

equilibrium 0.045-x      x    x

Ka = [H+][F-]/[HF]

Ka of HF = 6.6*10^-4

6.6*10^-4 = (x^2/(0.045-x))

x = 0.00513 M

[H+] = 0.00513 M

pH = -log[H+]

    = -log(0.00513)

    = 2.3

part c

              HF <---> H+ + F-

initial      0.02 M     0M   0M

change         x         x    x

equilibrium 0.02-x      x    x

Ka = [H+][F-]/[HF]

Ka of HF = 6.6*10^-4

6.6*10^-4 = (x^2/(0.02-x))

x = 0.00332 M

[H+] = 0.00332 M

pH = -log[H+]

    = -log(0.00332)

pH = 2.5

part C

as x is small in (b) and (c) it can not be neglected.