Question

A) What is the pH of a solution with a fluoride concentration of 0.272 M? The Kb for F- is 1.47 x 10^-11.

B) Calculate the pH value of of a 0.015 M solution of pyridine, C5H5N. The Kb of pyridine is 1.7 x 10^-9. Enter your values for a, b, and c for the quadratic equation along with the final answer.

Answer 1

Kb for F- is 1.47 x 10^-11

concentration = 0.272 M

F- + H2O   --------------> HF +   OH-

0.272                                  0         0

0.272-x                               x          x

Kb = x^2 / 0.272-x

1.47 x 10^-11 = x^2 / 0.272-x

x = 2 x 10^-6

[OH-] = 2 x 10^-6 M

pOH = -log[OH-] = -log(2 x 10^-6)

        = 5.7

pH =8.30

B)

C5H5N    +   H2O --------------> C5H5NH+ + OH-

0.015                                               0                 0

0.015 - x                                           x                 x

Kb = x^2 / 0.015-x

1.7 x 10^-9 = x^2 / 0.015 -x

x^2 + 1.7 x 10^-9 x - 2.55 x 10^-11 = 0

a x^2 + bx + c = 0

x = 5.05 x 10^-6

[OH-] = 5.05 x 10^-6 M

pOH = -log[OH-] = -log(

pOH = -log[OH-] = -log(2 x 10^-6)

        = 5.30

pH =8.70

        = 5.7

pH =8.30