Question

1. What is the pH of pure water at 40.0C if the Kw at this temperature is 2.92X10^-14?

2. Determine the pH of a 0.227M C5H5N solution at 25C. The Kb of C5H5N is 1.7X10^-9

Please show work.

Answer 1

1)

pH = -log(H+)

by definition

and

Kw = [H+][OH-]

always

due to stoichiomerty = [H+] = [OH-] for pure water

2.92*10^-14 = Kw

2.92*10^-14 = [H+][Oh-]

assume x = [H+] = [OH-]

2.92*10^-14 = x*x

x = sqrt(2.92*10^-14) = 1.70880*10^-7 M

pH = -log(x) = -log(1.70880*10^-7) = 6.76730876465

2)

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

Kb = [C5H5NH+][OH-]/[C5H5N]

assume in equilbirium that

[C5H5NH+] = x = [OH-]

and

[C5H5N] = M-x = 0.227-x

(1.7*10^-9) = x*x/(0.227x)

x = 1.964*10^-5

[OH-] = 1.964*10^-5

pOH = -log(OH-) = -log(1.964*10^-5) = 4.706858

pH = 14-pOH = 14-4.706858

pH = 9.293142