Question

Show how to calculate the pH of pure water or a neutral solution.

Answer 1

H2O ==> H^+ + OH^-
(H^+)(OH^-) = Kw = 1E-14
The equation tells you (H^+) = (OH^-)
(H^+)(H^+) = 1E-14 = (H^)^2
(H^+) = sqrt (1E-14) = 1E-7 M
pH = -log(H^+) = -log (1E-7) = 7
Since H^+ = OH^-, then pOH is 7 also and OH^- =1E-7 M
You will find pH + pOH = 14 helpful also.

Answer 2

Keq=1.8 x 10^-16 ====> determined experimentally and is accepted as the value at 25.0 C

Assuming 1L of water is 1000g and the molecular weight of water is 18.02g/mole.......

1000g/L*(mole/18.02g)=55.5 M


H2O ------------> H^+ + OH^-


Let X =H^+=OH^-

Keq=1.8 x 10^-16=[X][X]/[55.5M]




1.8 x 10^-16*[55.5M]= X^2

9.99 x 10^-15= 1 x 10^-14= X^2

√(1 x 10^-14)= X^2

1 x 10^-7=X

Back to the following relationship:

9.99 x 10^-15= 1 x 10^-14= X^2

1 x 10^-14= [H^+][OH^-]

-log[1 x 10^-14]=-log [H^+]*-log[OH^-]

14=pH+pOH