Suppose Kw is 4.0 × 10-14 at some temperature.

Question

Suppose K_w is 4.0 × 10^-14 at some temperature.

a. Calculate pH and pOH for pure water at this mystery temperature.

pH =

pOH =

b. Suppose you have a solution with a pH of 6.75 at this mystery temperature. Is that solution acidic,

basic, or pH-neutral?

c. Calculate the pH and pOH of a 0.0010 M NaOH solution at this different temperature.

pH =

pOH =

Answer 1

Kw = 4.0x10^-14

(a). Autoionization of H₂O: H₂O ==> H⁺ + OH^-

Kw = 4.0x10^-14 = [H⁺][OH-]

4.0x10^-14 = (x)(x) = x²

x = [H⁺] = [OH-] = 2.0x10^-7

pH = -log [H⁺] = 6.699

pOH = -log [OH-] = 6.699

(b). pH = 6.75

This represents a solution that is basic since neutral pH = pOH = 6.699

(c). 0.0010 M NaOH ==> 0.0010 M Na+ + 0.0010 M OH-

pOH = -log 0.0010

pOH = 3.00

pH + pOH = 6.699 + 6.699 = 13.398

pH = 13.398 - 3

pH = 10.398