Question

Calculate the pH of a solution that has a hydroxide ion concentration [OH-], 5.03 x 10^-9 M?

Answer 1

In order to be able to calculate pH, pOH and [H3O+]/[H+] and [OH-] we must understand the relationship between each of them

Remember that water will ionize as follows:

H2O(l) + H2O(l) <-> H3O+(aq) + OH-(aq)

so

Kw = [H3O+][OH-]

at T = 25°C, the Kw value is given as 10^-14

then,

10^-14 = [H3O+][OH-]

can be used always to relate H3O+ and OH-

also, note that

pH = -log([H+])

pOH = -log([OH-])

and

pH + pOH = 14 (stated above)

for this exercise:

get pOH first

pOH = -log([OH-])

pOH = -log(5.03*10^-9) = 8.298

pH = 14-pOH = 14-8.298 = 5.70