How do you find pOh and expected pH?

Question

1.) An aqueous solution at 25°C with [OH-] = (5.916x10^-5) M. What is the pOH of this solution?

2.) What is expected pH of a 0.222 M solution of Ca(OH)2(aq)?

Answer 1

1).

pOH = -log [OH-]

pOH = -log 5.916x10^-5

pOH = 4.23

2).

pH + pOH = 14

Ca(OH)₂ ==> Ca²⁺ + 2OH-

0.222 M Ca(OH)₂ ==> 2 x 0.222 M OH- = 0.444 M OH-

pOH = -log [OH-] = -log 0.444

pOH = 0.353

pH = 14 - pOH

pH = 13.6