find the pH and pOH

Question

a.) Find the pH of 9.87x10^-1 M of Sr(OH)₂

b.) Find the pOH of 5.71x10^-2 M of K₂O

Answer 1

These are the relationships you MUST know:

pH = -log [H⁺]

pOH = -log [OH^-]

pH + pOH = 14

[H⁺][OH^-] = 1x10^-14 = Kw

a). 9.87x10-1 M Sr(OH2)

Sr(OH)2 ==> Sr+2+ + 2OH-

[Sr2+] = 9.87x10-1 M and [OH-] = 2 x 9.87x10-1 M = 1.974 M

pOH = -log [OH-] = -log 1.974 = -0.295

pH + pOH = 14

pH = 14 - (-0.295) = 14.295 = 14 (to 2 significant figures)

(b). 5.71x10-2 M K2O

K2O + H2O ==> 2KOH

5.71x10-2 M K2O ==> 2 x 5.71x10-2 M KOH = 0.114 M OH-, because 1 mol K2O produces 2 moles KOH

pOH = -log[OH-] = -log 0.114

pOH = -1.058 = -1.1 (to 2 significant figures)