Choose the correct answer

Question

A chemist mixes 0.00040 mol/l solution of Ca(OH)2, where kb=5.0*10^-11. The PH of this solution is

a)between 6 and 9

b)between 3 and 6

c)between 12 and 14

d)between 9 and 12

Answer 1

Ca(OH)2 ==> Ca²⁺ + 2OH^-

[OH-] = 2 x 0.00040 mol/L = 0.00080 M

pOH = -log [OH-] = 3.1

pH = 14 - pOH

pH = 10.9 (answer d)