Please choose the write answer

Question

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

a)between 3 and 6

b) between 9 and 12

c) between 6 and 9

d) between 12 and 14

Answer 1

Ca(OH)₂ ==> Ca⁺ + 2OH^-

Calcium hydroxide is soluble in water and dissociates fully and so does NOT have a Kb. But if we assume that it did, and that the Kb was 5x110, then...

Kb = [Ca+][OH-]/[Ca(OH)2]

5x10-11 = (x)(2x)/0.0004 - x

Solve for x and multiply by 2 to get [OH-]. Take negative log to get pOH and subtract from 14 to get pH.

But since it is a strong base and is soluble, we actually have the following situation:

4.0x10^-4 mol/L --> 4.0x10^-4 M Ca⁺ and 8.0x10^-4 M OH^-

pOH = -log 8x10-4 = 3.1

pH = 14 - 3.1 = 10.9 (answer b would be correct)