Question

The solubility rules say that Ba(OH)2, Sr(OH)2, and Ca(OH)2 are marginally soluble hydroxides. Calculate the pH of a saturated solution of Sr(OH)2. Ksp = 4.4 x10-4 for Sr(OH)2.

Answer 1

Solution :-

Sr(OH)2 ---- > Sr^2+   + 2OH-

                           X              2x

Ksp= [x][2x]^2

Ksp = 4x^3

4.4*10^-4 = 4x^3

4.4*10^-4 / 4 = x^3

1.1*10^-4 = x^3

Taking cube root of both sides we get

3.667*10^-5 = x

So the concentration of the OH- = 2 * 3.667*10^-5 M = 7.334*10^-5 M

Now lets calculate the pOH

pOH= -log [OH-]

pOH= -log [7.334*10^-5 ]

pOH= 4.13

pH + pOH = 14

pH= 14 – pOH

pH= 14 – 4.13

pH= 9.87