Question

Using the appropriate ksp value from appendix d in the textbook, calculate the ph of a saturated solution of ca(oh)2 .

Answer 1

The ph of a saturated solution of Ca(OH)2 is 12.35

CALCULATION:
For the reaction 
     Ca(OH)2 → Ca2+ + 2OH- 
we will use the Ksp expression to solve for the concentration [OH-] and then use the acid base concepts to get the pH:
     Ksp = [Ca2+][OH-]^2

The listed Ksp value is 5.5 x 10^-6. Substituting this to the Ksp expression, we have
     Ksp = 5.5 x 10^-6 = (s) (2s)^2 = 4s^3
     s3 = 5.5x10^-6 / 4

Taking the cube root, we now have
     s = cube root of (5.5x10^-6 / 4)s
        = 0.01112

We know that the value of [OH-] is actually equal to 2s:
     [OH-] = 2s = 2 * 0.01112 = 0.02224 M

We can now calculate for pOH:          
     pOH = - log [OH-]
              = -log(0.02224)
              = 1.65

Therefore, the pH is
     pH = 14 - pOH
           = 14 - 1.65
           = 12.35

Answer 2

The pH of the solution of calcium hydroxide is 12.3

From that standard tables, we know that the Ksp of Ca(OH)2 is  5.5 x 10^-6. Recall, that the Ksp is obtained from;

Ca(OH)2(s) ⇄ Ca^2+(aq) + 2OH^-(aq)

Hence;

Ksp = s * (2s)^2

Ksp = 4s^3

s = ∛Ksp/4

s = ∛5.5 x 10^-6/4

s= 0.01 M

But [OH^-] = 2s

[OH^-] = 2(0.01 M)

[OH^-] = 0.02 M

pOH = - log(0.02 M)

pOH = 1.7

pH = 14 - 1.7 = 12.3

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