Am I doing this right?

[sotto voce]

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Question from mcatquestionaday.com:

Ca(OH)2 has Ksp = 10-6. If 0.1 mol of Ca(OH)2 was added to 1 L of pure water, what would be the approximate pH of the resulting solution?

A. 13
B. 12
C. 8
D. 2


Here's what I did:

1 x 10^-6 = 4x^3

x^3 = 0.25 x 10^-6

x = 0.6 x 10^-2 (approximation)

[OH] = 2x = 1.2 x 10^-2

pOH = ~2

pH = ~12


Is this how you guys would have done it? The answer is, in fact, B: 12. But I didn't utilize the 0.1 that they gave. I'm confused now. 🙁

 

[shoehornlettuce]

Question from mcatquestionaday.com:

Ca(OH)2 has Ksp = 10-6. If 0.1 mol of Ca(OH)2 was added to 1 L of pure water, what would be the approximate pH of the resulting solution?

A. 13
B. 12
C. 8
D. 2


Here's what I did:

1 x 10^-6 = 4x^3

x^3 = 0.25 x 10^-6

x = 0.6 x 10^-2 (approximation)

[OH] = 2x = 1.2 x 10^-2

pOH = ~2

pH = ~12


Is this how you guys would have done it? The answer is, in fact, B: 12. But I didn't utilize the 0.1 that they gave. I'm confused now. 🙁

You did it correctly. You don't have to worry about the 0.01 because you are dealing with Ksp (the starting material is always a solid so its concentration does not change). The equilibrium expression is:

Ksp = [Ca2+] [OH-]^2