[rleung]

Hi,

I have another multiple choice question whose answer is a little strange.  Suppose a student wants to make a basic solution by adding Ca(OH)2 to pure water.  But because Ca(OH)2 has a low solubility (Ksp=6.5e-6), he only adds a small amount of Ca(OH)2 to water.  What is the pH of a solution if he adds only 6.5e-9 mol of Ca(OH)2 to 10-L of water?

a. 4.81
b. 5.11
c. 7 * the right answer according to the key
d. 8.89
e. 9.19

I started by solving for the solubility of Ca(OH)2...

Ca(OH)2    =>   Ca 2+   +   2HO-
                       x               2x

Ksp = x(2x)^2
6.5e-6 =4x^3
x= 1.17e-2 M

This means that the solubility of HO- should be 2(1.17e-2 M) = 2.34e-2 M.

The person only adds 6.5e-10 mol / 10-L = 6.5e-10 M Ca(OH)2, which means that [OH-] = 2(6.5e-10 M) = 1.3e-9 M [OH-]

Since [HO-] is much less than its solubility, I assume all the [OH-] dissolves.  Therefore, pOH= -log[1.3e-9 M] = 8.886

pH = 14 - 8.886 = 5.11, or answer choice b.  Am I making some analytical error (because I know it's not a math error), or is the answer key's answer wrong?  Thanks.

Ryan
                   

[Borek]

Am I making some analytical error (because I know it's not a math error), or is the answer key's answer wrong?

What about water autodissociation?

[sdekivit]

yes it's the water autodissociation. But when i see this answer and see that 1,3 x 10^-9 mol OH(-) is dissolved in 10 L water, what influence would that have on the pH ... such a small quantity in such a huge amount of water.

[Borek]

7.03

[rleung]

::light goes off in my head::  Thanks so much.  It's amazing how much of this really simple stuff I overlook sometimes...