A professor wanted to set up a similar experiment as the one you performed in lab. The professor wanted to use Al(OH)3 in place of Ca(OH)2. Calculate how many mL of saturated Al(OH)3 solution it would take to titrate against 12.00mL of 0.0542M HCl solution? The Ksp of Al(OH)3 is 3.0x10-34. Do you think this would be reasonable experiment for a general chemistry lab?

Question

The Lab involved titrating a saturated solution of Ca(OH)2 with HCl to determine the Ksp and molar solubility, we were given the volume of Ca(OH)2 to use and in this situation I solved for it instead because they gave us Ksp so I could find M instead of V. Please check if I did this all right because I think the Volume is extraordinarily high and it does not make sense to me for it to be that high. Sorry!! Thank you!

12.00mL * 0.0542M HCl = 0.6504mmols H+ = 0.6504 mmols OH-
Ksp = [Al][OH]^3
3.0E-34 = x*(3x)^3 = 9x^4
x=2.4E-9
[OH-]=3x= 7.2E-9
(12.00mL * 0.0542) = ( [OH-] * V Al(OH)3 )
(12.00mL*0.0542)/7.2E-9=V Al(OH)3 = 9.0e7

Answer 1

The only error I see is you squared 3 instead of cubing it so it should be 27X^4 instead of 9x^4.
That changes the volume somewhat but the number still is extraordinarily large (about 10^5 L is what I calculated). Your professor may want to point out that this would not be a successful experiment because of the large volume of saturated Al(OH)3 required. Of course you could reduce that 12 mL too.

I don't know what volumes were used in your experiment but I looked up the Ksp value of Ca(OH)2 on the web (I used 5.5E-6 [a much more soluble compound than Al(OH)3] and calculated how much of it would be required. My calculation was approx 30 mL. If you used about that much in your experiment it means our calculations for Al(OH)3 are right, that the large volume is unreasonable and Al(OH)3 would not be a good substitute unless the experiment was changed somewhat. Hope this helps. I would be interested in knowing what volume you used.