2) A professor wanted to set up a similar experiment as the one you performed in lab. He 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 12.00 mL of 0.0542 M HCl solution. The Ksp of Al(OH)3 is 3.0x10-34. Show your work to receive credit. Finally, do you think this would be reasonable experiment for a general chemistry lab
The only flaw I can find is you squared 3 instead of cubing it and it will be 27X^4 instead of 9x^4.
This reduces the amount slightly, but the number is still incredibly high (about 10 ^ 5 L is what I've calculated). Your professor might want to point out that this will not be a effective experiment due to the large volume of saturated
The Ksp value of Ca(OH)2 on the site (I used 5.5E-6 [a far more soluble compound than Al(OH)3]) and estimated how much of it will be needed. My calculation was approximately 30 ml. If you were using that much in the experiment, it implies so our estimates for Al(OH)3 are right, that the high amount is unreasonably big and that Al(OH)3 will not be a suitable replacement unless the procedure was modified slightly.
