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?
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
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.
To calculate the volume of saturated Al(OH)3 solution needed to titrate against 12.00 mL of 0.0542 M HCl solution, you followed these steps:
1. You first determined the moles of HCl present in the 12.00 mL of the solution:
Moles of HCl = Volume (in L) * Concentration = 12.00 mL * 0.0542 M = 0.6504 mmol
2. Since Al(OH)3 is a strong base, it fully dissociates in water to produce three hydroxide ions (OH-) for every one aluminum ion (Al). Thus, you used the Ksp expression for Al(OH)3:
Ksp = [Al][OH-]^3
3. You assumed that the concentration of Al ions ([Al]) would be equal to the concentration of OH- ions ([OH-]) in the saturated Al(OH)3 solution.
4. You substituted the value of Ksp and the concentration of OH- ions into the Ksp expression:
3.0E-34 = x * (3x)^3 = 9x^4
5. You solved the equation for x, which represents the concentration of OH- ions:
x = 2.4E-9
6. Since concentration = moles/volume, you calculated the concentration of OH- ions in the saturated Al(OH)3 solution:
[OH-] = 3x = 7.2E-9 M
7. Finally, you used the concentration of OH- ions and the volume of HCl solution to calculate the volume of Al(OH)3 solution needed:
(12.00 mL * 0.0542) / 7.2E-9 = V Al(OH)3 = 9.0E7 mL
Now, regarding whether this would be a reasonable experiment for a general chemistry lab, it seems that the calculated volume of Al(OH)3 solution is extremely high (9.0E7 mL). This volume is unusually large and may not be practically feasible in a lab setting. It is recommended to double-check your calculations and consider whether there might be any errors or inconsistencies in your approach or the data given.
To calculate the volume of saturated Al(OH)3 solution needed to titrate against 12.00 mL of 0.0542 M HCl solution, you can follow these steps:
1. Calculate the number of moles of HCl used:
12.00 mL * 0.0542 M = 0.6504 mmol HCl
2. Since Al(OH)3 dissociates to form 3 OH- ions, the number of moles of OH- ions is equal to the number of moles of HCl used:
0.6504 mmol OH- = 0.6504 mmol Al(OH)3
3. Use the Ksp expression for Al(OH)3 to find the concentration of Al(OH)3:
Ksp = [Al][OH]^3
3.0 × 10^-34 = x * (3x)^3
x = 1.2 × 10^-9 M
4. Calculate the concentration of OH- ions from the Al(OH)3 concentration:
[OH-] = 3 * x = 3 * 1.2 × 10^-9 M = 3.6 × 10^-9 M
5. Use the concentration of OH- ions and the volume of HCl used to find the volume of Al(OH)3:
(12.00 mL * 0.0542 M) / 3.6 × 10^-9 M = 1.8 × 10^8 mL = 180,000 L
So, it would take approximately 180,000 liters of saturated Al(OH)3 solution to titrate against 12.00 mL of 0.0542 M HCl solution.
As for whether this would be a reasonable experiment for a general chemistry lab, the extremely large volume required suggests that it may not be practical or feasible. It would be challenging to obtain and work with such a large volume of solution in a laboratory setting. Additionally, the values obtained may have a large margin of error due to the difficulties involved in accurately measuring and handling such a large volume.