What was your experimental Ksp from Part 2? What was your experimental Ksp from
ID: 856680 • Letter: W
Question
What was your experimental Ksp from Part 2?
What was your experimental Ksp from Part 3?
Ideally the results from Part 2 and Part 3 should be the same. What could account for any
difference in results?
How many significant figures can you accurately report in the values for Ksp that you calculated in Parts 2 and 3 of the experiment? Are the two values different from each other to that number of significant figures?
Data Table 1:
Well 1
Well 2
Well 3
Well 4
Well 5
Well 6
Ca(NO?)?
5 drops
5 drops
Water
5 drops
5 drops
5 drops
5 drops
Solution from pervious well
5 drops
5 drops
5 drops
5 drops
[Ca??]
0.10M
0.05 M
0.025 M
0.0125 M
0.00625 M
0.0031 M
1*10-1
5*10-2
2.5*10-2
1.3*10-2
6.3*10-3
3.1*10-3
Data Table 2: Concentrations after mixing
Well 1
Well 2
Well 3
Well 4
Well 5
Well 6
[Ca??]
0.10M
0.05 M
0.025 M
0.0125 M
0.00625 M
0.0031 M
0.15 M NaOH
5 drops
5 drops
5 drops
5 drops
5 drops
5 drops
Final [Ca??]
0.05 M
0.025 M
0.0125 M
0.00625 M
0.0031 M
0.0016 M
Final [OH?]
0.05M
0.05M
0.05M
0.05M
0.05M
0.05M
Data Table 3:
Well 1
Well 2
Well 3
Well 4
Well 5
Well 6
Initial [OH?]
0.10M
0.05 M
0.025 M
0.0125 M
0.00625 M
0.0031 M
0.1 M Ca(NO?)?
5 drops
5 drops
5 drops
5 drops
5 drops
5 drops
Final [OH?]
0.05 M
0.025 M
0.0125 M
0.00625 M
0.0031 M
0.0016 M
Final [Ca??]
0.05M
0.05M
0.05M
0.05M
0.05M
0.05M
Data Table 1:
Well 1
Well 2
Well 3
Well 4
Well 5
Well 6
Ca(NO?)?
5 drops
5 drops
Water
5 drops
5 drops
5 drops
5 drops
Solution from pervious well
5 drops
5 drops
5 drops
5 drops
[Ca??]
0.10M
0.05 M
0.025 M
0.0125 M
0.00625 M
0.0031 M
1*10-1
5*10-2
2.5*10-2
1.3*10-2
6.3*10-3
3.1*10-3
Explanation / Answer
When carrying out the titration you calculated the concentration of Ca(OH)2 in solution:
Write a balanced equation
Ca(OH)2 + 2HCl ? CaCl2 + 2H2O
1molCa(OH)2 reacts with 2mol HCl
You have the volumes and molarity of the acid, now calculate the molarity of the base:
Use the equation:
Ma*Va = Mb*Vb *2
0.05345*7.50 = Mb * 10.0*2
Mb = 0.0200M Ca(OH)2 This is the answer that you require if you are solving for the molar concentarion of the Ca(OH)2
If you wish to have the solubility expressed as gram Ca(OH)2 per litre:
Molar mass Ca(OH)2 = 74.0932 g/mol
0.02mol = 74.0932*0.02 = 1.48g/litre Ca(OH)2
I do not follow your reasoning: I paste from your attempt:
Ma x Va = Mc x Vc
(.05345 M HCL) ( 7.50 mL HCL) = ( x M OH) (10.00mL OH)
turns out to be 3.2x10^-5 = Ksp
I do not follow how, from these data you can calculate Ksp = 3.2*10^-5
If you multiply out:(.05345 M HCL) ( 7.50 mL HCL) = ( x M OH) (10.00mL OH) you get a value of 0.0400 for (M OH-) which is correct. If you take into account that the base is Ca(OH)2 = 2 mol OH- per molecule, which gives [Ca(OH)2] = 0.0200, which is what I got.
You can now go on to calculate the Ksp from your data:
Ksp = [Ca 2+] * [OH-]
Ksp = (0.02)* (0.04)
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