Question

is this graph right? and how to get the ksp?

Adding Ca(NO3)2 to NaOH 9. On the report, number twelve (12) wells on the spot plate consecutively 1 - 12. In wells 2- 12, carefully count 10 drops of purified water into each well. All wells 2-12 will now contain 10 drops of water. 10. In well 1. place 10 drops of the 0.10 M NaOH solution. 11. In well 2, add 10 drops of the 0.10 M NaOH solution. Mix the well 2 contents by jiggling the spot plate or stirring with the dropper/plastic dropper. Pull the contents of well 2 into the dropper once or twice. Fill the dropper with the contents of well 2 and proceed to step 12. 12. In well 3, add 10 drops of the solution from well 2. Put the remaining solution from well 2 back into well 2. Mix the contents of well 3 by jiggling the spot plate and pull the contents of well 3 into the dropper once or twice. Fill the dropper with the contents of well 3 and proceed to step 13. 13. In well 4, add 10 drops of the solution from well 3. Put the remaining solution from well 3 back into well 3. Mix the contents of well 4 by jiggling the spot plate and pull the contents of well 4 into the dropper once or twice. Add 10 drops from well 4 to well 5 and put the remaining solution from well 4 back into well 4. Repeat this process until well 12. Remove 10 drops of well 12 solution and discard. 14. Use a new dropper and add 10 drops of the 0.1 M Ca(NO3)2 to each of the 12 wells. Mix all solutions by jiggling the well plate back and forth on the bench top. Let the spot plate sit undisturbed for at least 5 minutes. Record which wells had a precipitate and which wells did not. A precipitate will form consecutively in the wells and the first well that had no precipitate is the saturated solution.

Answer 1

I see a mistake concerning the concentration of Ca(II) and OH^- after the last dilution. In wells 2-12 your start with 10 drops of water, to which 10 drops of the previous well are added (20 drops total), but then 10 drops of the mixture are used for the next well, so you end up having 10 drops in each well prior to the addition of Ca(II). Since you the add 10 drops of Ca(II) solution, the dilution of both Ca(II) and OH^- is 1:2, not 1:3 or 2:3, respectively. This means that the concentration of Ca(II) is 0.05 M in all wells, not only in the first (even in well 12, where you are told to discard 10 drops after mixing the water with the solution from well 11) and that the concentration of hydroxide ions is half the concentration after dilution with water.

Regarding K_sp, this constant is defined (for Ca(OH)₂) as:

$K_{sp}=[Ca^{2+}][OH^{-}]^{2}$

Where the concentrations are the concentrations in equilibrium with the precipitate. When our solution is saturated, the concentrations of the species are practically those in equilibrium, so we can use them in the calculation. In your experiment, well 6 was the first in which no precipitate was formed. The concentrations in the well were 0.05 M for Ca(II) and 1.57x10^-3 for OH^- (taking into account the error in the calculation of these concentrations). K_sp is then:

$K_{sp}=0.05\cdot (1.57x10^{-3})^{2}=1.22x10^{-7}$