Calcium concentration of solution = Ca (IO3-) (S ) → Ca2+ (aq) + 2IO- 3 (aq) 5.367 x 10-3 x ½ = 2.684 x 10-3 M of Ca2+ Ksp1

Ahmed AL Balushi
11564611
Ksp post lab
Calculations of post lab: –
1) solution Titration 1: 0.13 mol/L x 7.41 x 10-3 L = 9.63 x 10-4 mol
9.63 x 10-4 x ( 1 / 6 ) = 1.61 x 10-4 mol IO3-
we used a 30 mL in solution
Concentration of the solution
= 1.61 x 10-4 / 0.03 = 5.367 x 10-3
Calcium concentration of solution
= Ca (IO3-) (S ) → Ca2+ (aq) + 2IO- 3 (aq)
5.367 x 10-3 x ½ = 2.684 x 10-3 M of Ca2+
Ksp1
= Ca2+ x IO3- = (2.684 x 10-3) x (5.367 x 10-3)2 = 7.73 x 10-8
solution Titration 2:
0.13 mol/L x 7.76 x 10-3 L = 1.008 x 10-3 mol
= 1.008 x 10-4 x ( 1 / 6 ) = 1.68 x 10-4 mol IO3-
we used a 30 mL in solution
Concentration of the solution
= 1.68 x 10-4 / 0.03 = 5.6 x 10-3
Calcium concentration of solution
= Ca (IO3- ) (S) → Ca2+ ( aq) + 2IO-3 (aq)
4.88 x 10-3 x 1/2 = 2.8 x 10-3 M of Ca2+
Ahmed AL Balushi
11564611
Ksp 2
= Ca2+ x IO3- = 2.8x 10-3 x 5.6 x 10-3 2 = 8.79x 10-8
2) Ksp solution average
= ( 7.73 x 10-8 + 8.79 x 10-8 ) / 2 = 8.26 x 10-8
3) 0.13 mol/L x 6.76×10-3 L = 8.78x 10-4 mol
=8.78x 10-4 x ( 1 / 6 ) = 1.46 x 10-4 mol IO3-
we used a 30 mL in solution
Concentration of the solution
= 1.46 x 10-4 / 0.03 = 4.88x 10-3
0.13 mol/L x 6.32×10-3 L = 8.216 x 10-4 mol
= 8.78x 10-4 x ( 1 / 6 ) = 1.37 x 1 0-4 mol of IO3-
we used a 30 mL in solution
Concentration of the solution
= 1.46 x 10-4 / 0.03 = 4.567 x 10-3
Discussion:
1- a) in this experiment the aim was to determine what is the ksp that the solution of calcium iodate have.
b) We started the experiment by getting the saturated calcium iodate and
start to filter it using funnel and filter paper and pour it into a beaker or flask,
then after filtring I will add 10 mL each in two ernylmer flasks and then I
will add 20 mL and afterwards I added 3 drops of 2% starch solution in each
of the two flasks and finally I added a scoop of potassium iodide, after
preparing the solution I started titrating them with Sodium thiosulfate using
burette to dispense it to the solution. The second part is tottaly the same but I
replaced adding the calcium iodate by the calcium chloride.
Ahmed AL Balushi
11564611
c) When I calculated the ksp average it equaled 8.26 x 10-8
2- the value that I had to get is 6.47 * 10–6 I got this value from
https://www.chm.ulaval.ca/gecha/chm1903/6_solubilite_solides/solubility_p
roducts.pdf
3- As it shows that there is a small difference between my average ksp and the ksp value that I’ve should’ve got, this happened because we made some
errors in burette reading and maybe also by not using clean apparatus and
adding more solutions.
4- The obtainment of this experiment is that when adding the calcium iodate the IO3- gets reduced and you can see this by the ksp values ive got in my
calculations.

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Author Since: November 30, 2020

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