## Concentration of the solution = 1.68 x 10-4 / 0.03 = 5.6 x 10-3

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

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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