Question description Identify three (3) new concepts you did not previously have any background on and state how this new information impacts your career or you personally. Discuss two (2)….

## Discuss Saturated Calcium iodate

Omer alzidi ksp post-lab

Calculations:-

1 Titration 1: 0.13 mol/L x 7.41×10-3 L = 9.63 x 10-4 mol 9.63 x 10-4 x ( 1 / 6 ) = 1.61 x 10-4 mol IO3- The solution has 30 mL so Concentration = 1.61 x 10-4 / 0.03 = 5.367 x 10-3

Calcium conc = Ca(IO3-)(S) → Ca2+(aq) + 2IO-3(aq) 5.367 x 10-3 * ½ = 2.684 x 10-3 M Ca2+ Ksp 1 = Ca2+ x IO3- = (2.684 x 10-3) x (5.367 x 10-3)2 = 7.73 x 10-8 Titration 2: 0.13 mol/L x 7.76×10-3 L = 1.008x 10-3 mol =1.008x 10-4 x ( 1 / 6 ) = 1.68x 10-4 mol IO3- The solution has 30 mL so Concentration = 1.68x 10-4 / 0.03 = 5.6x 10-3

Calcium conc = Ca(IO3-)(S) → Ca2+(aq) + 2IO-3(aq) 4.88x 10-3 * 1/2 = 2.8 x 10-3 M Ca2+ Ksp 2 = Ca2+ x IO3- = (2.8x 10-3) x (5.6 x 10-3)2 = 8.79x 10-8

2 Ksp average= (7.73 x 10-8 + 8.79x 10-8 )/2 =8.26*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- The solution has 30 mL so Concentration = 1.46 x 10-4 / 0.03 = 4.88x 10-3

0.13 mol/L x 6.32×10-3 L = 8.216x 10-4 mol =8.78x 10-4 x ( 1 / 6 ) = 1.37x 10-4 mol IO3- The solution has 30 mL so Concentration = 1.46 x 10-4 / 0.03 = 4.567 x 10-3

Omer alzidi ksp post-lab

Results Discussion:- 1)

a- In this experiment, the goal is to find the ksp of the calcium iodate. b- Firstly, I had a 70 mL of the Saturated Calcium iodate and I had to filter it and

pour it to the beaker im using. Then when I have the two flasks ready I prepare to do the two solution, where I must add 10 mL of the calcium iodate and 20 mL of the DI water in each flask and the I will add 3-5 drops of starch and a scoop of the Ki. After this we will titrate by using the Sodium thiosulfate that is placed in the burrete and start titrating and then we did the same thing but we replaced adding the calcium iodate with calcium chloride.

c- My calculations shows that my ksp in my first trials = 7.73 x 10-8 and in other trials of titration 2 it = 8.79x 10-8 and my average = 8.26*10-8

2) according to this website it supposed to be 6.47×10-6

http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf

3) comparing my average with the supposed answer I should get is quite different and this is made by many things such as human error not accurate measurements not clean equipment.

4) What I found is that whenever I add calcium iodate the more Ca2+ ions produced and less IO3- and this is what I’m supposed to get because I have more ksp in trial 1 and then trial 2.