Chem 12 help

[cause]

Hi guys, doing a chem lab report and I feel a little overwhelmed (skipped chem 11 lol). I'll try to include as much info as possible so as to not miss anything.

Experiment: Equilibrium - the determination of Ksp values for Calcium Iodate

Formulas used:

Ca(IO3)2(s) <-> Ca2+(aq) + 2IO3-(aq)

Ksp = [Ca2+][IO3-]2

5I-(aq) + IO3-(aq) + 6H+(aq) -> 3I2(aq) + 3H2O(l)


1. Obtain 200ml of Standardized 0.050 M Na2S2O3 in a burette
2. Measure out 10 mL of 20% KI solution into a Erlenmeyer flask
3. Obtain 50 mL of room temp (temp value recorded) saturated Ca(IO3)2 solution. Pipette 10 mL into the Erlenmeyer flask. Add 5 mL of 2 M HCl (note colour changes)
4. Titrate the above solution with the Na2S2O until a yellow colour is produced, then add 50 mL of deionized water then 5 mL starch indicator. Continue to titrate until blue-black solution goes clear. (Titre values recorded and an average is achieved within 0.10 mL)
5. Do the same thing with a 5 Celcius saturated Ca(IO3)2 solution.
6. Do it again with further samples of room temp Ca(IO3)2 in 0.10 M Ca(NO3)2

Now the fun part, calculations:
1. Using the average titre of S2O32-, (7.57 mL) calculate the moles of S2O32-, and hence the moles of I2 present in the Erlenmeyer flask.

So, fairly easy calculation? We are given the molarity of Na2S2O3 to be 0.050M so * 7.57 mL = 0.3785 moles of S2O32- (correct?)

Oh and also our chemical formula for this operation is:
2S2O32-(aq) + I2(aq) -> 2I-(aq) + S4O62-

So, is this a 2:1 ratio? So the moles of I is 0.3785/2 = 0.189 moles (correct?)

2. Calculate the moles of IO3- required to produce this amount of I2

At this point I am lost...the lab continues with a few more questions:

3. Calculate the molarity of the IO3- solution

4. Calculate the molarity of the Ca2+ solution

5. Calculate the solubility product (Ksp) for Ca(IO3)2

I'm hoping this is easy for someone to just take a look at and point me in the correct direction. Thanks for your time.

[cause]

Oh also, some info that may help:

Prelab question

A 10.00 mL sample of saturated Ca(IO3)2 was collected from a solution at 10.2oC. After adding excess KI and H+, the resulting solution was titrated with 0.0543 M Na2S2O3. If 8.52 mL of Na2S2O3 was used, calculate the Ksp of Ca(IO3)2 following the steps in the Treatment of Data section. (the steps I listed in the OP)

Answers:

Moles of IO3- = 7.71 x 10-5
Ksp = 2.29 x 10-7

[Pkingd3vil]

Oh also, some info that may help:

Prelab question

A 10.00 mL sample of saturated Ca(IO3)2 was collected from a solution at 10.2oC. After adding excess KI and H+, the resulting solution was titrated with 0.0543 M Na2S2O3. If 8.52 mL of Na2S2O3 was used, calculate the Ksp of Ca(IO3)2 following the steps in the Treatment of Data section. (the steps I listed in the OP)

Answers:

Moles of IO3- = 7.71 x 10-5
Ksp = 2.29 x 10-7

Explains A Mole:
https://www.youtube.com/watch?v=AsqEkF7hcII

Mole Conversions:
https://www.youtube.com/watch?v=xPdqEX_WMjo

Calculating Molarity:
https://www.youtube.com/watch?v=T1h20ArYipo


Here's a start hard thing about Chem is it is easier shown I think. So start at these if you don't know these by the back of your hand.

[King]

Hi guys, doing a chem lab report and I feel a little overwhelmed (skipped chem 11 lol). I'll try to include as much info as possible so as to not miss anything.

Experiment: Equilibrium - the determination of Ksp values for Calcium Iodate

Formulas used:

Ca(IO3)2(s) <-> Ca2+(aq) + 2IO3-(aq)

Ksp = [Ca2+][IO3-]2

5I-(aq) + IO3-(aq) + 6H+(aq) -> 3I2(aq) + 3H2O(l)


1. Obtain 200ml of Standardized 0.050 M Na2S2O3 in a burette
2. Measure out 10 mL of 20% KI solution into a Erlenmeyer flask
3. Obtain 50 mL of room temp (temp value recorded) saturated Ca(IO3)2 solution. Pipette 10 mL into the Erlenmeyer flask. Add 5 mL of 2 M HCl (note colour changes)
4. Titrate the above solution with the Na2S2O until a yellow colour is produced, then add 50 mL of deionized water then 5 mL starch indicator. Continue to titrate until blue-black solution goes clear. (Titre values recorded and an average is achieved within 0.10 mL)
5. Do the same thing with a 5 Celcius saturated Ca(IO3)2 solution.
6. Do it again with further samples of room temp Ca(IO3)2 in 0.10 M Ca(NO3)2

Now the fun part, calculations:
1. Using the average titre of S2O32-, (7.57 mL) calculate the moles of S2O32-, and hence the moles of I2 present in the Erlenmeyer flask.

So, fairly easy calculation? We are given the molarity of Na2S2O3 to be 0.050M so * 7.57 mL = 0.3785 moles of S2O32- (correct?)

Oh and also our chemical formula for this operation is:
2S2O32-(aq) + I2(aq) -> 2I-(aq) + S4O62-

So, is this a 2:1 ratio? So the moles of I is 0.3785/2 = 0.189 moles (correct?)

2. Calculate the moles of IO3- required to produce this amount of I2

At this point I am lost...the lab continues with a few more questions:

3. Calculate the molarity of the IO3- solution

4. Calculate the molarity of the Ca2+ solution

5. Calculate the solubility product (Ksp) for Ca(IO3)2

I'm hoping this is easy for someone to just take a look at and point me in the correct direction. Thanks for your time.

IO3 is a 1:2 ratio with I2 and oxygen is a 1:2 ratio as well, so get the mass of the compound by adding X2 Iodine + X2 Oxygen from the compound, then divide by the molar mass of IO3.

Kinda Grave dig, I wish I had saw this earlier, I'm Taking AP chem this year and we did this lab, did you get it? Have anymore chem questions feel free to ask me!