Based on the average number of moles of IO3- in your samples, and the initial volume of those samples, what is the solubility of Ca(IO3)2 in mol/L of the saturated solution in 0.01 M KIO3?
My average molarity was: .000092
My sample volume was: 5ml or .005 L
The solubility of Ca(IO3)2 in mol/L of the saturated solution in 0.01 M KIO3 is 0.00046 mol/L.
To calculate the solubility of Ca(IO3)2 in mol/L (M) of the saturated solution in 0.01 M KIO3, you will need to use the average number of moles of IO3- and the initial volume of the samples.
First, convert the sample volume from milliliters to liters. You mentioned that the sample volume is 5 mL, so it would be 0.005 L.
Next, use the formula for molarity (Molarity = moles/volume) to calculate the number of moles of Ca(IO3)2 in the sample. You have the average molarity, which is 0.000092 M, and the volume, which is 0.005 L. Multiply the average molarity by the sample volume:
Average moles of Ca(IO3)2 = 0.000092 M * 0.005 L = 4.6 x 10^-7 moles
Since Ca(IO3)2 dissociates into three moles of IO3- ions for every one mole of Ca(IO3)2, the number of moles of IO3- ions is three times the number of moles of Ca(IO3)2:
Moles of IO3- ions = 3 * 4.6 x 10^-7 moles = 1.38 x 10^-6 moles
Now, divide the number of moles of IO3- ions by the initial volume of the samples (0.005 L) to get the solubility in mol/L:
Solubility of Ca(IO3)2 = 1.38 x 10^-6 moles / 0.005 L ≈ 2.76 x 10^-4 M
Therefore, the solubility of Ca(IO3)2 in the saturated solution would be approximately 2.76 x 10^-4 M.
To find the solubility of Ca(IO3)2 in mol/L of the saturated solution in 0.01 M KIO3, you need to use the stoichiometry of the reaction and the given values.
The balanced equation for the reaction between Ca(IO3)2 and KIO3 is:
2 Ca(IO3)2 + 3 KIO3 → Ca(IO3)2 · 3 KIO3
From the balanced equation, you can see that two moles of Ca(IO3)2 react with three moles of KIO3 to form one mole of Ca(IO3)2 · 3 KIO3.
Given that the average molarity of IO3- in your samples is 0.000092 M and the sample volume is 0.005 L, you can calculate the number of moles of IO3- present in the sample:
moles of IO3- = molarity × volume
= 0.000092 M × 0.005 L
= 0.00000046 moles
Since there is a 1:1 stoichiometric ratio between IO3- and Ca(IO3)2 in the reaction, the number of moles of Ca(IO3)2 present is also 0.00000046 moles.
Now, you can calculate the solubility of Ca(IO3)2 in mol/L of the saturated solution by dividing the moles of Ca(IO3)2 by the sample volume:
solubility = moles of Ca(IO3)2 / volume
= 0.00000046 moles / 0.005 L
= 0.000092 mol/L
Therefore, the solubility of Ca(IO3)2 in mol/L of the saturated solution in 0.01 M KIO3 is 0.000092 mol/L.