Ca(IO3)2(s) ⇌ Ca^(2+)(aq) + 2IO3^(-)(aq)
First, you poured 30 mL of 0.100 M Ca^(2+) solution over a 2.000 g sample of solid Ca(IO3)2 and allowed the mixture to sit for 30 minutes. Some of the solid dissolved, causing the calcium ion concentration to increase by a certain quantity and the iodate concentration to increase by two times that certain quantity. Next, you isolate the filtrate through gravity filtration. What do you expect the iodate concentration to be in the filtered solution? How would the iodate concentration differ if only a 0.500 g sample of the solid was used? What if the solvent was water instead?
This is a solubility product problem with Ca^+2 as a common ion.
Ca(IO3)2 ==> Ca^+2 + 2IO3^-
Ksp = (Ca^+2)(IO3^-)^2
If we call the solubility of Ca(IO3)2 = X, then (Ca^+) = X in the equilibrated solution and (IO3^-) = 2X.
Now plug those values into Ksp.
(0.1 + X)(2X)^2 = Ksp.
Solve for X and multiply by 2 to find IO3^-.
Next, no difference if 0.5 g sample of Ca(IO3)2 used. Finally, your last question doesn't make any sense? The solvent IS water isn't it? instead of what? It dissolves in water.
To determine the iodate concentration in the filtered solution, we need to consider the stoichiometry of the reaction and the initial concentrations of the reactants.
From the balanced equation of the reaction:
Ca(IO3)2(s) ⇌ Ca^2+(aq) + 2IO3^-(aq)
We can see that for every 1 mole of Ca(IO3)2 that dissolves, it produces 1 mole of Ca^2+(aq) and 2 moles of IO3^-(aq).
Given that the initial calcium ion concentration (Ca^2+) increased by a certain quantity, let's denote this increase as "x". According to the stoichiometry of the reaction, the iodate ion concentration (IO3^-) will increase by two times that quantity. Therefore, the iodate ion concentration will increase by 2x.
Now, let's calculate the iodate concentration in the filtered solution.
First, determine the moles of Ca(IO3)2 that dissolved:
Moles of Ca(IO3)2 = (mass of Ca(IO3)2) / (molar mass of Ca(IO3)2)
Given that the mass of Ca(IO3)2 is 2.000 g and the molar mass of Ca(IO3)2 is calculated as follows:
Molar mass of Ca(IO3)2 = (molar mass of Ca) + 2 * (molar mass of IO3-)
= (40.08 g/mol) + 2 * (126.90 g/mol)
= 294.88 g/mol
Thus, Moles of Ca(IO3)2 = 2.000 g / 294.88 g/mol = 0.00677 mol
Since one mole of Ca(IO3)2 produces 2 moles of IO3^-, the moles of IO3^- produced will be twice the moles of Ca(IO3)2, i.e., 2 * 0.00677 mol = 0.01354 mol
Next, let's calculate the volume of the filtered solution:
Given that you poured 30 mL of a 0.100 M Ca^2+ solution:
Volume of filtered solution = 30 mL = 0.030 L
Now, we can calculate the iodate concentration in the filtered solution using the formula:
Iodate concentration (in mol/L) = moles of IO3^- / volume of filtered solution
Iodate concentration = 0.01354 mol / 0.030 L = 0.45133 mol/L ≈ 0.45 M
Therefore, we expect the iodate concentration in the filtered solution to be approximately 0.45 M.
Now let's consider two different scenarios:
1. What if only a 0.500 g sample of the solid was used?
In this case, we need to repeat the calculations above using the new mass of Ca(IO3)2. The molar mass and moles of Ca(IO3)2 will be different, but we will still determine that the iodate concentration will be approximately 0.45 M.
2. What if the solvent was water instead?
If the solvent was water, the reaction between Ca(IO3)2 and water would look like this:
Ca(IO3)2(s) + H2O(l) ⇌ Ca^2+(aq) + 2IO3^-(aq) + H2O(l)
The presence of excess water in the reaction would not affect the stoichiometry of the reaction or the concentration of the iodate ions in the filtered solution. Therefore, the iodate concentration in the filtered solution would remain approximately 0.45 M.
Note: The calculations provided here assume that the reaction reaches equilibrium and that there are no other factors affecting the dissolution of Ca(IO3)2. The actual iodate concentration in the filtered solution may be affected by various factors such as temperature, pressure, and solubility.