Determine the maximum mass of calcium iodate that can be dissolved in a 10.0L jug of 3.25x10^-2 molar calcium sulphide solution.

Hopefully, Dr. Bob222 corrects this if this is wrong.

The maximum mass of calcium iodate that can dissolve in a 10L solution of calcium sulfide can not exceed the number of moles of calcium sulfide in the 10L solution. Therefore, solve for the number of moles in the solution.

3.25x10^-2 M=moles/L, solve for moles

moles=Molarity * Volume (L)=(10.0L)*(3.25x10^-2 M)= moles of calcium sulfide

moles of calcium sulfide = moles of calcium iodate

moles of calcium iodate * 389.88 g/mol =maximum mass of calcium iodate that can dissolve in a 10L solution of calcium sulfide

I would do it this way. This is a Ksp problem with a common ion, Ca^2+ from CaS.

Ca(IO3)2 ==> Ca^2+ + 2IO3^-
...s..........s........2s

Ksp = (Ca^2+)(I03^-)^2
Substitute Ksp --look it up in a text or the web.
(Ca^2+) = s from Ca(IO3)2 + 0.00325M from CaS
(IO3^-) = 2s.
Substitute Ca^2+ and IO3^- and solve for s = solubility Ca(IO3)2 in M = mols/L.
Convert to g/L. g = mols/L x molar mass Ca(IO3)2.
Then multiply by 10 to find grams in 10 L.
You can get some of this information from Devron's post.

I thought about doing it that way, but the Ksp wasn't provided in the initial post, so I thought the way I initially did it would work, but I still wasn't that sure.

To determine the maximum mass of calcium iodate (Ca(IO3)2) that can be dissolved in a 10.0L jug of 3.25x10^-2 molar calcium sulphide (CaS) solution, we need to consider the solubility product constant (Ksp) for calcium iodate.

The solubility product constant is an equilibrium constant that represents the degree to which a sparingly soluble compound dissociates into its ions in a solution. In this case, the solubility product constant for calcium iodate can be expressed as:

Ksp = [Ca2+][IO3-]^2

To find the maximum mass of calcium iodate that can dissolve, we need to calculate the concentration of calcium ions (Ca2+) and iodate ions (IO3-) in the 3.25x10^-2 M calcium sulphide solution.

First, let's write the balanced chemical equation for the dissociation of calcium sulphide:

CaS(s) → Ca2+(aq) + S2-(aq)

From the balanced equation, we can see that one mole of calcium sulphide produces one mole of calcium ions (Ca2+). Therefore, the concentration of calcium ions in the 3.25x10^-2 M calcium sulphide solution is also 3.25x10^-2 M.

Next, we need to consider the stoichiometry of the calcium iodate dissolution reaction. One mole of calcium iodate produces one mole of calcium ions (Ca2+) and two moles of iodate ions (IO3-). Therefore, the concentration of iodate ions in the solution is twice the concentration of calcium ions.

Concentration of iodate ions (IO3-) = 2 * concentration of calcium ions (Ca2+)
= 2 * 3.25x10^-2 M
= 6.50x10^-2 M

Now that we know the concentrations of calcium ions (3.25x10^-2 M) and iodate ions (6.50x10^-2 M), we can calculate the maximum amount of calcium iodate that can dissolve using its solubility product constant (Ksp).

Ksp = [Ca2+][IO3-]^2

Let's assume that x moles of calcium iodate dissolve. Then the concentration of calcium ions ([Ca2+]) and iodate ions ([IO3-]) will be x moles/L.

Ksp = (3.25x10^-2 M)(6.50x10^-2 M)^2
= 1.40x10^-4

Since Ksp is a constant, we can use it to solve for x (the amount of calcium iodate that can dissolve):

Ksp = [Ca2+][IO3-]^2
1.40x10^-4 = (x moles/L)(x moles/L)^2
1.40x10^-4 = x^3 moles^3/L^3

Now we solve for x:

x^3 = 1.40x10^-4
x = (1.40x10^-4)^(1/3)
x ≈ 0.052 moles/L

Since the solution volume is 10.0L, we can find the maximum mass of calcium iodate that can dissolve by multiplying the moles per liter by the solution volume:

Maximum mass of calcium iodate = (0.052 moles/L)(10.0 L)
= 0.52 moles

To convert moles to grams, we need to know the molar mass of calcium iodate. The molar mass of calcium iodate (Ca(IO3)2) can be calculated by summing the atomic masses of its constituent elements from the periodic table.

Ca(IO3)2 molar mass = (1 * atomic mass of Ca) + (2 * atomic mass of I) + (6 * atomic mass of O)

Calculating the molar mass gives:
Ca(IO3)2 molar mass ≈ 293.89 g/mol

Now we can calculate the maximum mass of calcium iodate:

Maximum mass of calcium iodate = (0.52 moles)(293.89 g/mol)
≈ 152.85 grams

Therefore, the maximum mass of calcium iodate that can be dissolved in the 10.0L jug of 3.25x10^-2 M calcium sulphide solution is approximately 152.85 grams.