Stuck on this one please help !

The solubility of barium sulphate, BaSO4, is 2.4 x 10-4 g/100 mL of solution. Calculate the Ksp of this solution.

Ag2CrO4/BaSO4/TlCl. All of the problems are done the same way.

I won't work these for you but I shall be happy to help you through them if you tell us exactly what is troubling you with this type problem.

To calculate the solubility product constant (Ksp) of a solution, we need to know the molar solubility of the compound. The molar solubility is the number of moles of the compound dissolved per liter of solution.

In this case, we are given the solubility of barium sulfate (BaSO4) in grams per 100 mL of solution. We can convert this into molar solubility by using the molar mass of BaSO4.

1. Calculate the molar mass of BaSO4:
- Barium (Ba) has a molar mass of 137.3 g/mol.
- Sulfur (S) has a molar mass of 32.1 g/mol.
- Oxygen (O) has a molar mass of 16.0 g/mol.
- BaSO4 consists of one barium atom, one sulfur atom, and four oxygen atoms.
- Therefore, the molar mass of BaSO4 is: (1 * 137.3) + (1 * 32.1) + (4 * 16.0) = 233.4 g/mol.

2. Convert the given solubility to molar solubility:
- The solubility is 2.4 x 10^(-4) g/100 mL.
- To convert this to molar solubility, divide by the molar mass and convert mL to L by dividing by 1000 (since 1 L = 1000 mL).
- Molar solubility = (2.4 x 10^(-4) g/100 mL) / (233.4 g/mol) / (1000 mL/L).

3. Calculate Ksp:
- The balanced chemical equation for the dissociation of BaSO4 is: BaSO4 ⇌ Ba^(2+) + SO4^(2-).
- The stoichiometry of the reaction tells us that the concentration of Ba^(2+) and SO4^(2-) are equal to each other.
- Therefore, the concentration of Ba^(2+) is equal to the molar solubility of BaSO4, and the concentration of SO4^(2-) is also equal to the molar solubility of BaSO4.
- Ksp = [Ba^(2+)] * [SO4^(2-)] = (molar solubility of BaSO4)^2.

Plug in the value for the molar solubility calculated in step 2 into the above equation to find the Ksp of the solution.