Calculate the solubility of CuSO4 5H2O at 0 C in grams of solute per 100 g water.

To calculate the solubility of CuSO4 5H2O at 0°C in grams of solute per 100 grams of water, we need to consult a solubility table or use the solubility product constant (Ksp) value for CuSO4 5H2O.

One way to find the solubility of CuSO4 5H2O at 0°C is by using a solubility table. Here's an example of a solubility table showing the solubility of different compounds at various temperatures:

Compound | Solubility at 0°C (g/100g H2O)
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CuSO4 5H2O | 31.6

According to this table, the solubility of CuSO4 5H2O at 0°C is 31.6 grams of solute per 100 grams of water.

If you don't have access to a solubility table, another way to find the solubility is by using the Ksp value for CuSO4 5H2O. The Ksp value represents the equilibrium constant for the dissolution of a sparingly soluble salt in water.

The Ksp expression for CuSO4 5H2O can be written as:

CuSO4 5H2O --> Cu2+ + SO4^2-

The Ksp value indicates how much Cu2+ and SO4^2- ions can be dissolved in water at equilibrium.

The Ksp value for CuSO4 5H2O at 0°C is approximately 3.6 x 10^-18.

To find the solubility from the Ksp value, we need to determine the concentration of Cu2+ ions or SO4^2- ions in moles per liter (M). Since we want the solubility in grams per 100 grams of water, we need to convert the final answer.

Let's assume 'x' represents the solubility of CuSO4 5H2O in moles per liter.

Since CuSO4 5H2O dissociates into one Cu2+ ion and one SO4^2- ion, the concentrations of Cu2+ and SO4^2- ions will be 'x' M each.

Therefore, using the Ksp expression, we can write:

Ksp = [Cu2+][SO4^2-]
3.6 x 10^-18 = x * x

Solving this equation, we find that x is approximately 6.0 x 10^-9 M.

To convert this to grams per 100 grams of water, we multiply by the molar mass of CuSO4 5H2O (249.69 g/mol):

(6.0 x 10^-9 M) * (249.69 g/mol) = 1.5 x 10^-6 g/100 g

Therefore, the solubility of CuSO4 5H2O at 0°C is approximately 1.5 x 10^-6 grams of solute per 100 grams of water.

did anyone solve this?

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