A precipitate is formed by adding 20.0 ml of 0.200M of sodium phosphate to 30.0 ml of 0.50M Copper sulfate.
b.) what mass of solid will be formed?
c.) Calculate the concentration of each ion in the solution after the precipitation is complete.
See the other post.
Balance the equation.
Determine the moles of precipate, then the mass.
To determine the mass of the solid formed, we need to balance the chemical equation first. The balanced equation for the reaction between sodium phosphate and copper sulfate is as follows:
3Na3PO4 + 2CuSO4 -> Cu3(PO4)2 + 3Na2SO4
The coefficients in the balanced equation indicate the stoichiometry of the reaction. From this equation, we can see that 2 moles of copper sulfate react with 3 moles of sodium phosphate to produce 1 mole of copper(II) phosphate.
Given that the volume of sodium phosphate solution is 20.0 ml, which is equal to 0.020 liters, and the concentration is 0.200 M (moles per liter), we can calculate the number of moles of sodium phosphate used:
Moles of sodium phosphate = Volume * Concentration
Moles of sodium phosphate = 0.020 L * 0.200 M
Moles of sodium phosphate = 0.004 moles
Using the stoichiometry of the balanced equation, we can determine the number of moles of copper(II) phosphate formed. Since the mole ratio between sodium phosphate and copper(II) phosphate is 3:1, we can determine the moles of copper(II) phosphate as follows:
Moles of copper(II) phosphate = (0.004 moles Na3PO4) / (3 moles Na3PO4/1 mole Cu3(PO4)2)
Moles of copper(II) phosphate = 0.00133 moles
Next, we need to calculate the molar mass of copper(II) phosphate, which is Cu3(PO4)2. Adding up the atomic masses, we find:
Cu: 63.55 g/mol
P: 30.97 g/mol
O: 16.00 g/mol
Molar mass of Cu3(PO4)2 = (3 * 63.55 g/mol) + (2 * (30.97 g/mol + 4 * 16.00 g/mol))
Molar mass of Cu3(PO4)2 = 380.61 g/mol
Now, we can calculate the mass of copper(II) phosphate formed:
Mass of copper(II) phosphate = Moles * Molar mass
Mass of copper(II) phosphate = 0.00133 moles * 380.61 g/mol
Mass of copper(II) phosphate = 0.508 g
Therefore, the mass of the solid formed (copper(II) phosphate) is approximately 0.508 grams.
For part c), to calculate the concentration of each ion in the solution after precipitation is complete, we need to consider the stoichiometry of the balanced equation.
From the equation, we can see that for every 2 moles of copper sulfate, we get 1 mole of copper(II) phosphate. So, the final concentration of copper(II) sulfate will be halved, and the concentration of sodium phosphate will remain the same.
The concentration of copper(II) sulfate after precipitation is complete is 0.50 M / 2 = 0.25 M.
The concentration of sodium phosphate after precipitation is complete is 0.200 M.
Therefore, the concentration of the copper(II) ion in the solution is 0.25 M, and the concentration of the phosphate ion is 0.200 M.