Answer :
The molarity of magnesium ion that remains when 69.5% has precipitated is 0.021 M.
What is the balanced equation for the precipitation reaction of magnesium phosphate?
The balanced equation is Mg2+(aq) + PO43-(aq) → Mg3(PO4)2(s).
How did we calculate the amount of magnesium ions that react with the added phosphate?
We used the balanced equation to determine that 1 mole of magnesium ion reacts with 2 moles of phosphate ion, and then multiplied the amount of added phosphate by the stoichiometric ratio to calculate the amount of magnesium ions that will react.
To solve this problem, we first need to write the balanced equation for the precipitation reaction of magnesium phosphate:
Mg2+(aq) + PO43-(aq) → Mg3(PO4)2(s)
We know that 69.5% of the magnesium ions have precipitated, which means that 30.5% remain in solution. Therefore, the concentration of magnesium ions in solution is:
0.305 x 1.291 M = 0.394 M
This is the molarity of magnesium ions before the phosphate has been added. However, we also need to take into account the fact that the phosphate will react with some of the magnesium ions, forming magnesium phosphate precipitate. We don't know how much of the magnesium ions will react, so we can assume that the reaction goes to completion and calculate the molarity of magnesium ions that remain in solution after the reaction has occurred.
To do this, we need to determine the amount of phosphate that has been added to the solution. Since we don't know the volume of the solution, we can assume that we have added 1 liter of phosphate solution. The amount of phosphate added is:
0.746 M x 1 L = 0.746 mol
According to the balanced equation, 1 mole of magnesium ion reacts with 2 moles of phosphate ion. Therefore, the amount of magnesium ions that will react with the added phosphate is:
0.746 mol PO43- x (1 mol Mg2+/2 mol PO43-) = 0.373 mol Mg2+
This means that 0.373 moles of magnesium ions will react, leaving 0.394 - 0.373 = 0.021 moles of magnesium ions remaining in solution. To calculate the molarity of these remaining magnesium ions, we divide the amount by the total volume of the solution:
Molarity of magnesium ions = 0.021 mol / 1 L = 0.021 M
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