Etermine the molar concentration of magnesium ions in a solution formed by mixing 100.0 ml of 0.100 m mgcl, solution with 100.0 ml of 0.100 m mg3(po4)2 solution.

Molar concentration of magnesium ions in the final solution is 0.2 M.

Explanation:

c = Concentration of the solution

n =   moles of the compound

V = volume of the solution in L

1) Molarity of solution = 0.100 M

Volume of solution = 100.0 mL = 0.1 L

Moles of in 100 mL solution= n

1 mole of magnesium chloride gives 1 mol of magnesium ions and 2 moles of chloride ions.

Then 0.01 moles of magnesium chloride will give:

magnesium ions.

2) Molarity of solution = 0.100 M

Volume of solution = 100.0 mL = 0.1 L

Moles of in 100 mL solution= n'

1 mole of magnesium phosphate gives 3 mol of magnesium ions and 2 moles of phosphate ions.

Then 0.01 moles of magnesium phosphate will give:

magnesium ions.

After mixing both solutions:

Moles of magnesium ions = x + y = 0.01 mol + 0.03 mol = 0.04 mol

Total volume after mixing = 0.1 L + 0.1 L = 0.2 L

Molar concentration of magnesium ions in the final solution:

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