Question

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.

Answer 1

Answer:

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

Explanation:

$c=\frac{n}{V}$

c = Concentration of the solution

n =   moles of the compound

V = volume of the solution in L

1) Molarity of $MgCl_2$ solution = 0.100 M

Volume of $MgCl_2$ solution = 100.0 mL = 0.1 L

Moles of $MgCl_2$ in 100 mL solution= n

$0.100 M=\frac{n}{0.1L}$

$n=0.01 mol$

$MgCl_2(aq)\rightarrow Mg^{2+}(aq)+2Cl^-(aq)$

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:

$x=\frac{1}{1}\times 0.01 mol=0.01 mol$ magnesium ions.

2) Molarity of $Mg_3(PO_4)_2$ solution = 0.100 M

Volume of $Mg_3(PO_4)_2$ solution = 100.0 mL = 0.1 L

Moles of $Mg_3(PO_4)_2$ in 100 mL solution= n'

$0.100 M=\frac{n'}{0.1L}$

$n'=0.01 mol$

$Mg_3(PO_4)_2(aq)\rightarrow 3Mg^{2+}(aq)+2PO_{4}^-(aq)$

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:

$y=\frac{3}{1}\times 0.01 mol=0.03 mol$ 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:$[Mg^{2+}]$

$[Mg^{2+}]=\frac{0.04 mol}{0.2 L}=0.2 mol/L$