Question

A solution is prepared by mixing 100.0 mL of 0.438 M sodium chloride, 100.0 mL of magnesium chloride, and 250.0 mL of water. Assuming the volumes are additive, what are the concentration of all species in the resultant solution?

Answer 1

Mole of NaCl= molarity * volume

= 0.438 M *0.1000 L

= 0.0438 moles

0.0438 moles NaCl= 0.0438 moles Na+   + 0.0438 moles Cl-

Mole of MgCl2= molarity * volume

= 0.438 M *0.1000 L

= 0.0438 moles

0.0438 moles MgCl2= 0.0438 moles Mg^2+      + 0.0438*2 moles Cl-

= 0.0438 moles Mg^2+      + 0.0876 moles Cl-

Total volume = 100.0+100.0++250.0

= 450.0ml

= 0.450 L

Total Cl- = 0.0876 moles Cl-+ 0.0438 moles Cl-

= 0.1314 mole Cl-

Molarity = number of moles / volume in L

Cl-:0.1314 mole Cl- / 0.450L

= 0.292 M

Na+:

0.0438 moles Na+   / 0.450L

=0.97 M

Mg2+:

0.0438 moles Mg2+   / 0.450L

=0.97 M