Concentrations Problem?

Question

Assuming the volumes are additive, what is the concentraion of Cl ion  in a solution obtained by mixing 250 ml of 0.600 M KCl and 605 mL of 0.385 MgCl2?

Lexi R · Accepted Answer

Work out the number of moles of Cl- ions in each solution:

250 ml 0.600 M KCl
Molarity = moles / litres
therefore moles = Molarity x Litres
moles KCl = 0.600 M x 0.250 L
= 0.150 moles of KCl

A solution of KCl dissociates as follows

KCl --------> K+(aq) + Cl-(aq)

So 1 moles of KCl yields 1 mole of Cl- ions
therefore there are 0.150 moles of Cl- ions in the solution

605 ml of 0.385 M MgCl2

moles = M x L
= 0.385 M x 0.605 L
= 0.2329 moles of MgCl2 in the solution of MgCL2

MgCl2 dissociates as follows

MgCl2 --------> Mg2+(aq)  + 2Cl-(aq)

thus 1 moles of MgCl2 gives 2 moles of Cl-, therefore Cl- ions in the MgCl2 solution = 2 x 0.2329 mole
= 0.4658 moles of Cl- ions.

Now, total number of Cl- ions in both solutions = 0.4658 mol + 0.150 mol
= 0.6158 moles of Cl-

total volume of solution = 250 ml + 608 ml = 858 ml = 0.858 L

Now molarity of Cl- = moles Cl- / litres total solution
= 0.6158 mol / 0.858 L
= 0.7177 M Cl-
= 0.718M (3 sig fig)

Concentrations Problem?

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