What will be the [Cl-] when equal volumes of 0.10MNaCl and 0.20MAlCl3 are combined?

Question

A. 0.35M
B. 0.15M
C. 0.30M
D. 0.70M

Answer 1

Two ways to do this.
The easy way is to assume any convenient volume, say 100 mL.
mols NaCl = M x L = ??
moles AlCl3 = M x L x 3 + ??
(the x3 comes because there are 3 moles Cl for every 1 mol AlCl3).
Then the definition of M = moles/L. Total moles/total L = ??

The second way is a dilution technique.
Cl form NaCl is 0.1 initially. If that is diluted by an equal volume of the other reagent, this one will be cut in half so it is 0.1/2 = ??

The second one is AlCl3. It is 0.2 M initially, it will be 3x that for Cl^- and that will be 0.6 M in Cl^-, then chop it in half for the dilution with an equal volume of the other reagent. That makes it 0.3 M.
Now add 0.3 M + 0.05 M = ??
You should get the same answer either way.

Answer 2

Your question: What will be the [Cl-] when equal volumes of 0.10M NaCl and 0.20M AlCl3 are combined?

This question is quite simple to solve once you notice that NaCl and AlCl3 are combined in EQUAL VOLUMES. We can apply what we understand about the mol ratio to help us solve this question.

I would begin by doing dissociation reactions for NaCl and AlCl3.

NaCl —> Na+ + Cl- (1:1 mole ratio)
AlCl3 —> Al+ + 3Cl- (1:3 mole ratio)

NaCl: (0.10 mol/litre NaCl)(1 mol Cl-/1 mole NaCl) = 0.10 M
AlCl3: (0.20 mol/litre AlCl3)(3 mol Cl-/1 mole AlCl3) = 0.60 M

(0.10M + 0.60M)/2 = 0.35M Cl-

Note: Once I add my concentrations I divide the sum by 2 because NaCl and AlCl3 are combined in EQUAL volumes.

For all Canadian Chemistry 12 students, this question was #19 from the August 2005 Provincial! I hope this helped!