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