What will be the [Cl-] when equal volumes of 0.10MNaCl and 0.20MAlCl3 are combined?
A. 0.35M
B. 0.15M
C. 0.30M
D. 0.70M
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.
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!
To find the concentration of [Cl-] when equal volumes of 0.10 M NaCl and 0.20 M AlCl3 are combined, we need to consider the stoichiometry of the reaction.
The balanced chemical equation for the reaction between NaCl and AlCl3 is:
3NaCl + AlCl3 → AlCl3 + 3NaCl
From the equation, we can see that for every 1 mole of AlCl3, 3 moles of Cl- ions are produced. Since we have equal volumes of NaCl and AlCl3, we can assume that we have an equal number of moles of both compounds.
Let's assume we have 1 L of solution each of 0.10 M NaCl and 0.20 M AlCl3. The number of moles of NaCl is:
moles of NaCl = concentration of NaCl × volume of NaCl solution
= 0.10 M × 1 L
= 0.10 moles
Similarly, the number of moles of AlCl3 is:
moles of AlCl3 = concentration of AlCl3 × volume of AlCl3 solution
= 0.20 M × 1 L
= 0.20 moles
Since every mole of AlCl3 produces 3 moles of Cl-, the total moles of Cl- ions in the solution are:
total moles of Cl- ions = 3 × moles of AlCl3
= 3 × 0.20 moles
= 0.60 moles
The total volume of the combined solution is 2 L (1 L of NaCl + 1 L of AlCl3). Therefore, the concentration of Cl- ions in the solution is:
[Cl-] = total moles of Cl- ions / total volume of solution
= 0.60 moles / 2 L
= 0.30 M
Therefore, the concentration of [Cl-] when equal volumes of 0.10 M NaCl and 0.20 M AlCl3 are combined is C. 0.30 M.
To determine the concentration of chloride ions ([Cl-]) when equal volumes of 0.10M NaCl and 0.20M AlCl3 are combined, we need to consider the stoichiometry of the reaction.
The balanced chemical equation for the reaction between NaCl and AlCl3 is:
2 NaCl + AlCl3 -> 2 NaCl + AlCl3
From the equation, we can see that one mole of NaCl produces two moles of Cl- ions.
Given that the initial concentration of NaCl is 0.10M and its volume is doubled, the final concentration of NaCl will still be 0.10M.
Next, let's consider the AlCl3 solution. The initial concentration of AlCl3 is 0.20M, and since its volume is also doubled, the final concentration of AlCl3 will be 0.10M.
Now, we need to calculate the total moles of Cl- ions present.
For NaCl: Moles of Cl- ions = (concentration of NaCl) x (volume of NaCl)
= 0.10M x (volume of NaCl)
For AlCl3: Moles of Cl- ions = (concentration of AlCl3) x (volume of AlCl3)
= 0.10M x (volume of AlCl3)
Since the volumes of NaCl and AlCl3 are equal, the two moles values will be the same.
Total moles of Cl- ions = Moles of Cl- ions from NaCl + Moles of Cl- ions from AlCl3
Now, let's substitute the values and solve for the concentration of Cl- ions:
Total moles of Cl- ions = (0.10M x volume) + (0.10M x volume)
= 0.20M x volume
To find the concentration, divide the moles by the total volume:
Concentration of Cl- ions = Total moles of Cl- ions / Total volume
= (0.20M x volume) / (2 x volume)
= 0.10M / 2
= 0.05M
Therefore, the concentration of Cl- ions when equal volumes of 0.10M NaCl and 0.20M AlCl3 are combined is 0.05M.
None of the given options (A, B, C, D) match this concentration value.