Question

How do I calculate the number of chloride ions in a solution?

Which of the following contains the most chloride ions?
A) `10cm^3` of `3.3xx10^-2 mol dm^-3` aluminium chloride solution
B) `20cm^3` of `5.0xx10^-2 mol dm^-3`calcium chloride solution
C)`30cm^3` of `3.3xx10^-2 mol dm^-3`hydrochloric acid
D)`40cm^3` of `2.5xx10^-2 mol dm^-3`sodium chloride solution

Answer 1

Here's how you can do that.

Explanation:

Your strategy here will be to use the molarity and volume of the solutions to figure out how many moles of salt were dissolved to make the solution.

Once you know that, you can use the chemical formula of the salt to figure out how many moles of chloride anions, `"Cl"^(-)`, will be produced in each solution.

I'll show you how to solve points A) and B) and leave the rest to you as practice.

So, molarity is defined as the number of moles of solute present in `"1 L"` of solution. For the first solution, you have aluminium chloride, `"AlCl"_3`, as the solute and a volume of

`10 color(red)(cancel(color(black)("cm"^3))) * "1 dm"^3/(10^3color(red)(cancel(color(black)("cm"^3)))) = 10^(-2)color(white)(.)"dm"^3`

This means that the first solution will contain

`10^(-2) color(red)(cancel(color(black)("dm"^3))) * (3.3 * 10^(-2)color(white)(.)"moles AlCl"_3)/(1color(red)(cancel(color(black)("dm"^3)))) = 3.3 * 10^(-4)color(white)(.)"moles AlCl"_3`

Now, every mole of aluminium chloride that dissociates in aqueous solution produces

• one mole of aluminium cations, `1 xx "Al"^(3+)`
• three moles of chloride anions, `3 xx "Cl"^(-)`

This means that for every `1` mole of aluminium chloride dissolved to make this solution, you will end up with

`3.3 * 10^(-2) color(red)(cancel(color(black)("moles AlCl"_3))) * "3 moles Cl"^(-)/(1color(red)(cancel(color(black)("mole AlCl"_3)))) = 9.9 * 10^(-4)color(white)(.)"moles Cl"^(-)`

Now do the same for the calcium chloride solution. You will have

`20 color(red)(cancel(color(black)("cm"^3))) * "1 dm"^3/(10^3color(red)(cancel(color(black)("cm"^3)))) = 2.0 * 10^(-2)color(white)(.)"dm"^3`

This means that the solution contains

`2.0 * 10^(-2)color(red)(cancel(color(black)("dm"^3))) * (5.0 * 10^(-2)color(white)(.)"moles CaCl"_2)/(1color(red)(cancel(color(black)("dm"^3)))) = 1.0 * 10^(-3)color(white)(.)"moles CaCl"_2`

Now, every mole of calcium chloride that dissociates in aqueous solution will produce

• one mole of calcium actions, `1 xx "Ca"^(2+)`
• two moles of chloride anions, `2 xx "Cl"^(-)`

This means that the second solution will contain

`1.0 * 10^(-3)color(red)(cancel(color(black)("moles CaCl"_2))) * "2 moles Cl"^(-)/(1color(red)(cancel(color(black)("mole CaCl"_2)))) = 2.0 * 10^(-3)color(white)(.)"moles Cl"^(-)`

Therefore, you can say the solution B) will contain the higher concentration of chloride anions, since

`2.0 * 10^(-3)color(white)(.)"moles Cl"^(-) > 9.9 * 10^(-4)color(white)(.)"moles Cl"^(-)`

Use the same approach to calculate the number of moles of chloride anions present in solutions C) and D) and compare all four solutions to find the answer.