Question

16) What is the aluminum ion concentration in a solution that is 0.646 M in aluminum sulfate

Answer 1

Aluminum ion has a charge of 3+, Al³⁺, and sulfate is SO₄²⁻, so the compound aluminum sulfate has to have a number of aluminum and sulfate such that the final charge is zero, so the proportion on aluminum sulfate is:

`Al_2(SO_4)_3`

That way we have 6+ and 6-, so neutral compound.

This means that for 1 mol of aluminum sulfate, we have 2 moles of aluminum ion. The molar concentration is the number of moles of solute divided by the volume of solution, so it is directly proportional to the number of moles.

So, we can use a rule of three as follows:

aluminum ion --- aluminum sulfate

2 --- 1

x --- 0.646 M

So:

`\begin{gathered} (2)/(x)=(1)/(0.646M) \\ 2\cdot0.646M=x\cdot1 \\ 1.292M=x \\ x=1.292M \end{gathered}`

So, the concentration of aluminum ion in this solution is 1.292 M.