Question

If a solution is `1.1*"ppm"` with respect to calcium ion, what are `[Ca^(2+)]`, and `[Cl^-]` if the solution is prepared from calcium chloride?

Answer 1

You gots `CaCl_2`.........and the concentration of `Cl^-` is `2*"ppm"`.

Explanation:

And in aqueous solution, calcium chloride speciates to give.....

`CaCl_2(s) stackrel(H_2O)rarrCa^(2+) + 2Cl^(-)`

Now if it is `1.1*"ppm"` with respect to `Ca^(2+)`, there are `1.1xx1*mg*L^-1` of solution WITH RESPECT to the calcium ion (and at these concentrations we really don't have to worry about density change).

And so....

`[Ca^(2+)]=(1.1xx10^-3*g)/(40.08*g*mol^-1)=2.745xx10^-5*mol*L^-1`.

And necessarily (why), `[Cl^-]=5.489xx10^-5*mol*L^-1`. Why so.....?

This corresponds to a mass concentration of `5.489xx10^-5*mol*L^-1xx35.45*g*mol^-1=1.945xx10^-3*g*L^-1`

`-=2*"ppm"`