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?

1 Answer
anor277
Sep 1, 2017

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"#

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