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?

Question

Wyatt Adkins · Accepted Answer

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

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"#

Wyatt Adkins · Answer

undefined To calculate the concentrations of ions in the solution prepared from calcium chloride, we need to know the molar mass of calcium chloride and the stoichiometry of its dissociation in water. Calcium chloride dissociates into one calcium ion (Ca^2+) and two chloride ions (Cl^-) in solution. Then, we can use the given ppm concentration to find the molarity of calcium ion ([Ca^2+]) and chloride ion ([Cl^-]).