When 2.00 L of 1.00 M of #AgNO_3# and 1.25 L of 1.00 M of #CaI_2# fully react, what is the molarity of #Ca^2+#?

1 Answer
anor277
Jun 24, 2017

For #2AgNO_3(aq) + CaI_2(aq) rarr 2AgI(s)darr + Ca(NO_3)_2(aq)#

Explanation:

The calcium ion, #Ca^(2+)#, which is present in solution as (probably) #[Ca(OH_2)_6]^(2+)#, remains in solution, and expresses its concentration fully because it does not precipitate....

#"Moles of"# #Ca^(2+)=1.25*Lxx1.00*mol*L^-1=1.25*mol#.

The volume of solution is #(2.00+1.25)*L=3.25*L#

And thus concentration of calcium ion is............

#(1.25*mol)/((2.00+1.25)*L)=0.385*mol*L^-1......#

Of course, the silver ion precipitates as a fine yellow powder of #AgI#.........

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