Question

`35*mL` of `0.200*mol*L^-1` sodium chloride were mixed with `65*mL` of `0.100*mol*L^-1` calcium chloride. What is the concentration of chloride ion in the resultant solution?

Answer 1

`[Cl^-]=0.200*mol*L^-1`

Explanation:

By definition `"concentration, molarity"` is given by the quotient...

`"Molarity"="Moles of solute"/"Volume of solution"`

And thus.......

`"moles of chloride anion"=35.0*mLxx10^-3*L*mL^-1xx0.200*mol*L^-1+2xx65.0*mLxx10^-3*L*mL^-1xx0.100*mol*L^-1`

And please note that I had to account for the makeup of the calcium salt in that each equiv calcium chloride contains TWO equivs chloride....

And reasonably we assume the volumes to be additive, so we use the sum `35.0*mL+65.0*mL` as the denominator....

And so...

`=(0.035*L*xx0.200*(mol)/L+2xx0.065*Lxx0.100*(mol)/L)/(0.100*L)`

`=[Cl^-]=0.200*mol*L^-1`