Question

What is chloride ion concentration when....?

.... a `20*mL` volume of `"aluminum sulfate"` at `0.20*mol*L^-1` concentration is mixed with a `20*mL` volume of `"barium chloride"` at `6.6*mol*L^-1` concentration?

Answer 1

Well we need a stoichiometric reaction........and it turns out that chloride is a spectator ion. We get `[Cl^-]=6.6*mol*L^-1`.

Explanation:

`Al_2(SO_4)_3(aq) + 3BaCl_2(aq) rarr 3BaSO_4(s)darr + 2AlCl_3(aq)`

Note that barium sulfate is insoluble in aqueous solution. This is an experimental fact that simply must be known.

`"Moles of aluminum sulfate"=20xx10^-3*Lxx0.20*mol*L^-1=4.0xx10^-3*mol.`

`"Moles of barium chloride"=20xx10^-3*Lxx6.6*mol*L^-1=0.132*mol.`

And thus `[Cl^-]=(2xx0.132*mol)/(20xx10^-3*L+20xx10^-3*L)`

`=??*mol*L^-1`.

Why did I double the numerator in the calculation?