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?

1 Answer
May 31, 2017

Answer:

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?