What is chloride ion concentration when....?

.... a 20*mL20⋅mL volume of "aluminum sulfate"aluminum sulfate at 0.20*mol*L^-10.20⋅mol⋅L−1 concentration is mixed with a 20*mL20⋅mL volume of "barium chloride"barium chloride at 6.6*mol*L^-16.6⋅mol⋅L−1 concentration?

1 Answer
anor277
May 31, 2017

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

Explanation:

Al_2(SO_4)_3(aq) + 3BaCl_2(aq) rarr 3BaSO_4(s)darr + 2AlCl_3(aq)Al2(SO4)3(aq)+3BaCl2(aq)→3BaSO4(s)⏐↓+2AlCl3(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 aluminum sulfate=20×10−3⋅L×0.20⋅mol⋅L−1=4.0×10−3⋅mol.

"Moles of barium chloride"=20xx10^-3*Lxx6.6*mol*L^-1=0.132*mol.Moles of barium chloride=20×10−3⋅L×6.6⋅mol⋅L−1=0.132⋅mol.

And thus [Cl^-]=(2xx0.132*mol)/(20xx10^-3*L+20xx10^-3*L)[Cl−]=2×0.132⋅mol20×10−3⋅L+20×10−3⋅L

=??*mol*L^-1=??⋅mol⋅L−1.

Why did I double the numerator in the calculation?

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