2.08 grams of anhydrous barium chloride was dissolved in water to make 50.00mL of solution and then added to 50.00mL of an aqueous solution containing 2.84 g of anhydrous sodium sulfate. A white precipitate of barium sulfate formed. How do you...?

a) Calculate the mass (in grams) of the precipitate that was formed
b) Calculate the concentration (in mol/L) of sulfate ions in the final solution

1 Answer
Narad T.
Jul 22, 2018

Please see the explanation below

Explanation:

The reaction is

color(white)(aaaaaaa)BaCl_2(aq)+Na_2SO_4(aq)rarrBaSO_4(s)+2NaCl(aq)

color(white)(aa)"Initial(g)"color(white)(aa)2.08gcolor(white)(aaaa)2.84g

color(white)(a)"Initial(mol)"color(white)(aa)0.01color(white)(aaaa)0.02

color(white)(aaa)"final(mol)"color(white)(aaaa)0color(white)(aaaa)0.01color(white)(aaaaaaaaaa)0.01color(white)(aaaaaaaa)0.02

The mass of precipitate (for a complete reaction) is

=0.01*233.4=2.33g

The sulfate ions left in solution is =0.01 mol

The concentration in sulfate ions is

c=0.01/(50+50)=0.01(mol)/(100 mL)

=0.1(mol)/(L)

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