Question

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

Answer 1

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)` `2.08g` `color(white)(aaaa)` `2.84g`

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

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