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swat32
4 years ago
5

A 40.0-mL solution contains 0.019 M barium chloride (BaCl2). What is the minimum concentration of sodium sulfate (Na2SO4) requir

ed in the solution to produce a barium sulfate (BaSO4) precipitate? The solubility product for barium sulfate is Ksp = 1.1 ✕ 10−10.
Chemistry
1 answer:
Blababa [14]4 years ago
3 0

Answer:

The minimum concentration of sodium sulfate required to producer precipitation is 5.790\times 10^{-9} M.

Explanation:

BaCl_2\rightarrwo Ba^{2+}+2Cl^-

Concentration of barium chloride= [BaCl_2]=0.019 M

Concentration of barium ions = [Ba^{2+}]

1 mol of barium chloride gives 1 mol of barium ions.

[Ba^{2+}]=[BaCl_2]=0.019 M

The solubility product for barium sulfate= K_{sp} = 1.1\times 10^{-10}

BaSO_4\rightleftharpoons Ba^{2+}+SO_4^{2-}

              0.019 M              S

The expression of solubility product for barium sulfate:

K_{sp}=[Ba^{2+}]\times S

K_{sp}=0.019 M\times S

S=\frac{1.1\times 10^{-10}}{0.019 M}=5.790\times 10^{-9} M

The minimum concentration of sulfate ions  =[SO_4^{2-}]=5.790\times 10^{-9} M

Na_2SO_4(aq)\rightarrow 2Na^+(aq)+SO_4^{2-}(aq)

1 mole of sulfate ions are proceed form 1 mole of sodium sulfate solution.

Then 5.790\times 10^{-9} M sulfate ions will be obtained from:

=1\times 5.790\times 10^{-9} M=5.790\times 10^{-9} M of sodium sulfate

The minimum concentration of sodium sulfate required to producer precipitation is 5.790\times 10^{-9} M.

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