Question

A 40.0-mL solution contains 0.019 M barium chloride (BaCl2). What is the minimum concentration of sodium sulfate (Na2SO4) required in the solution to produce a barium sulfate (BaSO4) precipitate? The solubility product for barium sulfate is Ksp = 1.1 ✕ 10−10.

Answer 1

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 $5.790\times 10^{-9} M$.