Respuesta :
Answer:
The minimum concentration of sodium sulfate required to producer precipitation is [tex]5.790\times 10^{-9} M[/tex].
Explanation:
[tex]BaCl_2\rightarrwo Ba^{2+}+2Cl^-[/tex]
Concentration of barium chloride= [tex][BaCl_2]=0.019 M[/tex]
Concentration of barium ions = [tex][Ba^{2+}][/tex]
1 mol of barium chloride gives 1 mol of barium ions.
[tex][Ba^{2+}]=[BaCl_2]=0.019 M[/tex]
The solubility product for barium sulfate= [tex] K_{sp} = 1.1\times 10^{-10}[/tex]
[tex]BaSO_4\rightleftharpoons Ba^{2+}+SO_4^{2-}[/tex]
0.019 M S
The expression of solubility product for barium sulfate:
[tex]K_{sp}=[Ba^{2+}]\times S[/tex]
[tex]K_{sp}=0.019 M\times S[/tex]
[tex]S=\frac{1.1\times 10^{-10}}{0.019 M}=5.790\times 10^{-9} M[/tex]
The minimum concentration of sulfate ions =[tex][SO_4^{2-}]=5.790\times 10^{-9} M[/tex]
[tex]Na_2SO_4(aq)\rightarrow 2Na^+(aq)+SO_4^{2-}(aq)[/tex]
1 mole of sulfate ions are proceed form 1 mole of sodium sulfate solution.
Then [tex]5.790\times 10^{-9} M[/tex] sulfate ions will be obtained from:
[tex]=1\times 5.790\times 10^{-9} M=5.790\times 10^{-9} M[/tex] of sodium sulfate
The minimum concentration of sodium sulfate required to producer precipitation is [tex]5.790\times 10^{-9} M[/tex].
