Respuesta :
This is an incomplete question, here is a complete question.
The concentration of Cu²⁺ ions in the water (which also contains sulfate ions) discharged from a certain industrial plant is determined by adding excess sodium sulfide (Na₂S)solution to 0.700 L of the water.
The molecular equation is:
[tex]Na_2S(aq)+CuSO4(aq)\rightarrow Na_2SO_4(aq)+CuS(s)[/tex]
Write the net ionic equation and calculate the molar concentration of Cu²⁺ in the water sample if 0.0177 g of solid CuS is formed.
Answer :
The net ionic equation will be,
[tex]Cu^{2+}(aq)+S^{2-}(aq)\rightarrow CuS(s)[/tex]
The concentration of [tex]Cu^{2+}[/tex] is, [tex]2.65\times 10^{-4}M[/tex]
Explanation :
In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
The molecular equation is:
[tex]Na_2S(aq)+CuSO4(aq)\rightarrow Na_2SO_4(aq)+CuS(s)[/tex]
The ionic equation in separated aqueous solution will be,
[tex]2Na^+(aq)+S^{2-}(aq)+Cu^{2+}(aq)+SO_4^{2-}(aq)\rightarrow CuS(s)+2Na^+(aq)+SO_4^{2-}(aq)[/tex]
In this equation, [tex]Na^+\text{ and }SO_4^{2-}[/tex] are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
[tex]Cu^{2+}(aq)+S^{2-}(aq)\rightarrow CuS(s)[/tex]
Now we have to calculate the mass of [tex]CuSO_4[/tex]
Molar mass of [tex]CuSO_4[/tex] = 159.5 g/mol
Molar mass of CuS is = 95.5 g/mol
From the balanced chemical reaction we conclude that,
As, 95.5 g of CuS produces from 159.5 g [tex]CuSO_4[/tex]
As, 0.0177 g of CuS produces from [tex]\frac{159.5}{95.5}\times 0.0177=0.0296g[/tex] [tex]CuSO_4[/tex]
Now we have to calculate the concentration of [tex]CuSO_4[/tex]
[tex]\text{Concentration}=\frac{\text{Mass of }CuSO_4}{\text{Molar mass of }CuSO_4\times \text{Volume of solution (in L)}}[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Concentration}=\frac{0.0296g}{159.5g/mol\times 0.700L}=2.65\times 10^{-4}M[/tex]
Concentration of [tex]Cu^{2+}[/tex] = [tex]2.65\times 10^{-4}M[/tex]
Therefore, the concentration of [tex]Cu^{2+}[/tex] is, [tex]2.65\times 10^{-4}M[/tex]
