Question

Be sure to answer all parts. 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

Answer 1

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:

`Na_2S(aq)+CuSO4(aq)\rightarrow Na_2SO_4(aq)+CuS(s)`

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,

`Cu^(2+)(aq)+S^(2-)(aq)\rightarrow CuS(s)`

The concentration of
`Cu^(2+)` is,
`2.65* 10^(-4)M`

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:

`Na_2S(aq)+CuSO4(aq)\rightarrow Na_2SO_4(aq)+CuS(s)`

The ionic equation in separated aqueous solution will be,

`2Na^+(aq)+S^(2-)(aq)+Cu^(2+)(aq)+SO_4^(2-)(aq)\rightarrow CuS(s)+2Na^+(aq)+SO_4^(2-)(aq)`

In this equation,
`Na^+\text{ and }SO_4^(2-)` 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,

`Cu^(2+)(aq)+S^(2-)(aq)\rightarrow CuS(s)`

Now we have to calculate the mass of
`CuSO_4`

Molar mass of
`CuSO_4` = 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
`CuSO_4`

As, 0.0177 g of CuS produces from
`(159.5)/(95.5)* 0.0177=0.0296g`
`CuSO_4`

Now we have to calculate the concentration of
`CuSO_4`

`\text{Concentration}=\frac{\text{Mass of }CuSO_4}{\text{Molar mass of }CuSO_4* \text{Volume of solution (in L)}}`

Now put all the given values in this formula, we get:

`\text{Concentration}=(0.0296g)/(159.5g/mol* 0.700L)=2.65* 10^(-4)M`

Concentration of
`Cu^(2+)` =
`2.65* 10^(-4)M`

Therefore, the concentration of
`Cu^(2+)` is,
`2.65* 10^(-4)M`