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 0.800L of the water.The molecular equation is:Na2S (aq) + CuSO4 (aq) ----> Na2SO4 (aq) + CuS (s)Write the net ionic equation and calculate the molar concentration of Cu2+ in the water sample if 0.0177g of solid CuS is formed.

Answer 1

Explanation:

(i)   As the given reaction equation is as follows.

$Na_{2}S(aq) + CuSO_{4}(aq) \rightarrow Na_{2}SO_{4}(aq) + CuS(s)$

Therefore, ionic equation for this reaction is as follows.

 $2Na^{+}(aq) + S^{2-}(aq) + Cu^{2+}(aq) + SO^{2-}_{4}(aq) \rightarrow 2Na^{+}(aq) + SO^{2-}_{4}(aq) + CuS(s)$

On cancelling the spectator ions, the net ionic equation will be as follows.

    $S^{2-}(aq) + Cu^{2+}(aq) + SO^{2-}_{4}(aq) \rightarrow SO^{2-}_{4}(aq) + CuS(s)$

(ii)  Now, we will calculate the moles of CuS in 0.0177 g as follows.

      Moles of CuS = $\frac{mass}{\text{molar mass}}$

                              = $0.0177 g CuS \times \frac{1 mol CuS}{95.611 g CuS}$

                             = $1.85 \times 10^{-4}$ mol

As dissociation of CuS will be as follows.

          $CuS \rightarrow Cu^{2+} + S^{2-}$

Now, we will calculate the moles of $Cu^{2+}$ as follows.

          $1.85 \times 10^{-4} mol CuS \times \frac{1 mol Cu^{2+}}{1 mol CuS}$

            = $1.85 \times 10^{-4} mol Cu^{2+}$

Hence, molar concentration of $Cu^{2+}$ in the water sample is as follows.

     $\frac{1.85 \times 10^{-4} mol}{0.800 L}$

        = $2.312 \times 10^{-4}$ M

Thus, we can conclude that molar concentration of given $Cu^{2+}$ ions is $2.312 \times 10^{-4}$ M.

Answer 2

A water sample, whose concentration in Cu²⁺ is 2.31 × 10⁻⁴ M, produces 0.0177 g of CuS, according to the following net ionic equation:

S²⁻(aq) + Cu²⁺(aq) ⇒ CuS(s)

What is the net ionic equation?

The net ionic equation is a chemical equation in which only those ions undergoing chemical changes during the course of the reaction are represented.

• Step 1: Write the net ionic equation

S²⁻(aq) + Cu²⁺(aq) ⇒ CuS(s)

• Step 2: Calculate the moles of Cu²⁺ that formed 0.0177 g of CuS

We will consider the following relationships:

• The molar mass of CuS is 95.61 g/mol.
• The molar ratio of CuS to Cu²⁺ is 1:1.

0.0177 g CuS × (1 mol CuS/95.61 g CuS) × (1 mol Cu²⁺/1 mol CuS) =

= 1.85 × 10⁻⁴ mol Cu²⁺

• Step 3: Calculate the molar concentration of Cu²⁺

1.85 × 10⁻⁴ mol of Cu²⁺ are in 0.800 L of water (solution).

[Cu²⁺] = 1.85 × 10⁻⁴ mol/0.800 L = 2.31 × 10⁻⁴ M

A water sample, whose concentration in Cu²⁺ is 2.31 × 10⁻⁴ M, produces 0.0177 g of CuS, according to the following net ionic equation:

S²⁻(aq) + Cu²⁺(aq) ⇒ CuS(s)

Learn more about net ionic equations here: https://brainly.com/question/25652890