To solve the problem, we need to analyze the situation step by step:
### Step 1: Understand the Reaction
When a zinc rod is placed in a copper sulfate solution, a redox reaction occurs. Zinc (Zn) gets oxidized to zinc ions (Zn²⁺), and copper ions (Cu²⁺) in the solution get reduced to solid copper (Cu). The reaction can be represented as:
\[ \text{Zn (s)} + \text{Cu}^{2+} (aq) \rightarrow \text{Zn}^{2+} (aq) + \text{Cu (s)} \]
### Step 2: Initial Concentration of Cu²⁺
The initial concentration of Cu²⁺ in the solution is given as 1 M. Since the volume of the solution is 100 mL (0.1 L), the initial number of moles of Cu²⁺ can be calculated as:
\[ \text{Moles of Cu}^{2+} = \text{Molarity} \times \text{Volume} = 1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.1 \, \text{mol} \]
### Step 3: Change in Concentration of Cu²⁺
After the reaction, the concentration of Cu²⁺ changes to 0.7 M. This means that some of the Cu²⁺ ions have been reduced to solid copper. We can calculate the moles of Cu²⁺ remaining:
\[ \text{Moles of Cu}^{2+} \text{ after reaction} = 0.7 \, \text{mol/L} \times 0.1 \, \text{L} = 0.07 \, \text{mol} \]
### Step 4: Calculate Moles of Cu²⁺ Consumed
The moles of Cu²⁺ that have reacted can be found by subtracting the remaining moles from the initial moles:
\[ \text{Moles of Cu}^{2+} \text{ consumed} = 0.1 \, \text{mol} - 0.07 \, \text{mol} = 0.03 \, \text{mol} \]
### Step 5: Moles of Zn²⁺ Produced
According to the stoichiometry of the reaction, 1 mole of Zn produces 1 mole of Zn²⁺. Therefore, the moles of Zn²⁺ produced will also be 0.03 mol.
### Step 6: Concentration of SO₄²⁻
The sulfate ions (SO₄²⁻) are not involved in the redox reaction and will remain unchanged in concentration. Initially, the concentration of SO₄²⁻ in the solution is the same as that of CuSO₄, which is 1 M. Since the volume of the solution has not changed, the concentration of SO₄²⁻ remains 1 M.
### Conclusion
Thus, the molarity of SO₄²⁻ at this stage is:
\[ \text{Molarity of SO}_4^{2-} = 1 \, \text{M} \]
### Final Answer
The molarity of SO₄²⁻ at this stage will be **1 M**.
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