Zn rod is placed in 100mL of 1M `CuSO_(4)` solution so that molarity of `Cu^(2+)` changes to 0.7M. The molarity of `SO_(4)^(-)` at this stage will be

Options

A. 0.8M
B. 1M
C. 0.7M
D. 1.8M

Answer

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 displacement reaction occurs. The reaction can be represented as:
\[ \text{Zn (s)} + \text{Cu}^{2+} \text{(aq)} \rightarrow \text{Zn}^{2+} \text{(aq)} + \text{Cu (s)} \]

### Step 2: Initial Conditions
Initially, we have:
- Volume of CuSO₄ solution = 100 mL = 0.1 L
- Molarity of CuSO₄ = 1 M

From the molarity of CuSO₄, we can determine the initial concentration of Cu²⁺ ions:
- Molarity of Cu²⁺ = 1 M

### Step 3: Calculate Initial Moles of Cu²⁺
Using the formula:
\[ \text{Moles} = \text{Molarity} \times \text{Volume} \]
\[ \text{Moles of Cu}^{2+} = 1 \, \text{M} \times 0.1 \, \text{L} = 0.1 \, \text{moles} \]

### Step 4: Final Conditions After Reaction
After the reaction occurs, the molarity of Cu²⁺ changes to 0.7 M. We need to find out how many moles of Cu²⁺ remain:
- Final Molarity of Cu²⁺ = 0.7 M

Using the same formula for moles:
\[ \text{Moles of Cu}^{2+} = 0.7 \, \text{M} \times 0.1 \, \text{L} = 0.07 \, \text{moles} \]

### Step 5: Calculate Moles of Cu²⁺ Reacted
The moles of Cu²⁺ that reacted can be calculated as:
\[ \text{Moles reacted} = \text{Initial moles} - \text{Final moles} \]
\[ \text{Moles reacted} = 0.1 \, \text{moles} - 0.07 \, \text{moles} = 0.03 \, \text{moles} \]

### Step 6: Determine Molarity of SO₄²⁻
The initial concentration of sulfate ions (SO₄²⁻) in the solution is equal to the initial concentration of CuSO₄, which is 1 M. Since sulfate ions do not participate in the reaction, their concentration remains unchanged.

Thus, the molarity of sulfate ions 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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