A current of 1.40 ampere is passed through 500 mL of 0.180 M solution of zinc sulphate for 200 seconds. What will be the molarity of `Zn^(2+)` ions after deposition of zinc?

Options

A. 0.154 M
B. 0.177 M
C. 2 M
D. 0.180 M

Explanation

To solve the problem step by step, we will follow the outlined procedure:

### Step 1: Gather Given Information
- Current (I) = 1.40 A
- Time (T) = 200 seconds
- Volume of ZnSO₄ solution = 500 mL = 0.5 L
- Molarity of ZnSO₄ solution = 0.180 M

### Step 2: Calculate the Total Charge (Q)
Using the formula:
\[ Q = I \times T \]
Substituting the values:
\[ Q = 1.40 \, \text{A} \times 200 \, \text{s} = 280 \, \text{C} \]

### Step 3: Convert Charge to Faraday
To convert the charge from coulombs to Faraday, we use the relation:
\[ 1 \, \text{Faraday} = 96500 \, \text{C} \]
Thus,
\[ \text{Charge in Faraday} = \frac{Q}{96500} = \frac{280}{96500} \approx 0.0029 \, \text{Faraday} \]

### Step 4: Determine Moles of Zinc Deposited
From the reaction:
\[ \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \]
We see that 2 Faraday is required to deposit 1 mole of Zn. Therefore, the moles of Zn deposited can be calculated as:
\[ \text{Moles of Zn} = \frac{0.0029 \, \text{Faraday}}{2} = 0.00145 \, \text{moles} \]

### Step 5: Calculate Initial Moles of Zn²⁺
Using the molarity formula:
\[ \text{Moles of Zn}^{2+} = \text{Molarity} \times \text{Volume} \]
Substituting the values:
\[ \text{Moles of Zn}^{2+} = 0.180 \, \text{mol/L} \times 0.5 \, \text{L} = 0.09 \, \text{moles} \]

### Step 6: Calculate Remaining Moles of Zn²⁺
After the deposition of Zn, the remaining moles of Zn²⁺ will be:
\[ \text{Remaining moles of Zn}^{2+} = \text{Initial moles} - \text{Moles deposited} \]
\[ \text{Remaining moles of Zn}^{2+} = 0.09 - 0.00145 = 0.08855 \, \text{moles} \]

### Step 7: Calculate Final Molarity of Zn²⁺
Finally, we calculate the molarity of Zn²⁺ after deposition:
\[ \text{Molarity of Zn}^{2+} = \frac{\text{Remaining moles}}{\text{Volume in L}} \]
\[ \text{Molarity of Zn}^{2+} = \frac{0.08855 \, \text{moles}}{0.5 \, \text{L}} = 0.1771 \, \text{M} \]

### Conclusion
The molarity of Zn²⁺ ions after the deposition of zinc is approximately **0.177 M**.

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