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 calculate the amount of zinc deposited during the electrolysis and then determine the molarity of `Zn^(2+)` ions remaining in the solution.

### Step 1: Calculate the total charge (Q) passed through the solution.
The formula to calculate charge is:
\[ Q = I \times t \]
Where:
- \( I \) = current in amperes (1.40 A)
- \( t \) = time in seconds (200 s)

Substituting the values:
\[ Q = 1.40 \, \text{A} \times 200 \, \text{s} = 280 \, \text{C} \]

### Step 2: Determine the moles of zinc deposited.
From Faraday's laws of electrolysis, we know that:
- 1 mole of zinc is deposited by 2 Faradays of charge.
- 1 Faraday = 96500 C.

Thus, the number of moles of zinc deposited can be calculated using:
\[ \text{Moles of Zn} = \frac{Q}{nF} \]
Where:
- \( n \) = number of electrons transferred (for zinc, \( n = 2 \))
- \( F \) = Faraday's constant (96500 C/mol)

Substituting the values:
\[ \text{Moles of Zn} = \frac{280 \, \text{C}}{2 \times 96500 \, \text{C/mol}} \]
\[ \text{Moles of Zn} = \frac{280}{193000} \approx 0.00145 \, \text{mol} \]

### Step 3: Calculate the initial moles of `Zn^(2+)` in the solution.
The initial concentration of the zinc sulfate solution is 0.180 M, and the volume is 500 mL (0.500 L).

Using the formula:
\[ \text{Moles of Zn}^{2+} = \text{Molarity} \times \text{Volume (L)} \]
\[ \text{Moles of Zn}^{2+} = 0.180 \, \text{mol/L} \times 0.500 \, \text{L} = 0.090 \, \text{mol} \]

### Step 4: Calculate the remaining moles of `Zn^(2+)` after deposition.
The moles of `Zn^(2+)` remaining after the deposition of zinc can be calculated as:
\[ \text{Remaining moles of Zn}^{2+} = \text{Initial moles} - \text{Moles deposited} \]
\[ \text{Remaining moles of Zn}^{2+} = 0.090 \, \text{mol} - 0.00145 \, \text{mol} = 0.08855 \, \text{mol} \]

### Step 5: Calculate the new molarity of `Zn^(2+)` ions.
To find the new molarity, we use the remaining moles and the volume of the solution:
\[ \text{New Molarity} = \frac{\text{Remaining moles}}{\text{Volume (L)}} \]
\[ \text{New Molarity} = \frac{0.08855 \, \text{mol}}{0.500 \, \text{L}} = 0.1771 \, \text{M} \]

### Final Answer
The molarity of `Zn^(2+)` ions after the deposition of zinc is approximately **0.177 M**.

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