A current of 0.193 amp is passed through 100 ml of 0.2M NaCl for an hour. Calculate pH of solution after electrolysis if current efficiency is 90%. Assume no volume change.

Answer

To solve the problem, we need to follow these steps:

### Step 1: Calculate Total Charge Transferred
The total charge (Q) transferred during electrolysis can be calculated using the formula:
\[ Q = I \times t \times \text{Efficiency} \]
Where:
- \( I = 0.193 \, \text{A} \) (current)
- \( t = 1 \, \text{hour} = 3600 \, \text{seconds} \)
- Efficiency = 90% = 0.90

Calculating the total charge:
\[ Q = 0.193 \, \text{A} \times 3600 \, \text{s} \times 0.90 = 626.88 \, \text{C} \]

### Step 2: Calculate Moles of NaCl Electrolyzed
Using Faraday's constant (\( F = 96500 \, \text{C/mol} \)), we can find the number of moles of NaCl electrolyzed:
\[ \text{Moles of NaCl} = \frac{Q}{F} = \frac{626.88 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.0065 \, \text{mol} \]

### Step 3: Calculate Molarity of NaOH Produced
Since the electrolysis of NaCl produces NaOH, the moles of NaOH produced will be equal to the moles of NaCl electrolyzed. The volume of the solution is 100 mL, which is 0.1 L. Therefore, the molarity (M) of NaOH is:
\[ \text{Molarity of NaOH} = \frac{\text{Moles of NaOH}}{\text{Volume in L}} = \frac{0.0065 \, \text{mol}}{0.1 \, \text{L}} = 0.065 \, \text{M} \]

### Step 4: Calculate pOH of the Solution
The pOH of the solution can be calculated using the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Where \([\text{OH}^-] = 0.065 \, \text{M}\):
\[ \text{pOH} = -\log(0.065) \approx 1.187 \]

### Step 5: Calculate pH of the Solution
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
We can find the pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 1.187 \approx 12.813 \]

### Final Answer
The pH of the solution after electrolysis is approximately **12.81**.

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