A 500 ml of aqueous NaCl solution is electrolyzed for 1 minute using a current of 8mA at `25^@C`. What is the pH of the final solution? (Use 1F = 96000 C)

Answer

To find the pH of the final solution after electrolyzing a 500 ml aqueous NaCl solution, we can follow these steps:

### Step 1: Calculate the total charge passed during electrolysis
The current (I) is given as 8 mA, which is equal to \(8 \times 10^{-3}\) A. The time (t) is 1 minute, which is equal to 60 seconds.

\[ \text{Total charge (Q)} = I \times t = 8 \times 10^{-3} \, \text{A} \times 60 \, \text{s} = 0.48 \, \text{C} \]

### Step 2: Calculate the number of moles of NaOH produced
Using Faraday's constant (1 F = 96000 C), we can find the number of moles of NaOH produced. Since 1 mole of NaCl produces 1 mole of NaOH and requires 96500 C:

\[ \text{Number of moles of NaOH} = \frac{Q}{F} = \frac{0.48 \, \text{C}}{96000 \, \text{C/mol}} = 0.000005 \, \text{mol} = 5 \times 10^{-6} \, \text{mol} \]

### Step 3: Calculate the molarity of NaOH in the solution
The volume of the solution is 500 ml, which is equal to \(0.5\) L. The molarity (M) is calculated as:

\[ \text{Molarity (M)} = \frac{\text{Number of moles}}{\text{Volume in L}} = \frac{5 \times 10^{-6} \, \text{mol}}{0.5 \, \text{L}} = 1 \times 10^{-5} \, \text{mol/L} \]

### Step 4: Calculate the pOH of the solution
Since NaOH is a strong base, we can find the pOH using the concentration of OH⁻ ions:

\[ \text{pOH} = -\log[OH^-] = -\log(1 \times 10^{-5}) = 5 \]

### Step 5: Calculate the pH of the solution
Using the relationship \(pH + pOH = 14\):

\[ pH = 14 - pOH = 14 - 5 = 9 \]

### Final Answer
The pH of the final solution after electrolysis is **9**.

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