One litre of `0.5 M KCl` is electrolysed by passing `9650` coulombs of electricity. The `pH of resulting solution is

Options

A. `1.0`
B. `2.0`
C. `7.0`
D. `13.0`

Explanation

To solve the problem step by step, we will analyze the electrolysis of a 0.5 M KCl solution and determine the pH of the resulting solution after passing 9650 coulombs of electricity.

### Step 1: Determine the number of moles of electrons
Using Faraday's law of electrolysis, we know that 1 mole of electrons corresponds to 96500 coulombs of charge. Therefore, we can calculate the number of moles of electrons (n) that correspond to 9650 coulombs:

\[ n = \frac{9650 \text{ C}}{96500 \text{ C/mol}} = 0.1 \text{ mol} \]

**Hint:** Remember that 1 mole of electrons corresponds to 96500 coulombs. Use this relationship to convert coulombs to moles.

### Step 2: Determine the moles of OH⁻ produced
From the electrolysis reactions, we know that for every mole of electrons, 1 mole of OH⁻ is produced at the cathode. Therefore, the moles of OH⁻ produced will also be 0.1 mol.

**Hint:** The stoichiometry of the reaction at the cathode indicates that 1 mole of electrons produces 1 mole of OH⁻.

### Step 3: Calculate the concentration of OH⁻
Since the volume of the solution is 1 liter, the concentration of OH⁻ ions in the solution will be:

\[ [\text{OH}^-] = \frac{0.1 \text{ mol}}{1 \text{ L}} = 0.1 \text{ M} \]

**Hint:** Concentration is calculated as moles of solute divided by the volume of solution in liters.

### Step 4: Calculate the concentration of H⁺ ions
Using the ion product of water (Kw), which is \(1.0 \times 10^{-14}\) at 25°C, we can find the concentration of H⁺ ions:

\[ [\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{0.1} = 1.0 \times 10^{-13} \text{ M} \]

**Hint:** The relationship between H⁺ and OH⁻ concentrations is given by the ion product of water.

### Step 5: Calculate the pH of the solution
Finally, we can calculate the pH of the solution using the formula:

\[ \text{pH} = -\log_{10} [\text{H}^+] \]

Substituting the concentration of H⁺ ions:

\[ \text{pH} = -\log_{10} (1.0 \times 10^{-13}) = 13 \]

**Hint:** The pH is the negative logarithm of the hydrogen ion concentration.

### Final Answer
The pH of the resulting solution after electrolysis is **13**.