Assuming no change in volume , the time required to obtain solution of ph = 12 by electrolysis of 50 mL 0.1 M NaCl (using current of 0.5 amp) will be

Options

A. 65 sec
B. 96.5 sec
C. 965 sec
D. 9.65 sec

Answer

To solve the problem step by step, we will follow the electrolysis process of NaCl and the calculations related to pH and time required for electrolysis.

### Step 1: Understanding the Electrolysis of NaCl
During the electrolysis of NaCl, the following reactions occur:
- At the anode (oxidation):
\[ 2Cl^- \rightarrow Cl_2 + 2e^- \]
- At the cathode (reduction):
\[ 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \]

### Step 2: Determining the pH and pOH
Given that the desired pH is 12, we can calculate the pOH:
\[ \text{pOH} = 14 - \text{pH} = 14 - 12 = 2 \]

### Step 3: Calculating the Concentration of OH⁻
Using the pOH, we can find the concentration of hydroxide ions (OH⁻):
\[ \text{pOH} = -\log[\text{OH}^-] \implies [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-2} = 0.01 \, \text{M} \]

### Step 4: Finding the Number of Equivalents
We need to calculate the number of equivalents of OH⁻ in 50 mL of 0.01 M solution:
\[ \text{Number of equivalents} = \text{Concentration} \times \text{Volume in L} \times \text{N factor} \]
Here, the volume in liters is \(0.050 \, \text{L}\) and the N factor for NaCl is 1 (since it produces one mole of Na⁺ ions):
\[ \text{Number of equivalents} = 0.01 \, \text{mol/L} \times 0.050 \, \text{L} \times 1 = 5 \times 10^{-4} \, \text{equivalents} \]

### Step 5: Calculating the Charge (Q)
Using Faraday's constant (approximately \(96500 \, \text{C/mol}\)), we can find the total charge:
\[ Q = \text{Number of equivalents} \times \text{Faraday's constant} = 5 \times 10^{-4} \times 96500 \, \text{C} = 48.25 \, \text{C} \]

### Step 6: Using the Formula \(Q = I \times T\)
We can now calculate the time (T) required for electrolysis using the formula \(Q = I \times T\):
\[ T = \frac{Q}{I} = \frac{48.25 \, \text{C}}{0.5 \, \text{A}} = 96.5 \, \text{s} \]

### Final Answer
The time required to obtain a solution of pH = 12 by electrolysis of 50 mL of 0.1 M NaCl using a current of 0.5 A is:
\[ \boxed{96.5 \, \text{seconds}} \]