100 c.c of N/10 NaOH solution is mixed with 100 c.c of N/5 HCI solution and the whole volume is made to 1 litre . The pH of the resulting solution will be

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of HCl
Given:
- Normality of HCl = N/5
- Volume of HCl = 100 c.c = 0.1 L

To find the molarity of HCl:
- Molarity = Normality (since n-factor for HCl = 1)
- Molarity of HCl = N/5 = 0.2 N (since N = 1)

Now, calculate the number of moles of HCl:
\[ \text{Number of moles of HCl} = \text{Molarity} \times \text{Volume in L} = 0.2 \times 0.1 = 0.02 \text{ moles} \]

### Step 2: Calculate the number of moles of NaOH
Given:
- Normality of NaOH = N/10
- Volume of NaOH = 100 c.c = 0.1 L

To find the molarity of NaOH:
- Molarity = Normality (since n-factor for NaOH = 1)
- Molarity of NaOH = N/10 = 0.1 N (since N = 1)

Now, calculate the number of moles of NaOH:
\[ \text{Number of moles of NaOH} = \text{Molarity} \times \text{Volume in L} = 0.1 \times 0.1 = 0.01 \text{ moles} \]

### Step 3: Determine the neutralization reaction
The reaction between NaOH and HCl is:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

From the stoichiometry of the reaction:
- 1 mole of NaOH reacts with 1 mole of HCl.

### Step 4: Calculate the remaining moles after neutralization
Initial moles:
- NaOH = 0.01 moles
- HCl = 0.02 moles

After neutralization:
- NaOH will completely react with 0.01 moles of HCl, leaving:
\[ \text{Remaining HCl} = 0.02 - 0.01 = 0.01 \text{ moles} \]
- NaOH will be completely consumed.

### Step 5: Calculate the concentration of HCl in the final solution
The total volume of the resulting solution is 1 L. Therefore, the concentration of HCl is:
\[ \text{Concentration of HCl} = \frac{\text{Number of moles of HCl}}{\text{Volume in L}} = \frac{0.01}{1} = 0.01 \text{ M} \]

### Step 6: Calculate the pH of the solution
Since HCl is a strong acid, it completely dissociates in solution:
\[ \text{[H}^+\text{]} = 0.01 \text{ M} \]
Now, calculate the pH:
\[ \text{pH} = -\log[\text{H}^+] = -\log(0.01) = 2 \]

### Final Answer
The pH of the resulting solution is **2**.

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