30 cc of `M/3` HCl, 20cc of `M/2` `HNO_(3)` and 40 cc of `M/4 NaOH` solutions are mixed and the volume was made upto `1 dm^(3)`. The pH of the resulting solution is :

Options

A. 8
B. 2
C. 1
D. 3

Explanation

To find the pH of the resulting solution after mixing the given volumes of HCl, HNO3, and NaOH, we can follow these steps:

### Step 1: Calculate the milliequivalents of HCl
- Given: 30 cc of M/3 HCl
- Molarity (M) = M/3 = 1/3 M
- Volume (V) = 30 cc = 30 mL

Using the formula for normality (N) and converting it to milliequivalents:
\[ \text{N} = \text{Molarity} \times \text{n-factor} \]
For HCl, the n-factor = 1.
\[ \text{N}_{\text{HCl}} = \frac{1}{3} \times 1 = \frac{1}{3} \text{ N} \]

Now, calculate the milliequivalents:
\[ \text{Milliequivalents of HCl} = \text{N} \times \text{Volume (in L)} \]
\[ = \frac{1}{3} \times 30 \text{ mL} = \frac{1}{3} \times 0.030 \text{ L} = 0.01 \text{ eq} = 10 \text{ milliequivalents} \]

### Step 2: Calculate the milliequivalents of HNO3
- Given: 20 cc of M/2 HNO3
- Molarity (M) = M/2 = 1/2 M
- Volume (V) = 20 cc = 20 mL

For HNO3, the n-factor = 1.
\[ \text{N}_{\text{HNO3}} = \frac{1}{2} \times 1 = \frac{1}{2} \text{ N} \]

Now, calculate the milliequivalents:
\[ \text{Milliequivalents of HNO3} = \text{N} \times \text{Volume (in L)} \]
\[ = \frac{1}{2} \times 20 \text{ mL} = \frac{1}{2} \times 0.020 \text{ L} = 0.01 \text{ eq} = 10 \text{ milliequivalents} \]

### Step 3: Calculate the milliequivalents of NaOH
- Given: 40 cc of M/4 NaOH
- Molarity (M) = M/4 = 1/4 M
- Volume (V) = 40 cc = 40 mL

For NaOH, the n-factor = 1.
\[ \text{N}_{\text{NaOH}} = \frac{1}{4} \times 1 = \frac{1}{4} \text{ N} \]

Now, calculate the milliequivalents:
\[ \text{Milliequivalents of NaOH} = \text{N} \times \text{Volume (in L)} \]
\[ = \frac{1}{4} \times 40 \text{ mL} = \frac{1}{4} \times 0.040 \text{ L} = 0.01 \text{ eq} = 10 \text{ milliequivalents} \]

### Step 4: Determine the total milliequivalents of H⁺ and OH⁻
- Total milliequivalents of H⁺ from HCl and HNO3:
\[ \text{Total H⁺} = 10 \text{ (HCl)} + 10 \text{ (HNO3)} = 20 \text{ milliequivalents} \]

- Total milliequivalents of OH⁻ from NaOH:
\[ \text{Total OH⁻} = 10 \text{ milliequivalents} \]

### Step 5: Calculate the remaining milliequivalents of H⁺
- Since NaOH neutralizes some of the H⁺:
\[ \text{Remaining H⁺} = 20 - 10 = 10 \text{ milliequivalents} \]

### Step 6: Calculate the concentration of H⁺ ions in the final solution
- The total volume of the solution after mixing is 1 dm³ = 1000 mL.
\[ \text{Concentration of H⁺} = \frac{\text{Remaining H⁺}}{\text{Total Volume}} = \frac{10 \text{ milliequivalents}}{1000 \text{ mL}} = 0.01 \text{ N} \]

### Step 7: Calculate the pH of the solution
- pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H⁺}] \]
Since 0.01 N H⁺ corresponds to \(10^{-2}\) M:
\[ \text{pH} = -\log(10^{-2}) = 2 \]

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

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