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 solve the problem of finding 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 number of moles of HCl
Given:
- Volume of HCl = 30 cc = 30 ml
- Molarity of HCl = M/3 = 1/3 M

Using the formula for moles:
\[ \text{Number of moles of HCl} = \text{Molarity} \times \text{Volume (in L)} \]
Convert volume from ml to L:
\[ 30 \text{ ml} = 0.030 \text{ L} \]
Now calculate the moles:
\[ \text{Number of moles of HCl} = \frac{1}{3} \times 0.030 = 0.010 \text{ moles} \]

### Step 2: Calculate the number of moles of HNO3
Given:
- Volume of HNO3 = 20 cc = 20 ml
- Molarity of HNO3 = M/2 = 1/2 M

Using the same formula:
\[ \text{Number of moles of HNO3} = \text{Molarity} \times \text{Volume (in L)} \]
Convert volume from ml to L:
\[ 20 \text{ ml} = 0.020 \text{ L} \]
Now calculate the moles:
\[ \text{Number of moles of HNO3} = \frac{1}{2} \times 0.020 = 0.010 \text{ moles} \]

### Step 3: Calculate the number of moles of NaOH
Given:
- Volume of NaOH = 40 cc = 40 ml
- Molarity of NaOH = M/4 = 1/4 M

Using the same formula:
\[ \text{Number of moles of NaOH} = \text{Molarity} \times \text{Volume (in L)} \]
Convert volume from ml to L:
\[ 40 \text{ ml} = 0.040 \text{ L} \]
Now calculate the moles:
\[ \text{Number of moles of NaOH} = \frac{1}{4} \times 0.040 = 0.010 \text{ moles} \]

### Step 4: Determine the total moles of H+ and OH-
Total moles of H+ from acids (HCl and HNO3):
\[ \text{Total moles of H+} = 0.010 + 0.010 = 0.020 \text{ moles} \]

Total moles of OH- from NaOH:
\[ \text{Total moles of OH-} = 0.010 \text{ moles} \]

### Step 5: Calculate the remaining moles of H+
Since OH- will react with H+, we need to find the remaining moles of H+ after the reaction:
\[ \text{Remaining moles of H+} = \text{Total moles of H+} - \text{Total moles of OH-} = 0.020 - 0.010 = 0.010 \text{ moles} \]

### Step 6: Calculate the concentration of H+ ions
The total volume of the solution is made up to 1 dm³ (1000 ml):
\[ \text{Concentration of H+} = \frac{\text{Remaining moles of H+}}{\text{Total volume in L}} = \frac{0.010}{1} = 0.010 \text{ M} \]

### Step 7: Calculate the pH
Using the formula for pH:
\[ \text{pH} = -\log[\text{H+}] \]
Substituting the concentration:
\[ \text{pH} = -\log(0.010) = -\log(10^{-2}) = 2 \]

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

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