20 ml of 0.4 `M H_(2)SO_(4)` , and 80 ml of 0.2 M NaOH are mixed. Then the `p^(H)` of the resulting solution is

Options

A. 7
B. 1.097
C. 12.903
D. 11.903

Answer

To find the pH of the resulting solution when 20 ml of 0.4 M H₂SO₄ is mixed with 80 ml of 0.2 M NaOH, we will follow these steps:

### Step 1: Calculate the Normality of H₂SO₄
H₂SO₄ is a diprotic acid, meaning it can donate two protons (H⁺ ions). Therefore, the normality (N) is twice the molarity (M):
\[ \text{Normality of H₂SO₄} = 2 \times \text{Molarity} = 2 \times 0.4 \, \text{M} = 0.8 \, \text{N} \]

### Step 2: Calculate the Equivalent of H₂SO₄
Next, we calculate the equivalents of H₂SO₄ in the solution:
\[ \text{Equivalents of H₂SO₄} = \text{Normality} \times \text{Volume (in L)} \]
Convert 20 ml to liters:
\[ 20 \, \text{ml} = 0.020 \, \text{L} \]
Now calculate:
\[ \text{Equivalents of H₂SO₄} = 0.8 \, \text{N} \times 0.020 \, \text{L} = 0.016 \, \text{equivalents} \]

### Step 3: Calculate the Normality of NaOH
NaOH is a monoprotic base, so its normality is equal to its molarity:
\[ \text{Normality of NaOH} = 0.2 \, \text{N} \]

### Step 4: Calculate the Equivalent of NaOH
Now calculate the equivalents of NaOH:
\[ \text{Equivalents of NaOH} = \text{Normality} \times \text{Volume (in L)} \]
Convert 80 ml to liters:
\[ 80 \, \text{ml} = 0.080 \, \text{L} \]
Now calculate:
\[ \text{Equivalents of NaOH} = 0.2 \, \text{N} \times 0.080 \, \text{L} = 0.016 \, \text{equivalents} \]

### Step 5: Compare the Equivalents
Now we compare the equivalents of H₂SO₄ and NaOH:
\[ \text{Equivalents of H₂SO₄} = 0.016 \]
\[ \text{Equivalents of NaOH} = 0.016 \]
Since the equivalents of H₂SO₄ and NaOH are equal, we have complete neutralization.

### Step 6: Determine the pH of the Resulting Solution
In a neutralization reaction where equal amounts of a strong acid and a strong base are mixed, the resulting solution will have a pH of 7 (neutral pH at 25°C).

### Final Answer
Thus, the pH of the resulting solution is:
\[ \text{pH} = 7 \]

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