If pH of solution of `NaOH` is `12.0` the pH of `H_(2)SO_(4)` solution of same molarity will be

Options

A. `2.0`
B. `12.0`
C. `1.7`
D. `10.0387`.

Explanation

To solve the problem, we need to determine the pH of a sulfuric acid (H₂SO₄) solution that has the same molarity as a sodium hydroxide (NaOH) solution with a pH of 12.0. Here’s a step-by-step solution:

### Step 1: Calculate the concentration of H⁺ ions in the NaOH solution.
Given that the pH of the NaOH solution is 12.0, we can find the concentration of H⁺ ions using the formula:
\[ \text{pH} = -\log[H^+] \]

Rearranging this gives:
\[ [H^+] = 10^{-\text{pH}} \]

Substituting the given pH:
\[ [H^+] = 10^{-12} \, \text{M} \]

### Step 2: Calculate the concentration of OH⁻ ions in the NaOH solution.
We know that the product of the concentrations of H⁺ and OH⁻ ions at 25°C is given by:
\[ K_w = [H^+][OH^-] = 10^{-14} \]

Using the concentration of H⁺ ions calculated in Step 1:
\[ [OH^-] = \frac{K_w}{[H^+]} = \frac{10^{-14}}{10^{-12}} = 10^{-2} \, \text{M} \]

### Step 3: Determine the molarity of the NaOH solution.
Since NaOH dissociates completely in solution:
\[ [NaOH] = [OH^-] = 10^{-2} \, \text{M} \]

### Step 4: Calculate the concentration of H⁺ ions in the H₂SO₄ solution.
H₂SO₄ is a strong acid and dissociates completely in the first step:
\[ H_2SO_4 \rightarrow H^+ + HSO_4^- \]

In the second step, HSO₄⁻ can also dissociate:
\[ HSO_4^- \rightarrow H^+ + SO_4^{2-} \]

For each mole of H₂SO₄, we get 2 moles of H⁺ ions. Therefore, if the molarity of H₂SO₄ is the same as that of NaOH (10⁻² M), the concentration of H⁺ ions will be:
\[ [H^+] = 2 \times [H_2SO_4] = 2 \times 10^{-2} \, \text{M} = 2 \times 10^{-2} \, \text{M} \]

### Step 5: Calculate the pH of the H₂SO₄ solution.
Using the concentration of H⁺ ions calculated in Step 4:
\[ \text{pH} = -\log[H^+] = -\log(2 \times 10^{-2}) \]

Using logarithmic properties:
\[ \text{pH} = -\log(2) - \log(10^{-2}) \]
\[ \text{pH} = -\log(2) + 2 \]

Approximating \(-\log(2) \approx -0.301\):
\[ \text{pH} \approx 2 - 0.301 = 1.699 \]

Thus, the pH of the H₂SO₄ solution will be approximately **1.7**.

### Final Answer:
The pH of the H₂SO₄ solution of the same molarity as the NaOH solution with pH 12.0 is approximately **1.7**.

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