To a `10mL` of `10^(-3) N H_(2)SO_(4)` solution water has been to make the total volume of one litre. Its `pOH` would be `:`

Options

A. 3
B. 12
C. 9
D. 5

Answer

To solve the problem, we need to find the pOH of a diluted solution of sulfuric acid (H₂SO₄). Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the number of equivalents of H₂SO₄ in the original solution
We have a 10 mL solution of \(10^{-3} N\) H₂SO₄.

\[ \text{Normality (N)} = \text{Number of equivalents} / \text{Volume in liters} \]

Since we have 10 mL, we convert it to liters:

\[ 10 \, \text{mL} = 0.010 \, \text{L} \]

Now, the number of equivalents in 10 mL of \(10^{-3} N\) H₂SO₄ is:

\[ \text{Number of equivalents} = 10^{-3} \, \text{N} \times 0.010 \, \text{L} = 10^{-5} \, \text{equivalents} \]

### Step 2: Calculate the concentration of H⁺ ions after dilution
When we dilute this solution to a total volume of 1 liter, the concentration of H⁺ ions can be calculated as follows:

\[ \text{Concentration of H}^+ = \frac{\text{Number of equivalents}}{\text{Total volume in liters}} = \frac{10^{-5} \, \text{equivalents}}{1 \, \text{L}} = 10^{-5} \, \text{M} \]

### Step 3: Calculate the pH of the solution
The pH is calculated using the formula:

\[ \text{pH} = -\log[\text{H}^+] \]

Substituting the concentration we found:

\[ \text{pH} = -\log(10^{-5}) = 5 \]

### Step 4: Calculate the pOH of the solution
We know that:

\[ \text{pH} + \text{pOH} = 14 \]

Now, substituting the pH value we calculated:

\[ 5 + \text{pOH} = 14 \]

Solving for pOH:

\[ \text{pOH} = 14 - 5 = 9 \]

### Final Answer
The pOH of the solution is **9**.

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