`pOH` of `0.002 MHNO_(3) `is `:`

Options

A. `11+log2`
B. `11-log2`
C. `-3 +log 2`
D. none of these

Explanation

To find the pOH of a 0.002 M HNO₃ solution, we can follow these steps:

### Step 1: Determine the concentration of H⁺ ions
Since HNO₃ is a strong acid, it dissociates completely in solution. Therefore, the concentration of H⁺ ions will be equal to the concentration of HNO₃.

\[ \text{[H⁺]} = \text{[HNO₃]} = 0.002 \, \text{M} \]

### Step 2: Calculate the pH
The pH is calculated using the formula:

\[ \text{pH} = -\log[\text{H⁺}] \]

Substituting the concentration of H⁺:

\[ \text{pH} = -\log(0.002) \]

To simplify this, we can express 0.002 in scientific notation:

\[ 0.002 = 2 \times 10^{-3} \]

Now, applying the logarithm:

\[ \text{pH} = -\log(2 \times 10^{-3}) = -\log(2) - \log(10^{-3}) = -\log(2) + 3 \]

### Step 3: Calculate the pOH
Using the relationship between pH and pOH:

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

We can rearrange this to find pOH:

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

Substituting the expression for pH:

\[ \text{pOH} = 14 - (3 - \log(2)) = 14 - 3 + \log(2) = 11 + \log(2) \]

### Final Answer
Thus, the pOH of the 0.002 M HNO₃ solution is:

\[ \text{pOH} = 11 + \log(2) \]

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