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

Options

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

Answer

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

### Step 1: Understand the dissociation of HNO3
HNO3 is a strong acid and it dissociates completely in water:
\[ \text{HNO}_3 \rightarrow \text{H}^+ + \text{NO}_3^- \]
Since the concentration of HNO3 is 0.002 M, the concentration of \(\text{H}^+\) ions will also be 0.002 M.

### Step 2: Calculate the pH
The pH can be calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the concentration of \(\text{H}^+\):
\[ \text{pH} = -\log(0.002) \]
This can be rewritten as:
\[ \text{pH} = -\log(2 \times 10^{-3}) \]
Using the logarithmic identity \(\log(AB) = \log A + \log B\):
\[ \text{pH} = -\log(2) - \log(10^{-3}) \]
\[ \text{pH} = -\log(2) + 3 \]

### Step 3: Calculate the pOH
We know that:
\[ \text{pH} + \text{pOH} = 14 \]
Thus, we can rearrange this to find pOH:
\[ \text{pOH} = 14 - \text{pH} \]
Substituting the expression for pH:
\[ \text{pOH} = 14 - (3 - \log(2)) \]
\[ \text{pOH} = 14 - 3 + \log(2) \]
\[ \text{pOH} = 11 + \log(2) \]

### Final Answer
The pOH of the 0.002 M HNO3 solution is:
\[ \text{pOH} = 11 + \log(2) \]

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