Calculate the `pOH` of solution at `25^(@)C` that contains `1xx10^(-10) M` of hydronium ions, i.e., `H_(3)O^(+)`

Options

A. `7.00`
B. `4.00`
C. `9.00`
D. `1.00`

Explanation

To calculate the pOH of a solution at 25°C that contains \(1 \times 10^{-10} \, M\) of hydronium ions (\(H_3O^+\)), we can follow these steps:

### Step 1: Identify the concentration of hydronium ions
We are given that the concentration of hydronium ions is:
\[ [H_3O^+] = 1 \times 10^{-10} \, M \]

### Step 2: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[H_3O^+] \]
Substituting the value of hydronium ion concentration:
\[ \text{pH} = -\log(1 \times 10^{-10}) \]
Using the logarithmic property \( \log(a \times 10^b) = b + \log(a) \), we can simplify:
\[ \text{pH} = -(-10) = 10 \]

### Step 3: Use the relationship between pH and pOH
At 25°C, the relationship between pH and pOH is given by:
\[ \text{pH} + \text{pOH} = 14 \]
We can rearrange this to find pOH:
\[ \text{pOH} = 14 - \text{pH} \]
Substituting the value of pH we calculated:
\[ \text{pOH} = 14 - 10 = 4 \]

### Conclusion
Thus, the pOH of the solution is:
\[ \text{pOH} = 4 \]

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