A monoprotic acid `(HA)` is `1%` ionised in its aqueous solution of `0.1M` strength. Its `pOH` will be

Options

A. `11`
B. `3`
C. `10`
D. `2`

Explanation

To solve the problem, we need to find the pOH of a monoprotic acid (HA) that is 1% ionized in a 0.1 M solution. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the degree of ionization (α)
The degree of ionization (α) is given as 1%. This can be expressed as:
\[ \alpha = \frac{1}{100} = 0.01 \]

### Step 2: Calculate the concentration of hydronium ions \([H_3O^+]\)
For a monoprotic acid, the concentration of hydronium ions at equilibrium can be calculated using the formula:
\[ [H_3O^+] = C \cdot \alpha \]
where \(C\) is the initial concentration of the acid. Given that \(C = 0.1 \, M\), we can substitute the values:
\[ [H_3O^+] = 0.1 \, M \cdot 0.01 = 0.001 \, M = 10^{-3} \, M \]

### Step 3: Calculate the pH
The pH can be calculated using the formula:
\[ pH = -\log[H_3O^+] \]
Substituting the concentration of hydronium ions:
\[ pH = -\log(10^{-3}) = 3 \]

### Step 4: Calculate the pOH
Using the relationship between pH and pOH:
\[ pH + pOH = 14 \]
We can rearrange this to find pOH:
\[ pOH = 14 - pH = 14 - 3 = 11 \]

### Final Answer
Thus, the pOH of the solution is:
\[ \text{pOH} = 11 \]

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