The `P^(H)` of `10^(-3)` M mono acidic base, if it is 1% ionised is

Options

A. 5
B. 8
C. 3
D. 9

Answer

To solve the question regarding the pH of a 10^(-3) M monoacidic base that is 1% ionized, we can follow these steps:

### Step 1: Determine the concentration of OH⁻ ions
Since the base is 1% ionized, we can calculate the concentration of OH⁻ ions produced from the base.

\[ \text{Ionization} = \frac{1}{100} \times \text{Concentration of base} \]
\[ \text{Concentration of OH⁻} = \frac{1}{100} \times 10^{-3} \, \text{M} = 10^{-5} \, \text{M} \]

### Step 2: Use the ionic product of water (K_w)
The ionic product of water (K_w) at 25°C is \(10^{-14}\). We can use this to find the concentration of H⁺ ions.

\[ K_w = [H^+][OH^-] \]
Substituting the known values:
\[ 10^{-14} = [H^+][10^{-5}] \]

### Step 3: Solve for [H⁺]
Rearranging the equation to find the concentration of H⁺ ions:

\[ [H^+] = \frac{10^{-14}}{10^{-5}} = 10^{-9} \, \text{M} \]

### Step 4: Calculate the pH
The pH is calculated using the formula:

\[ \text{pH} = -\log[H^+] \]
Substituting the value we found:

\[ \text{pH} = -\log(10^{-9}) = 9 \]

### Final Answer
The pH of the 10^(-3) M monoacidic base that is 1% ionized is **9**.

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