The pH of decinormal solution of KOH is

Options

A. 1
B. 4
C. 10
D. 13

Explanation

To find the pH of a decinormal (0.1 N) solution of KOH, we can follow these steps:

### Step 1: Understand the nature of KOH
KOH (potassium hydroxide) is a strong base, which means it completely dissociates in water. The dissociation can be represented as:
\[ \text{KOH} \rightarrow \text{K}^+ + \text{OH}^- \]

### Step 2: Determine the concentration of OH⁻ ions
Since KOH dissociates completely, the concentration of hydroxide ions (OH⁻) in a decinormal solution (0.1 N) is equal to the normality of KOH:
\[ [\text{OH}^-] = 0.1 \, \text{N} = 0.1 \, \text{mol/L} \]

### Step 3: Calculate the pOH of the solution
The pOH can be calculated using the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Substituting the concentration of OH⁻:
\[ \text{pOH} = -\log(0.1) \]
\[ \text{pOH} = 1 \]

### Step 4: Calculate the pH from pOH
The relationship between pH and pOH is given by:
\[ \text{pH} + \text{pOH} = 14 \]
Using the pOH calculated in the previous step:
\[ \text{pH} = 14 - \text{pOH} \]
\[ \text{pH} = 14 - 1 \]
\[ \text{pH} = 13 \]

### Final Answer
The pH of a decinormal solution of KOH is **13**.

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