The pH of a 0.1 M solution of `NH_4Cl` is 5:13. Find the dissociation constant of `NH_(4)OH`.

Answer

To find the dissociation constant of \( NH_4OH \) from the given information, we can follow these steps:

### Step 1: Understand the relationship between pH, pKa, and pKb
The dissociation of \( NH_4OH \) can be represented as:
\[ NH_4OH \rightleftharpoons NH_4^+ + OH^- \]

The relationship between \( pK_a \) and \( pK_b \) is given by:
\[ pK_a + pK_b = 14 \]

### Step 2: Calculate \( pK_b \) using the given pH
Given that the pH of the \( 0.1 \, M \) solution of \( NH_4Cl \) is \( 5.13 \), we can find the \( pK_b \) of \( NH_4OH \).

First, we calculate the \( pOH \):
\[ pOH = 14 - pH = 14 - 5.13 = 8.87 \]

### Step 3: Calculate the concentration of hydroxide ions
Using the \( pOH \), we can find the concentration of hydroxide ions (\( [OH^-] \)):
\[ [OH^-] = 10^{-pOH} = 10^{-8.87} \]

Calculating this gives:
\[ [OH^-] \approx 1.35 \times 10^{-9} \, M \]

### Step 4: Set up the expression for \( K_b \)
The dissociation constant \( K_b \) for \( NH_4OH \) can be expressed as:
\[ K_b = \frac{[NH_4^+][OH^-]}{[NH_4OH]} \]

Assuming \( [NH_4^+] \) is approximately equal to the concentration of \( NH_4Cl \) (which is \( 0.1 \, M \)), and since \( NH_4OH \) is a weak base, we can assume its concentration does not change significantly from \( 0.1 \, M \).

Thus, we can write:
\[ K_b = \frac{(0.1)(1.35 \times 10^{-9})}{0.1} \]

### Step 5: Calculate \( K_b \)
This simplifies to:
\[ K_b = 1.35 \times 10^{-9} \]

### Step 6: Calculate \( pK_b \)
Now, we can find \( pK_b \):
\[ pK_b = -\log K_b = -\log(1.35 \times 10^{-9}) \]

Calculating this gives:
\[ pK_b \approx 8.87 \]

### Step 7: Calculate \( pK_a \)
Finally, using the relationship between \( pK_a \) and \( pK_b \):
\[ pK_a = 14 - pK_b = 14 - 8.87 = 5.13 \]

### Final Answer
The dissociation constant \( K_b \) of \( NH_4OH \) is approximately:
\[ K_b = 1.35 \times 10^{-9} \]