A buffer solution with`pH = 9` is to be prepared by mixing `NH_(4)Cl" in " 1.8` mole of `NH_(4)OH`. What will be the number of moles of salt that should be added to one litre `(K_(b) = 10^(-5))`

Options

A. `3.4`
B. `2.6`
C. `1.5`
D. `1.8`

Answer

To solve the problem of preparing a buffer solution with a pH of 9 by mixing ammonium chloride (NH₄Cl) with 1.8 moles of ammonium hydroxide (NH₄OH), we can follow these steps:

### Step 1: Understand the Buffer System
A buffer solution consists of a weak base and its conjugate acid. In this case, NH₄OH is the weak base and NH₄Cl is the salt that provides the conjugate acid (NH₄⁺).

### Step 2: Convert pH to pOH
Since we are dealing with a base, we first convert the pH to pOH:
\[ \text{pOH} = 14 - \text{pH} = 14 - 9 = 5 \]

### Step 3: Calculate pKₐ from Kb
We are given the base dissociation constant (Kₐ) for NH₄OH:
\[ K_b = 10^{-5} \]
To find pKₐ, we use the formula:
\[ pK_b = -\log(K_b) = -\log(10^{-5}) = 5 \]

### Step 4: Use the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation for a buffer solution is given by:
\[ \text{pH} = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
For our case, we can rearrange it to find the ratio of salt to base:
\[ \text{pOH} = pK_b + \log\left(\frac{[NH₄^+]}{[NH₄OH]}\right) \]
Substituting the known values:
\[ 5 = 5 + \log\left(\frac{[NH₄^+] }{[NH₄OH]}\right) \]

### Step 5: Simplify the Equation
From the equation above, we can simplify:
\[ 0 = \log\left(\frac{[NH₄^+]}{[NH₄OH]}\right) \]
This implies:
\[ \frac{[NH₄^+]}{[NH₄OH]} = 1 \]

### Step 6: Calculate the Concentration of Salt
Since the concentration of NH₄OH is given as 1.8 moles in 1 liter, we have:
\[ [NH₄OH] = 1.8 \, \text{mol/L} \]
Thus, for the ratio to be 1:
\[ [NH₄^+] = 1.8 \, \text{mol/L} \]

### Step 7: Conclusion
Therefore, the number of moles of NH₄Cl (the salt) that should be added to achieve this concentration is:
\[ \text{Number of moles of } NH₄Cl = 1.8 \, \text{moles} \]

### Final Answer
The number of moles of NH₄Cl that should be added is **1.8 moles**.

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