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 using ammonium chloride (NH4Cl) and ammonium hydroxide (NH4OH), we will 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 (conjugate acid) of that weak base.

### Step 2: Use the Henderson-Hasselbalch Equation
For a basic buffer, the Henderson-Hasselbalch equation is given by:
\[ \text{pOH} = \text{pK}_b + \log \left( \frac{[\text{Salt}]}{[\text{Base}]} \right) \]

### Step 3: Calculate pOH
Given that pH = 9, we can find pOH using the relationship:
\[ \text{pH} + \text{pOH} = 14 \]
Thus,
\[ \text{pOH} = 14 - 9 = 5 \]

### Step 4: Calculate pK_b
We are given \( K_b = 10^{-5} \). To find \( pK_b \):
\[ pK_b = -\log(K_b) = -\log(10^{-5}) = 5 \]

### Step 5: Substitute Values into the Henderson-Hasselbalch Equation
Now we substitute the values into the equation:
\[ 5 = 5 + \log \left( \frac{[\text{Salt}]}{[\text{Base}]} \right) \]

### Step 6: Simplify the Equation
From the equation:
\[ 5 - 5 = \log \left( \frac{[\text{Salt}]}{[\text{Base}]} \right) \]
This simplifies to:
\[ 0 = \log \left( \frac{[\text{Salt}]}{[\text{Base}]} \right) \]

### Step 7: Solve for the Ratio of Salt to Base
The equation \( \log \left( \frac{[\text{Salt}]}{[\text{Base}]} \right) = 0 \) implies:
\[ \frac{[\text{Salt}]}{[\text{Base}]} = 1 \]
This means that the concentration of salt is equal to the concentration of the base.

### Step 8: Calculate the Moles of Salt
We know the concentration of the base (NH4OH) is 1.8 moles in 1 liter. Therefore, the concentration of salt (NH4Cl) must also be 1.8 moles in 1 liter to maintain the ratio of 1:1.

### Conclusion
The number of moles of NH4Cl that should be added to prepare the buffer solution is **1.8 moles**.

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