The weight of `CH_(3)COONa` needed to add in 1 litre ` ( N )/( 10 ) CH_(3) COOH` to obtain a buffer solution of pH value 4 is ( Given `K_(a) = 1.8 xx 10^(-5)` )

Options

A. 1.2 g
B. 1.47 g
C. 1.8 g
D. 4 g

Answer

To solve the problem of how much weight of sodium acetate (CH₃COONa) is needed to add to 1 liter of 0.1 N acetic acid (CH₃COOH) to obtain a buffer solution with a pH of 4, we can follow these steps:

### Step 1: Understand the Buffer Solution
A buffer solution is made from a weak acid and its conjugate base. In this case, the weak acid is acetic acid (CH₃COOH) and the conjugate base is sodium acetate (CH₃COONa).

### Step 2: Use the Henderson-Hasselbalch Equation
The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

### Step 3: Calculate pKa
We are given the dissociation constant (Kₐ) for acetic acid:
\[ K_a = 1.8 \times 10^{-5} \]
To find pKa:
\[ \text{pKa} = -\log(K_a) \]
Calculating:
\[ \text{pKa} = -\log(1.8 \times 10^{-5}) \approx 4.75 \]

### Step 4: Set Up the Equation
We know the desired pH is 4. Using the Henderson-Hasselbalch equation:
\[ 4 = 4.75 + \log\left(\frac{[\text{Salt}]}{0.1}\right) \]

### Step 5: Solve for the Ratio of Salt to Acid
Rearranging the equation:
\[ 4 - 4.75 = \log\left(\frac{[\text{Salt}]}{0.1}\right) \]
\[ -0.75 = \log\left(\frac{[\text{Salt}]}{0.1}\right) \]

### Step 6: Convert Logarithm to Exponential Form
To eliminate the logarithm, we convert to exponential form:
\[ 10^{-0.75} = \frac{[\text{Salt}]}{0.1} \]
Calculating \(10^{-0.75}\):
\[ 10^{-0.75} \approx 0.177 \]

### Step 7: Find the Concentration of Salt
Now we can find the concentration of the salt:
\[ 0.177 = \frac{[\text{Salt}]}{0.1} \]
\[ [\text{Salt}] = 0.177 \times 0.1 = 0.0177 \, \text{M} \]

### Step 8: Calculate the Number of Moles
Since we have 1 liter of solution, the number of moles of salt needed is:
\[ \text{Number of moles} = \text{Concentration} \times \text{Volume} = 0.0177 \, \text{mol/L} \times 1 \, \text{L} = 0.0177 \, \text{mol} \]

### Step 9: Calculate the Molar Mass of Sodium Acetate
The molar mass of CH₃COONa (sodium acetate) is calculated as follows:
- Carbon (C): 12 g/mol × 2 = 24 g/mol
- Hydrogen (H): 1 g/mol × 3 = 3 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol
- Sodium (Na): 23 g/mol × 1 = 23 g/mol

Total molar mass:
\[ 24 + 3 + 32 + 23 = 82 \, \text{g/mol} \]

### Step 10: Calculate the Weight of Sodium Acetate
Now we can calculate the weight of sodium acetate needed:
\[ \text{Weight} = \text{Number of moles} \times \text{Molar mass} \]
\[ \text{Weight} = 0.0177 \, \text{mol} \times 82 \, \text{g/mol} \approx 1.45 \, \text{g} \]

### Final Answer
The weight of CH₃COONa needed to add to 1 liter of 0.1 N CH₃COOH to obtain a buffer solution of pH 4 is approximately **1.45 grams**.