How much sodium acetate should be added to a 0.1 M solution of `CH_(3)COOH` to give a solution of pH = 5.5 (`pK_(a)` of `CH_(3)COOH = 4.5`)

Options

A. 0.1 M
B. 0.2 M
C. `1.0 M`
D. `10.0 M`

Explanation

To solve the problem of how much sodium acetate should be added to a 0.1 M solution of acetic acid to achieve a pH of 5.5, we can use the Henderson-Hasselbalch equation, which is given by:

\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]

Where:
- \([\text{A}^-]\) is the concentration of the acetate ion (sodium acetate).
- \([\text{HA}]\) is the concentration of acetic acid.

### Step 1: Identify the known values
- pH = 5.5
- pK_a of acetic acid = 4.5
- Concentration of acetic acid \([\text{HA}] = 0.1 \, \text{M}\)

### Step 2: Rearrange the Henderson-Hasselbalch equation
We can rearrange the equation to solve for \([\text{A}^-]\):

\[ 5.5 = 4.5 + \log \left( \frac{[\text{A}^-]}{0.1} \right) \]

Subtract 4.5 from both sides:

\[ 5.5 - 4.5 = \log \left( \frac{[\text{A}^-]}{0.1} \right) \]

\[ 1 = \log \left( \frac{[\text{A}^-]}{0.1} \right) \]

### Step 3: Convert the logarithmic equation to exponential form
To eliminate the logarithm, we convert it to exponential form:

\[ \frac{[\text{A}^-]}{0.1} = 10^1 \]

\[ \frac{[\text{A}^-]}{0.1} = 10 \]

### Step 4: Solve for \([\text{A}^-]\)
Now, multiply both sides by 0.1:

\[ [\text{A}^-] = 10 \times 0.1 = 1 \, \text{M} \]

### Step 5: Calculate the amount of sodium acetate needed
To prepare a 1 M solution of sodium acetate in a total volume of 1 L (assuming we want to make 1 L of buffer solution):

1. Calculate the number of moles of sodium acetate needed:

\[ \text{Moles of sodium acetate} = 1 \, \text{M} \times 1 \, \text{L} = 1 \, \text{mol} \]

2. The molar mass of sodium acetate (CH₃COONa) is approximately 82.03 g/mol.

3. Calculate the mass of sodium acetate required:

\[ \text{Mass} = \text{moles} \times \text{molar mass} = 1 \, \text{mol} \times 82.03 \, \text{g/mol} = 82.03 \, \text{g} \]

### Final Answer:
You need to add **82.03 grams** of sodium acetate to the 0.1 M solution of acetic acid to achieve a pH of 5.5.

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