What is the concentration of `CH_(3)COOH(aq.)` in a solution prepared by dissolving 0.01 mole of `NH_(4)^(+)CH_(3)COO^(-)` in 1 L `H_(2)O`? `[K_(a) (CH_(3)COOH))=1.8xx10^(-5)),K_(b) (NH_(4)OH)=1.8xx10^(-5))]`

Options

A. `5.55xx10^(-5)`
B. `0.10`
C. `6.4xx10^(-4)`
D. `5.55xx10^(-3)`

Answer

To find the concentration of acetic acid (`CH₃COOH`) in a solution prepared by dissolving 0.01 mole of ammonium acetate (`NH₄CH₃COO`) in 1 L of water, we will follow these steps:

### Step 1: Understanding the Hydrolysis Reaction
Ammonium acetate is a salt formed from a weak acid (acetic acid) and a weak base (ammonium hydroxide). When dissolved in water, it hydrolyzes according to the following reaction:

\[ NH_4^+ (aq) + CH_3COO^- (aq) \rightleftharpoons CH_3COOH (aq) + NH_4OH (aq) \]

### Step 2: Setting Up the Initial Concentrations
Initially, when 0.01 moles of ammonium acetate are dissolved in 1 L of water, the concentrations of the ions are:

- \([NH_4^+]\) = 0.01 M
- \([CH_3COO^-]\) = 0.01 M
- \([CH_3COOH]\) = 0 M
- \([NH_4OH]\) = 0 M

### Step 3: Establishing the Change at Equilibrium
Let the degree of hydrolysis be represented by \(\alpha\). At equilibrium, the concentrations will be:

- \([NH_4^+]\) = \(0.01 - \alpha\)
- \([CH_3COO^-]\) = \(0.01 - \alpha\)
- \([CH_3COOH]\) = \(\alpha\)
- \([NH_4OH]\) = \(\alpha\)

### Step 4: Writing the Hydrolysis Constant Expression
The hydrolysis constant \(K_H\) can be expressed as:

\[ K_H = \frac{[CH_3COOH][NH_4OH]}{[NH_4^+][CH_3COO^-]} \]

### Step 5: Substituting the Equilibrium Concentrations
Substituting the equilibrium concentrations into the expression gives:

\[ K_H = \frac{\alpha \cdot \alpha}{(0.01 - \alpha)(0.01 - \alpha)} = \frac{\alpha^2}{(0.01 - \alpha)^2} \]

### Step 6: Calculating \(K_H\)
The hydrolysis constant \(K_H\) can be calculated using the formula:

\[ K_H = \frac{K_w}{K_a \cdot K_b} \]

Where:
- \(K_w = 1.0 \times 10^{-14}\)
- \(K_a = 1.8 \times 10^{-5}\)
- \(K_b = 1.8 \times 10^{-5}\)

Calculating \(K_H\):

\[ K_H = \frac{1.0 \times 10^{-14}}{(1.8 \times 10^{-5})^2} = \frac{1.0 \times 10^{-14}}{3.24 \times 10^{-10}} \approx 3.09 \times 10^{-5} \]

### Step 7: Setting Up the Equation
Now we can set up the equation:

\[ 3.09 \times 10^{-5} = \frac{\alpha^2}{(0.01 - \alpha)^2} \]

Assuming \(\alpha\) is small compared to 0.01, we can approximate \(0.01 - \alpha \approx 0.01\):

\[ 3.09 \times 10^{-5} = \frac{\alpha^2}{(0.01)^2} \]

### Step 8: Solving for \(\alpha\)
Rearranging gives:

\[ \alpha^2 = 3.09 \times 10^{-5} \times (0.01)^2 = 3.09 \times 10^{-7} \]

Taking the square root:

\[ \alpha = \sqrt{3.09 \times 10^{-7}} \approx 5.55 \times 10^{-4} \]

### Step 9: Finding the Concentration of Acetic Acid
The concentration of acetic acid is given by:

\[ [CH_3COOH] = \alpha = 5.55 \times 10^{-4} \text{ M} \]

### Final Answer
The concentration of acetic acid in the solution is approximately:

\[ \boxed{5.55 \times 10^{-4} \text{ M}} \]