Calculate the amount of acetic acid present in `1L` of solution having `alpha = 1%` and `K_(a) = 1.8 xx 10^(-5)`.

Explanation

To calculate the amount of acetic acid present in 1 liter of solution with a dissociation constant (α) of 1% and \( K_a = 1.8 \times 10^{-5} \), we can follow these steps:

### Step 1: Write the Dissociation Reaction
The dissociation of acetic acid (CH₃COOH) can be represented as:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{COO}^- \]

### Step 2: Define Initial Concentration
Assume the initial concentration of acetic acid is \( C \) (in mol/L). Since we are considering 1 liter of solution, the initial amount of acetic acid is \( C \) moles.

### Step 3: Define Change in Concentration
At equilibrium, the concentration of acetic acid will decrease by \( C \alpha \) due to dissociation. Therefore, the equilibrium concentrations will be:
- Concentration of CH₃COOH: \( C - C \alpha = C(1 - \alpha) \)
- Concentration of H⁺: \( C \alpha \)
- Concentration of CH₃COO⁻: \( C \alpha \)

### Step 4: Write the Expression for the Equilibrium Constant
The expression for the equilibrium constant \( K_a \) is given by:
\[ K_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \]
Substituting the equilibrium concentrations, we get:
\[ K_a = \frac{(C \alpha)(C \alpha)}{C(1 - \alpha)} = \frac{C^2 \alpha^2}{C(1 - \alpha)} \]

### Step 5: Simplify the Equation
Since \( C \) is in both the numerator and denominator, we can cancel one \( C \):
\[ K_a = \frac{C \alpha^2}{1 - \alpha} \]

### Step 6: Substitute Known Values
Given \( K_a = 1.8 \times 10^{-5} \) and \( \alpha = 0.01 \) (1%):
\[ 1.8 \times 10^{-5} = \frac{C (0.01)^2}{1 - 0.01} \]
This simplifies to:
\[ 1.8 \times 10^{-5} = \frac{C \times 0.0001}{0.99} \]

### Step 7: Solve for C
Rearranging gives:
\[ C = \frac{1.8 \times 10^{-5} \times 0.99}{0.0001} \]
Calculating this:
\[ C = \frac{1.782 \times 10^{-5}}{0.0001} = 0.1782 \, \text{mol/L} \]

### Step 8: Calculate the Mass of Acetic Acid
Now, to find the mass of acetic acid in grams:
- Molar mass of acetic acid (CH₃COOH) = 12 (C) * 2 + 1 (H) * 4 + 16 (O) * 2 = 60 g/mol
- Amount of acetic acid in grams = Concentration (mol/L) × Volume (L) × Molar Mass (g/mol)
\[ \text{Mass} = 0.1782 \, \text{mol/L} \times 1 \, \text{L} \times 60 \, \text{g/mol} = 10.692 \, \text{g} \]

### Final Answer
The amount of acetic acid present in 1 liter of solution is approximately **10.69 grams**.

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