A solution of acetic acid is `1.0%` ionised. Determine the molar concentration of acid `(K_(a)=1.8xx10^(-5))` and also the `[H^(+)]`.

Options

A. `1.8xx10^(-1) M` and `1.8xx10^(-3) M`
B. `0.18xx10^(-1) M` and `1.8xx10^(-4) M`
C. `0.18xx10^(-2) M` and `1.8xx10^(-2) M`
D. `0.18xx10^(-3) M` and `1.8xx10^(-1) M`

Explanation

To solve the problem, we need to determine the molar concentration of acetic acid and the concentration of hydrogen ions \([H^+]\) in a solution where acetic acid is 1% ionized. Given that the dissociation constant \(K_a\) for acetic acid is \(1.8 \times 10^{-5}\), we can follow these steps:

### Step 1: Understand the Ionization of Acetic Acid
Acetic acid (\(CH_3COOH\)) ionizes in water as follows:

\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]

Let \(C\) be the initial concentration of acetic acid. If it is \(1\%\) ionized, then the degree of ionization \(\alpha\) is:

\[ \alpha = \frac{1}{100} = 0.01 \]

### Step 2: Set Up the Equilibrium Expression
At equilibrium, the concentrations will be:

- Concentration of \(CH_3COOH\) = \(C(1 - \alpha)\)
- Concentration of \(CH_3COO^-\) = \(C\alpha\)
- Concentration of \(H^+\) = \(C\alpha\)

### Step 3: Write the Expression for \(K_a\)
The expression for the acid dissociation constant \(K_a\) is given by:

\[ K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} \]

Substituting \(\alpha = 0.01\):

\[ K_a = \frac{(C \cdot 0.01)(C \cdot 0.01)}{C(1 - 0.01)} = \frac{C^2 \cdot 0.01^2}{C \cdot 0.99} \]

### Step 4: Simplify the Expression
This simplifies to:

\[ K_a = \frac{C \cdot 0.01^2}{0.99} \]

### Step 5: Substitute the Value of \(K_a\)
Now, substituting the value of \(K_a\):

\[ 1.8 \times 10^{-5} = \frac{C \cdot (0.01)^2}{0.99} \]

### Step 6: Solve for \(C\)
Rearranging the equation to solve for \(C\):

\[ C = \frac{1.8 \times 10^{-5} \cdot 0.99}{(0.01)^2} \]

Calculating this gives:

\[ C = \frac{1.8 \times 10^{-5} \cdot 0.99}{1 \times 10^{-4}} = \frac{1.782 \times 10^{-5}}{1 \times 10^{-4}} = 0.1782 \text{ M} \]

### Step 7: Calculate \([H^+]\)
Since \([H^+] = C\alpha\):

\[ [H^+] = C \cdot 0.01 = 0.1782 \cdot 0.01 = 1.782 \times 10^{-3} \text{ M} \]

### Final Results
- Molar concentration of acetic acid, \(C \approx 0.1782 \text{ M}\)
- Concentration of hydrogen ions, \([H^+] \approx 1.782 \times 10^{-3} \text{ M}\)