Calculate the degree of dissociation of 0.2 N of a monobasic acid at `25^C.` The dissociation constant of acetic acid at this temperature is `1.8 xx 10^(-5).` What will be the `H^+` concentration?

Answer

To calculate the degree of dissociation (α) of a monobasic acid at 25°C, we can use the dissociation constant (K_a) and the concentration (C) of the acid. The formula we will use is:

\[ K_a = C \cdot \alpha^2 \]

Where:
- \( K_a \) is the dissociation constant,
- \( C \) is the concentration of the acid,
- \( \alpha \) is the degree of dissociation.

Given:
- \( K_a = 1.8 \times 10^{-5} \)
- \( C = 0.2 \, N \)

### Step 1: Set up the equation

We start with the equation for the dissociation constant:

\[ K_a = C \cdot \alpha^2 \]

Substituting the known values:

\[ 1.8 \times 10^{-5} = 0.2 \cdot \alpha^2 \]

### Step 2: Solve for α²

Rearranging the equation to isolate \( \alpha^2 \):

\[ \alpha^2 = \frac{1.8 \times 10^{-5}}{0.2} \]

### Step 3: Calculate α²

Now, perform the division:

\[ \alpha^2 = \frac{1.8 \times 10^{-5}}{0.2} = 9.0 \times 10^{-5} \]

### Step 4: Calculate α

Taking the square root of both sides to find α:

\[ \alpha = \sqrt{9.0 \times 10^{-5}} \]

Calculating the square root:

\[ \alpha \approx 0.00949 \]

### Step 5: Calculate the H⁺ concentration

Since the acid is monobasic, the concentration of \( H^+ \) ions will be equal to the degree of dissociation multiplied by the initial concentration:

\[ [H^+] = C \cdot \alpha = 0.2 \cdot 0.00949 \]

Calculating this gives:

\[ [H^+] \approx 0.001898 \, N \]

### Final Results

- Degree of dissociation (α) ≈ 0.00949
- \( H^+ \) concentration ≈ 0.001898 N

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