Calculate the degree of ionisation and concentration of `H_(3)O^(+)` ions in a solution of 0.01 M formic acid `(K_(a)= 2.1 xx 10^(-4))`

Explanation

To solve the problem of calculating the degree of ionization and the concentration of \( H_3O^+ \) ions in a 0.01 M solution of formic acid with a given \( K_a = 2.1 \times 10^{-4} \), we can follow these steps:

### Step 1: Write the ionization equation for formic acid
Formic acid (\( HCOOH \)) ionizes in water as follows:
\[ HCOOH \rightleftharpoons H^+ + HCOO^- \]
This means that for every mole of formic acid that ionizes, one mole of \( H^+ \) (or \( H_3O^+ \)) and one mole of formate ion (\( HCOO^- \)) is produced.

### Step 2: Set up the expression for the equilibrium constant \( K_a \)
The expression for the acid dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[H^+][HCOO^-]}{[HCOOH]} \]
Let \( \alpha \) be the degree of ionization. Initially, we have:
- Concentration of \( HCOOH \) = 0.01 M
- Change in concentration due to ionization = \( \alpha \)

At equilibrium:
- Concentration of \( HCOOH \) = \( 0.01 - \alpha \)
- Concentration of \( H^+ \) = \( \alpha \)
- Concentration of \( HCOO^- \) = \( \alpha \)

Substituting these into the \( K_a \) expression gives:
\[ K_a = \frac{\alpha \cdot \alpha}{0.01 - \alpha} = \frac{\alpha^2}{0.01 - \alpha} \]

### Step 3: Substitute the known values into the \( K_a \) expression
We know \( K_a = 2.1 \times 10^{-4} \), so we can set up the equation:
\[ 2.1 \times 10^{-4} = \frac{\alpha^2}{0.01 - \alpha} \]

### Step 4: Make an assumption to simplify the equation
Assuming that \( \alpha \) is small compared to 0.01, we can approximate \( 0.01 - \alpha \approx 0.01 \). Thus, the equation simplifies to:
\[ 2.1 \times 10^{-4} = \frac{\alpha^2}{0.01} \]

### Step 5: Solve for \( \alpha^2 \)
Rearranging gives:
\[ \alpha^2 = 2.1 \times 10^{-4} \times 0.01 = 2.1 \times 10^{-6} \]

### Step 6: Calculate \( \alpha \)
Taking the square root of both sides:
\[ \alpha = \sqrt{2.1 \times 10^{-6}} \approx 0.00145 \]

### Step 7: Calculate the concentration of \( H_3O^+ \) ions
The concentration of \( H_3O^+ \) ions is equal to \( \alpha \) times the initial concentration of formic acid:
\[ [H_3O^+] = \alpha \times 0.01 = 0.00145 \times 0.01 = 1.45 \times 10^{-3} \, M \]

### Final Answers
- Degree of ionization \( \alpha \approx 0.00145 \) or \( 0.145\% \)
- Concentration of \( H_3O^+ \) ions \( \approx 1.45 \times 10^{-3} \, M \)