A 0.010M solution of maleic acid, a monoprotic organic acid is 14% ionised. What is `K_(a)` for maleic acid ?

Options

A. `2.3xx10^(-3)`
B. `2.3xx10^(-4)`
C. `2.0xx10^(-4)`
D. `2.0xx10^(-6)`

Answer

To find the \( K_a \) for maleic acid, we can follow these steps:

### Step 1: Identify the given values
- Concentration of maleic acid (\( C \)) = 0.010 M
- Percentage ionization (\( \alpha \)) = 14% = 0.14

### Step 2: Use the formula for \( K_a \)
The formula for the acid dissociation constant (\( K_a \)) for a weak acid is given by:
\[ K_a = \frac{C \cdot \alpha^2}{1 - \alpha} \]

### Step 3: Substitute the known values into the formula
Substituting \( C = 0.010 \) M and \( \alpha = 0.14 \) into the formula:
\[ K_a = \frac{0.010 \cdot (0.14)^2}{1 - 0.14} \]

### Step 4: Calculate \( \alpha^2 \)
Calculate \( \alpha^2 \):
\[ (0.14)^2 = 0.0196 \]

### Step 5: Calculate \( 1 - \alpha \)
Calculate \( 1 - \alpha \):
\[ 1 - 0.14 = 0.86 \]

### Step 6: Substitute these values back into the equation
Now substitute these values back into the equation for \( K_a \):
\[ K_a = \frac{0.010 \cdot 0.0196}{0.86} \]

### Step 7: Perform the multiplication in the numerator
Calculate the numerator:
\[ 0.010 \cdot 0.0196 = 0.000196 \]

### Step 8: Divide by the denominator
Now divide by the denominator:
\[ K_a = \frac{0.000196}{0.86} \approx 0.000228 \]

### Step 9: Convert to scientific notation
Convert \( K_a \) to scientific notation:
\[ K_a \approx 2.28 \times 10^{-4} \]

### Final Answer
Thus, the \( K_a \) for maleic acid is approximately:
\[ K_a \approx 2.3 \times 10^{-4} \]