The degree of dissociation of a weak monoprotic acid of concentration `1.2xx10^(-3)"M having "K_(a)=1.0xx10^(-4` is

Options

A. 1
B. 10
C. 15
D. 25

Answer

To find the degree of dissociation (α) of a weak monoprotic acid with a given concentration and dissociation constant, we can follow these steps:

### Step 1: Write the dissociation equation
For a weak monoprotic acid (HA), the dissociation can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]

### Step 2: Set up the initial and equilibrium concentrations
Let the initial concentration of the acid (HA) be \( c = 1.2 \times 10^{-3} \, M \). At equilibrium, if \( \alpha \) is the degree of dissociation, then:
- Concentration of \( HA \) at equilibrium = \( c(1 - \alpha) \)
- Concentration of \( H^+ \) at equilibrium = \( c\alpha \)
- Concentration of \( A^- \) at equilibrium = \( c\alpha \)

### Step 3: Write the expression for \( K_a \)
The acid dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
Substituting the equilibrium concentrations:
\[ K_a = \frac{(c\alpha)(c\alpha)}{c(1 - \alpha)} \]
This simplifies to:
\[ K_a = \frac{c^2 \alpha^2}{c(1 - \alpha)} = \frac{c\alpha^2}{1 - \alpha} \]

### Step 4: Substitute known values
Given \( K_a = 1.0 \times 10^{-4} \) and \( c = 1.2 \times 10^{-3} \):
\[ 1.0 \times 10^{-4} = \frac{(1.2 \times 10^{-3})\alpha^2}{1 - \alpha} \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ 1.0 \times 10^{-4} (1 - \alpha) = 1.2 \times 10^{-3} \alpha^2 \]
Expanding this:
\[ 1.0 \times 10^{-4} - 1.0 \times 10^{-4} \alpha = 1.2 \times 10^{-3} \alpha^2 \]

### Step 6: Rearrange into standard quadratic form
Rearranging leads to:
\[ 1.2 \times 10^{-3} \alpha^2 + 1.0 \times 10^{-4} \alpha - 1.0 \times 10^{-4} = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1.2 \times 10^{-3} \), \( b = 1.0 \times 10^{-4} \), and \( c = -1.0 \times 10^{-4} \).

Calculating the discriminant:
\[ b^2 - 4ac = (1.0 \times 10^{-4})^2 - 4(1.2 \times 10^{-3})(-1.0 \times 10^{-4}) \]
\[ = 1.0 \times 10^{-8} + 4.8 \times 10^{-7} = 4.9 \times 10^{-7} \]

Now, substituting back into the quadratic formula:
\[ \alpha = \frac{-1.0 \times 10^{-4} \pm \sqrt{4.9 \times 10^{-7}}}{2 \times 1.2 \times 10^{-3}} \]

Calculating the square root:
\[ \sqrt{4.9 \times 10^{-7}} \approx 7.0 \times 10^{-4} \]
So,
\[ \alpha = \frac{-1.0 \times 10^{-4} \pm 7.0 \times 10^{-4}}{2.4 \times 10^{-3}} \]

Calculating the two possible values for α:
1. \( \alpha = \frac{6.0 \times 10^{-4}}{2.4 \times 10^{-3}} \approx 0.25 \)
2. The negative root is not physically meaningful.

### Step 8: Convert to percentage
To express α as a percentage:
\[ \alpha \times 100 = 0.25 \times 100 = 25\% \]

### Final Answer
The degree of dissociation of the weak monoprotic acid is **25%**.

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