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 α is the degree of dissociation, the concentrations will be:
- \([HA] = C(1 - \alpha)\)
- \([H^+] = C\alpha\)
- \([A^-] = C\alpha\)

### Step 3: Write the expression for the acid dissociation constant (Ka)
The expression for the acid dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
Substituting the equilibrium concentrations into this expression gives:
\[ K_a = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} \]

### Step 4: Simplify the equation
This simplifies to:
\[ K_a = \frac{C^2 \alpha^2}{C(1 - \alpha)} \]
Cancelling \( C \) from the numerator and denominator, we have:
\[ K_a = \frac{C \alpha^2}{1 - \alpha} \]

### Step 5: Substitute the 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 6: 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 7: Form a quadratic equation
Rearranging the equation results in:
\[ 1.2 \times 10^{-3} \alpha^2 + 1.0 \times 10^{-4} \alpha - 1.0 \times 10^{-4} = 0 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where:
- \( a = 1.2 \times 10^{-3} \)
- \( b = 1.0 \times 10^{-4} \)
- \( c = -1.0 \times 10^{-4} \)

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:
\[ \alpha = \frac{-1.0 \times 10^{-4} \pm \sqrt{4.9 \times 10^{-7}}}{2(1.2 \times 10^{-3})} \]
Calculating the square root:
\[ \sqrt{4.9 \times 10^{-7}} \approx 7.0 \times 10^{-4} \]

Now substituting:
\[ \alpha = \frac{-1.0 \times 10^{-4} + 7.0 \times 10^{-4}}{2.4 \times 10^{-3}} \]
\[ = \frac{6.0 \times 10^{-4}}{2.4 \times 10^{-3}} = 0.25 \]

### Step 9: Convert to percentage
To find the degree of dissociation in percentage:
\[ \alpha \times 100 = 0.25 \times 100 = 25\% \]

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