Weak acids dissociate partially in aqueous medium. If any common ion is present in the solution, the degree of dissociation of the acid is suppressed however the dissociation constant value remains constant.
What is the degree of dissociation of water is 0.0IM HCOOH `(K_(a) = 10^(-6))` at `25^(@)` C?

Options

A. `1.8xx10^(-12)`
B. `1.8xx10^(-14)`
C. `0.8xx10^(-12)`
D. `3.6xx10^(-10)`

Answer

To find the degree of dissociation of water in a 0.01 M acetic acid solution at 25°C, we can follow these steps:

### Step 1: Understand the dissociation of acetic acid
Acetic acid (HCOOH) is a weak acid that partially dissociates in water:
\[ \text{HCOOH} \rightleftharpoons \text{H}^+ + \text{HCOO}^- \]

### Step 2: Write the expression for the dissociation constant (Ka)
The dissociation constant \( K_a \) for acetic acid is given by:
\[ K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]} \]
Given that \( K_a = 10^{-6} \).

### Step 3: Set up the initial concentrations
Let the initial concentration of acetic acid be \( C = 0.01 \, \text{M} \). Initially, the concentrations are:
- \([\text{HCOOH}] = 0.01 \, \text{M}\)
- \([\text{H}^+] = 0\)
- \([\text{HCOO}^-] = 0\)

### Step 4: Define the degree of dissociation (α)
Let \( \alpha \) be the degree of dissociation of acetic acid. At equilibrium, the concentrations will be:
- \([\text{HCOOH}] = 0.01 - \alpha\)
- \([\text{H}^+] = \alpha\)
- \([\text{HCOO}^-] = \alpha\)

### Step 5: Substitute into the Ka expression
Substituting the equilibrium concentrations into the \( K_a \) expression:
\[ K_a = \frac{[\alpha][\alpha]}{[0.01 - \alpha]} \]
\[ 10^{-6} = \frac{\alpha^2}{0.01 - \alpha} \]

### Step 6: Assume α is small compared to 0.01
Since acetic acid is a weak acid, we can assume that \( \alpha \) is small compared to 0.01 M, so \( 0.01 - \alpha \approx 0.01 \):
\[ 10^{-6} = \frac{\alpha^2}{0.01} \]

### Step 7: Solve for α
Rearranging gives:
\[ \alpha^2 = 10^{-6} \times 0.01 \]
\[ \alpha^2 = 10^{-8} \]
\[ \alpha = 10^{-4} \]

### Step 8: Calculate [H+] from water dissociation
At 25°C, the dissociation of water can be represented as:
\[ \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \]
The ion product of water \( K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} \).

### Step 9: Calculate [OH-]
Using the \( [\text{H}^+] \) from acetic acid:
\[ [\text{H}^+] = \alpha + 10^{-4} \approx 10^{-4} \]
Then,
\[ K_w = [\text{H}^+][\text{OH}^-] \]
\[ 10^{-14} = (10^{-4})([\text{OH}^-]) \]
\[ [\text{OH}^-] = \frac{10^{-14}}{10^{-4}} = 10^{-10} \]

### Step 10: Calculate the degree of dissociation of water
The degree of dissociation of water can be calculated as:
\[ \text{Degree of dissociation} = \frac{[\text{OH}^-]}{C_{H_2O}} \]
Where \( C_{H_2O} \approx 55.5 \, \text{M} \):
\[ \text{Degree of dissociation} = \frac{10^{-10}}{55.5} \approx 1.8 \times 10^{-12} \]

### Final Answer:
The degree of dissociation of water in 0.01 M acetic acid at 25°C is approximately \( 1.8 \times 10^{-12} \).

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