Sure, let's discuss the problem and find the percentage of acid dissociated at equilibrium using Ostwald's dilution law.
Ostwald's Dilution Law:
Ostwald's dilution law relates the dissociation constant of a weak acid (or base) to its degree of dissociation and concentration. For a weak acid HA dissociating in water:
\(\[ \text{HA} \leftrightharpoons \text{H}^+ + \text{A}^- \]\)
where:
- \(\( \text{HA} \)\) is the weak acid,
- \(\( \text{H}^+ \)\) is the hydrogen ion,
- \(\( \text{A}^- \)\) is the conjugate base.
The acid dissociation constant \(\( K_a \)\) for the weak acid can be expressed as:
\(\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]\)
Given the data:
- \(\( K_a = 1.0 \times 10^{-5} \)\)
- Initial concentration of the weak acid \(\( [\text{HA}]_0 = 0.1 \)\) moles per liter
Let \(\( x \)\) be the concentration of hydrogen ions \(\( [\text{H}^+] \)\) and acetate ions \(\( [\text{A}^-] \)\) at equilibrium. Therefore, at equilibrium:
- \(\( [\text{H}^+] = x \)\)
- \(\( [\text{A}^-] = x \)\)
- Remaining \(\( [\text{HA}] = 0.1 - x \)\)
The equilibrium expression becomes:
\(\[ K_a = \frac{x^2}{0.1 - x} \]\)
We can solve for \(\( x \)\) by rearranging the equation:
\(\[ 1.0 \times 10^{-5} = \frac{x^2}{0.1 - x} \]\)
By solving this quadratic equation for \(\( x \)\):
\(\[ x = \text{concentration of } [\text{H}^+] \text{ at equilibrium} \]\)
The numerical results show that:
- \(\( x \approx 0.000995 \)\) moles/liter
To find the percentage of dissociation, we use the formula:
\(\[ \text{Percentage dissociation} = \left( \frac{\text{concentration of } [\text{H}^+]}{\text{initial concentration of } \text{HA}} \right) \times 100 \]\)
Substituting the values:
\(\[ \text{Percentage dissociation} = \left( \frac{0.000995}{0.1} \right) \times 100 \]\)
\(\[ \text{Percentage dissociation} \approx 0.995\% \]\)
So, the percentage of the acid dissociated at equilibrium is approximately \(\( 0.995\% \)\).
