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