Calculate the percent dissociation of acetic acid (Ka=1.8×10^-5) in: A) 1.00 M HC2H3O2 B) 0.100 M C2H3O2H C) For solutions of any weak acid HA, H+ decreases as HA decreases. But the percent dissociation increases as HA decreases. Explain why this is the case.
The Correct Answer and Explanation is:
To calculate the percent dissociation of acetic acid (HC₂H₃O₂), we use the acid dissociation constant (Ka), which is given as 1.8 × 10⁻⁵. We also need to apply an ICE table (Initial, Change, Equilibrium) for each concentration of acetic acid.
A) For 1.00 M HC₂H₃O₂:
- Write the dissociation equation:
HC₂H₃O₂ ⇌ H⁺ + C₂H₃O₂⁻ - Set up the ICE table:
| HC₂H₃O₂ | H⁺ | C₂H₃O₂⁻ | |
|---|---|---|---|
| Initial | 1.00 M | 0 | 0 |
| Change | -x | +x | +x |
| Equilibrium | 1.00 – x | x | x |
- Write the expression for Ka:
Ka=[H+][C2H3O2−][HC2H3O2]Ka = \frac{[H⁺][C₂H₃O₂⁻]}{[HC₂H₃O₂]}Ka=[HC2H3O2][H+][C2H3O2−]
Substitute the equilibrium concentrations:1.8×10−5=x21.00−x1.8 \times 10^{-5} = \frac{x^2}{1.00 – x}1.8×10−5=1.00−xx2
Since KaKaKa is small, we can assume that xxx will be much smaller than 1.00, so 1.00−x≈1.001.00 – x \approx 1.001.00−x≈1.00. Therefore:1.8×10−5=x21.001.8 \times 10^{-5} = \frac{x^2}{1.00}1.8×10−5=1.00×2
Solve for xxx:x2=1.8×10−5x^2 = 1.8 \times 10^{-5}x2=1.8×10−5x=1.8×10−5≈4.24×10−3 Mx = \sqrt{1.8 \times 10^{-5}} \approx 4.24 \times 10^{-3} \, \text{M}x=1.8×10−5≈4.24×10−3M
The percent dissociation is given by:Percent dissociation=x[HC2H3O2]×100=4.24×10−31.00×100≈0.424%\text{Percent dissociation} = \frac{x}{[HC₂H₃O₂]} \times 100 = \frac{4.24 \times 10^{-3}}{1.00} \times 100 \approx 0.424\%Percent dissociation=[HC2H3O2]x×100=1.004.24×10−3×100≈0.424%
B) For 0.100 M HC₂H₃O₂:
- Set up the ICE table:
| HC₂H₃O₂ | H⁺ | C₂H₃O₂⁻ | |
|---|---|---|---|
| Initial | 0.100 M | 0 | 0 |
| Change | -x | +x | +x |
| Equilibrium | 0.100 – x | x | x |
- Write the expression for Ka:
Ka=x20.100−xKa = \frac{x^2}{0.100 – x}Ka=0.100−xx2
Assume xxx is small compared to 0.100, so 0.100−x≈0.1000.100 – x \approx 0.1000.100−x≈0.100:1.8×10−5=x20.1001.8 \times 10^{-5} = \frac{x^2}{0.100}1.8×10−5=0.100×2
Solve for xxx:x2=1.8×10−5×0.100=1.8×10−6x^2 = 1.8 \times 10^{-5} \times 0.100 = 1.8 \times 10^{-6}x2=1.8×10−5×0.100=1.8×10−6x=1.8×10−6≈1.34×10−3 Mx = \sqrt{1.8 \times 10^{-6}} \approx 1.34 \times 10^{-3} \, \text{M}x=1.8×10−6≈1.34×10−3M
The percent dissociation is:Percent dissociation=x[HC2H3O2]×100=1.34×10−30.100×100=1.34%\text{Percent dissociation} = \frac{x}{[HC₂H₃O₂]} \times 100 = \frac{1.34 \times 10^{-3}}{0.100} \times 100 = 1.34\%Percent dissociation=[HC2H3O2]x×100=0.1001.34×10−3×100=1.34%
C) Explanation of Why Percent Dissociation Increases as HA Decreases:
As the concentration of a weak acid HA decreases, its dissociation becomes more significant in terms of the total concentration of acid. In a solution of weak acid, the dissociation is governed by an equilibrium where the concentration of dissociated ions is very small compared to the undissociated acid. At higher concentrations of HA, the dissociation is suppressed because the equilibrium favors the undissociated acid.
However, as the concentration of HA decreases (such as in more dilute solutions), the concentration of undissociated acid decreases, so the relative amount of dissociation increases. In other words, with less HA present, a larger proportion of the acid dissociates to maintain the equilibrium, leading to a higher percent dissociation. This explains why percent dissociation increases as the concentration of the acid decreases.
