Chemistry What is the percent ionization of a monoprotic weak acid solution that is 0.154 M? The acid-dissociation (or ionization) constant, Ka, of this acid is 4.92×10-10 percent ionization:
The Correct Answer and Explanation is :
The percent ionization of the monoprotic weak acid solution is 0.00565%.
Explanation:
The percent ionization of a weak acid refers to the fraction of the acid molecules that ionize in solution. It is calculated using the formula:
[
\% \text{Ionization} = \left(\frac{[\text{H}^+]}{C} \right) \times 100
]
where:
- ([\text{H}^+]) is the concentration of hydrogen ions at equilibrium.
- (C) is the initial concentration of the acid.
For a weak acid, the equilibrium concentration of (\text{H}^+) is approximated using:
[
[\text{H}^+] \approx \sqrt{K_a \times C}
]
Substituting the given values:
- (K_a = 4.92 \times 10^{-10})
- (C = 0.154)
[
[\text{H}^+] = \sqrt{(4.92 \times 10^{-10}) \times (0.154)}
]
[
[\text{H}^+] \approx 9.29 \times 10^{-6} \text{ M}
]
Now, calculating the percent ionization:
[
\% \text{Ionization} = \left(\frac{9.29 \times 10^{-6}}{0.154} \right) \times 100
]
[
\% \text{Ionization} \approx 0.00565\%
]
Conclusion:
Since the acid is weak, only a very small fraction of its molecules ionize in solution. The extremely low (K_a) value confirms that the acid is only minimally ionized, leading to a percent ionization of 0.00565%. This shows that in a relatively concentrated solution, weak acids ionize even less compared to more diluted solutions.
Here is a scientific illustration of a weak acid solution undergoing ionization, showing partial dissociation into H⁺ and A⁻ ions with equilibrium notations. Let me know if you need any modifications or further explanations!
