Calculate the percent ionization of arsenous acid (H3AsO3) in solutions of each of the following concentrations (Ka = 5.1e-10). (a) 0.239 M (b) 0.431 M (c) 0.827 M
The Correct Answer and Explanation is:
To calculate the percent ionization of arsenous acid (H₃AsO₃), we use the expression for a weak acid dissociation:
Step 1: Write the dissociation equation
H₃AsO₃ ⇌ H⁺ + H₂AsO₃⁻
Given:
- Ka = 5.1 × 10⁻¹⁰
Let the initial concentration be [HA] = C, and x be the amount ionized at equilibrium.
Then:
- [H⁺] = x
- [H₂AsO₃⁻] = x
- [H₃AsO₃] = C – x
The Ka expression:Ka=x2CKa = \frac{x^2}{C}Ka=Cx2
Since x is small relative to C, we approximate:x=Ka⋅Cx = \sqrt{Ka \cdot C}x=Ka⋅C
Finally, the percent ionization is:Percent Ionization=(xC)×100\text{Percent Ionization} = \left( \frac{x}{C} \right) \times 100Percent Ionization=(Cx)×100
(a) C = 0.239 M
x=5.1×10−10×0.239=1.2189×10−10≈1.104×10−5x = \sqrt{5.1 \times 10^{-10} \times 0.239} = \sqrt{1.2189 \times 10^{-10}} \approx 1.104 \times 10^{-5}x=5.1×10−10×0.239=1.2189×10−10≈1.104×10−5Percent Ionization=(1.104×10−50.239)×100≈0.00462%\text{Percent Ionization} = \left( \frac{1.104 \times 10^{-5}}{0.239} \right) \times 100 \approx 0.00462\%Percent Ionization=(0.2391.104×10−5)×100≈0.00462%
(b) C = 0.431 M
x=5.1×10−10×0.431=2.1981×10−10≈1.4826×10−5x = \sqrt{5.1 \times 10^{-10} \times 0.431} = \sqrt{2.1981 \times 10^{-10}} \approx 1.4826 \times 10^{-5}x=5.1×10−10×0.431=2.1981×10−10≈1.4826×10−5Percent Ionization=(1.4826×10−50.431)×100≈0.00344%\text{Percent Ionization} = \left( \frac{1.4826 \times 10^{-5}}{0.431} \right) \times 100 \approx 0.00344\%Percent Ionization=(0.4311.4826×10−5)×100≈0.00344%
(c) C = 0.827 M
x=5.1×10−10×0.827=4.2177×10−10≈2.053×10−5x = \sqrt{5.1 \times 10^{-10} \times 0.827} = \sqrt{4.2177 \times 10^{-10}} \approx 2.053 \times 10^{-5}x=5.1×10−10×0.827=4.2177×10−10≈2.053×10−5Percent Ionization=(2.053×10−50.827)×100≈0.00248%\text{Percent Ionization} = \left( \frac{2.053 \times 10^{-5}}{0.827} \right) \times 100 \approx 0.00248\%Percent Ionization=(0.8272.053×10−5)×100≈0.00248%
Explanation
Arsenous acid (H₃AsO₃) is a weak acid, meaning it only partially ionizes in solution. The degree of ionization is governed by its acid dissociation constant (Ka), which for H₃AsO₃ is quite small (5.1 × 10⁻¹⁰). A smaller Ka value indicates less ionization in water.
Percent ionization is the ratio of ionized acid to the initial acid concentration, expressed as a percentage. It tells us how much of the acid has donated protons to form hydronium ions. This can be calculated using the approximation valid for weak acids:x=Ka⋅Cx = \sqrt{Ka \cdot C}x=Ka⋅C
where x is the hydrogen ion concentration at equilibrium, and C is the initial acid concentration.
As the initial concentration increases, percent ionization decreases. This happens because as the concentration of the acid increases, the value of x (the amount ionized) increases more slowly than the total amount of acid present. This inverse relationship between concentration and percent ionization is a hallmark of weak acids and is due to the common ion effect and Le Châtelier’s Principle.
In all three scenarios above, the percent ionization values are very small, reflecting the weakly acidic nature of H₃AsO₃. At 0.239 M, about 0.00462 percent of the acid ionizes. As the concentration increases to 0.431 M and 0.827 M, the percent ionization drops to 0.00344 percent and 0.00248 percent respectively.
This behavior is typical of weak acids and demonstrates that the acid becomes even less ionized at higher concentrations, even though the total amount of ionized acid may be slightly higher in absolute terms.
