Question

Arsenic acid \(\left(\mathrm{H}_{3} \mathrm{AsO}_{4}\right)\) is a triprotic acid with \(K_{\mathrm{a}_{1}}=5 \times 10^{-3}\) \(K_{\mathrm{a}_{2}}=8 \times 10^{-8}\), and \(K_{\mathrm{a}_{3}}=6 \times 10^{-10} \cdot\) Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\) \(\left[\mathrm{H}_{3} \mathrm{AsO}_{4}\right],\left[\mathrm{H}_{2} \mathrm{AsO}_{4}^{-}\right],\left[\mathrm{HAsO}_{4}^{2-}\right]\), and \(\left[\mathrm{AsO}_{4}^{3-}\right]\) in a \(0.20 \mathrm{M}\) arsenic acid solution.

Short answer

In a \(0.20\,M\) arsenic acid solution, the concentrations of each species are: \(\left[\mathrm{H}^{+}\right] \approx 0.0316\, \mathrm{M}\) \(\left[\mathrm{OH}^{-}\right] \approx 3.16 \times 10^{-14}\, \mathrm{M}\) \(\left[\mathrm{H}_{3} \mathrm{AsO}_{4}\right] \approx 0.1684\, \mathrm{M}\) \(\left[\mathrm{H}_{2} \mathrm{AsO}_{4}^{-}\right] \approx 0.0316\, \mathrm{M}\) \(\left[\mathrm{HAsO}_{4}^{2-}\right] \approx 2.53 \times 10^{-6}\, \mathrm{M}\) \(\left[\mathrm{AsO}_{4}^{3-}\right] \approx 1.90 \times 10^{-8}\, \mathrm{M}\)

Step-by-step solution

Write the dissociation reactions for arsenic acid

First, let's write out the three dissociation reactions for arsenic acid: \(1^{st}\) dissociation: \[H_{3}AsO_{4}(aq) \rightleftharpoons H^{+}(aq) + H_{2}AsO_{4}^{-}(aq)\] \(2^{nd}\) dissociation: \[H_{2}AsO_{4}^{-}(aq) \rightleftharpoons H^{+}(aq) + HAsO_{4}^{2-}(aq)\] \(3^{rd}\) dissociation: \[HAsO_{4}^{2-}(aq) \rightleftharpoons H^{+}(aq) + AsO_{4}^{3-}(aq)\]

Calculate the \(\left[\mathrm{H}^{+}\right]\) from the first dissociation

Since \(K_{a_{1}}\) is much larger than \(K_{a_{2}}\) and \(K_{a_{3}}\), we can make the assumption that the first dissociation of arsenic acid contributes the most to the \(\left[\mathrm{H}^{+}\right]\) concentration. We can therefore ignore the contributions from the subsequent dissociations initially. Let the change in concentration of \(H^{+}\) be x, then: \[\frac{[H^{+}][H_{2}AsO_{4}^{-}]}{[H_{3}AsO_{4}]} = K_{a_{1}}\] Substitute known values and variables in the expression: \[\frac{(x)(x)}{(0.20 - x)} = 5 \times 10^{-3}\] Considering the large difference in equilibrium constants, we can assume that \(x << 0.20\), thus the equation becomes: \[\frac{(x)(x)}{0.20} = 5 \times 10^{-3}\] Now, solve for x: \[x = \sqrt{(5 \times 10^{-3})(0.20)}\] \[x \approx 0.0316 \;\mathrm{M}\] So, \(\left[\mathrm{H}^{+}\right] \approx 0.0316 \;\mathrm{M}\).

Calculate the concentrations of other species

Now that we have found the \(\left[\mathrm{H}^{+}\right]\) concentration, we can use it to find the concentrations of the other species in the solution: 1. \(\left[\mathrm{H}_{3} \mathrm{AsO}_{4}\right] = 0.20 - 0.0316 \approx 0.1684 \;\mathrm{M}\) (considering only the first dissociation) 2. \(\left[\mathrm{H}_{2} \mathrm{AsO}_{4}^{-}\right] \approx 0.0316 \;\mathrm{M}\) (since it was produced in the first dissociation) Now, for the second dissociation, consider the change in concentration of \(H^{+}\) to be y: \[\frac{([H^{+}+y][HAsO_{4}^{2-}])}{[H_{2}AsO_{4}^{-} - y]} = K_{a_{2}}\] Since we already determined that the second and third dissociation constants are much smaller than the first one, we can safely assume that y is negligible compared to \(0.0316 \;\mathrm{M}\): \[\frac{([0.0316][HAsO_{4}^{2-}])}{[0.0316 - y]} \approx K_{a_{2}}\] which gives: \[HAsO_{4}^{2-} = \frac{K_{a_{2}}([0.0316 - y])}{0.0316} \approx \frac{K_{a_{2}}}{0.0316}\] we have: \(\left[\mathrm{HAsO}_{4}^{2-}\right] \approx \frac{8 \times 10^{-8}}{0.0316} \approx 2.53 \times 10^{-6} \;\mathrm{M}\) Finally, for the third dissociation, we can similarly calculate: \(\left[\mathrm{AsO}_{4}^{3-}\right] \approx \frac{K_{a_{3}}}{0.0316} \approx 1.90 \times 10^{-8} \;\mathrm{M}\)

Calculate the \(\left[\mathrm{OH}^{-}\right]\) concentration using the \(\left[\mathrm{H}^{+}\right]\) concentration

The relationship between \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) is given by the ion product of water: \[K_{w} = [H^{+}][OH^{-}]\] Using \(K_{w} = 1 \times 10^{-14}\) at 25°C, we can solve for \(\left[\mathrm{OH}^{-}\right]\): \[\left[\mathrm{OH}^{-}\right] = \frac{K_{w}}{\left[\mathrm{H}^{+}\right]} = \frac{1 \times 10^{-14}}{0.0316}\] \[\left[\mathrm{OH}^{-}\right] \approx 3.16 \times 10^{-14} \;\mathrm{M}\]

Final Results

Thus, the concentrations of each species in the 0.20 M arsenic acid solution are: \(\left[\mathrm{H}^{+}\right] \approx 0.0316 \;\mathrm{M}\) \(\left[\mathrm{OH}^{-}\right] \approx 3.16 \times 10^{-14} \;\mathrm{M}\) \(\left[\mathrm{H}_{3} \mathrm{AsO}_{4}\right] \approx 0.1684 \;\mathrm{M}\) \(\left[\mathrm{H}_{2} \mathrm{AsO}_{4}^{-}\right] \approx 0.0316 \;\mathrm{M}\) \(\left[\mathrm{HAsO}_{4}^{2-}\right] \approx 2.53 \times 10^{-6} \;\mathrm{M}\) \(\left[\mathrm{AsO}_{4}^{3-}\right] \approx 1.90 \times 10^{-8} \;\mathrm{M}\)

Key concepts

Triprotic Acid

When we refer to an acid as being 'triprotic,' we're talking about its ability to donate three protons (hydrogen ions, H+) per molecule to other substances, usually water when dissolved. Arsenic acid, symbolized as \(H_{3}AsO_{4}\), is a perfect example of a triprotic acid. Each of its hydrogen atoms can dissociate, or separate from the molecule, in a step-wise fashion. This step-wise process is important because it indicates that the dissociation occurs in stages, with each stage having its own characteristic equilibrium constant, represented as \(K_{a}\). These constants are crucial for understanding the extent to which each proton dissociation occurs in a given acid concentration and are fundamental in performing advanced calculations related to the acid's behavior in solution.

Understanding the nature of triprotic acids is essential, especially when intending to solve exercises involving pH calculations and when studying the acid's ionic equilibrium. Each dissociation stage can potentially contribute to the overall hydrogen ion concentration, which, in turn, affects the solution's pH. It's also why, in the case of arsenic acid, we typically focus on the first dissociation, since it's the most significant due to its relatively larger equilibrium constant.

Acid Dissociation Constant

The acid dissociation constant, or \(K_{a}\), is a quantitative measure of the strength of an acid in solution. It's the equilibrium constant for the chemical reaction where an acid donates a proton to water, forming its conjugate base and a hydronium ion (H3O+). The larger the value of \(K_{a}\), the stronger the acid and the more it will dissociate at equilibrium. In the case of arsenic acid, we are given three constants, one for each dissociation step: \(K_{a1} = 5 \times 10^{-3}\), \(K_{a2} = 8 \times 10^{-8}\), and \(K_{a3} = 6 \times 10^{-10}\).

These values reflect the decreasing tendency to donate protons with each successive dissociation step. Initially, arsenic acid readily gives up the first proton, indicated by the relatively higher \(K_{a1}\). The following protons are held more tightly, as evidenced by the much smaller \(K_{a2}\) and \(K_{a3}\). When computing the pH or analyzing the ionic equilibrium of arsenic acid, these constants are invaluable as they directly influence how we calculate the concentrations of each ion at equilibrium.

Ionic Equilibrium

Ionic equilibrium refers to the state of balance where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of the reactants and products. This is particularly important in the context of acids and bases, where the equilibrium lies between the undissociated acid or base and the ions produced by their dissociation in water. When discussing arsenic acid, the ionic equilibrium is described by three separate equilibrium equations, corresponding to each of the dissociations.

Understanding ionic equilibrium allows chemists to predict how the acid will behave under different conditions, such as changes in concentration or temperature. In our calculations for arsenic acid, we made the assumption that due to the respective magnitudes of the dissociation constants, the primary contribution to the free hydrogen ion concentration comes from the first dissociation step, thus simplifying the overall equilibrium consideration and highlighting the sequential nature of triprotic acid dissociation.

pH Calculation

Calculating the pH of a solution involves determining the power of hydrogen, or the hydrogen ion concentration \(\left[\mathrm{H}^{+}\right]\). It's a logarithmic scale measurement of the acidity or basicity of an aqueous solution. pH is calculated using the formula \(-\log\left[\mathrm{H}^{+}\right]\), where \(\left[\mathrm{H}^{+}\right]\) is the concentration of hydrogen ions in moles per liter. In the case of arsenic acid, the concentration of hydrogen ions primarily comes from the first dissociation because it has the highest \(K_{a}\), thus contributing the most to the solution's acidity.

After finding the \(\left[\mathrm{H}^{+}\right]\), we can calculate the hydroxide ion concentration \(\left[\mathrm{OH}^{-}\right]\) using the ion product of water (\(K_{w}\)), and ultimately determine the pH of the solution. The knowledge of pH is crucial as it affects various chemical and biological processes, helping in understanding the behavior of the solution in different contexts, such as environmental or biological systems.