Arsenic acid (H3AsO4) is a triprotic acid with Ka1 = 5 ✕ 10-3, Ka2 = 8 ✕ 10-8, and Ka3 = 6 ✕ 10-10. Calculate [H+], [OH ‾ ], [H3AsO4], [H2AsO4‾ ], [HAsO42 ‾ ], and [AsO43 ‾ ] in a 0.16 M arsenic acid solution.
To solve this problem, we will use the concept of acid-base equilibrium and the equations provided for the dissociation of arsenic acid.
Let's first define the variables for the concentrations of each species in the solution:
[H+] = concentration of H+ ions
[OH-] = concentration of OH- ions
[H3AsO4] = concentration of H3AsO4
[H2AsO4-] = concentration of H2AsO4-
[HAsO4^2-] = concentration of HAsO4^2-
[AsO4^3-] = concentration of AsO4^3-
Given:
Ka1 = 5 × 10^-3
Ka2 = 8 × 10^-8
Ka3 = 6 × 10^-10
Initial concentration of arsenic acid (H3AsO4) = 0.16 M
To solve this problem, we'll follow these steps:
Step 1: Calculate the concentration of [H+] using Ka1.
Step 2: Apply the concept of water dissociation to calculate the concentration of [OH-].
Step 3: Calculate the concentrations of [H3AsO4] and [H2AsO4-] using the [H+] and the dissociation constants Ka1 and Ka2.
Step 4: Calculate the concentrations of [HAsO4^2-] and [AsO4^3-] using the [H+] and the dissociation constants Ka2 and Ka3.
Let's calculate each step:
Step 1: Calculate [H+] using Ka1.
[H+] = √(Ka1 × [H3AsO4])
[H+] = √(5 × 10^-3 × 0.16)
[H+] = √(8 × 10^-5)
[H+] ≈ 8.94 × 10^-3 M
Step 2: Calculate [OH-] using the concept of water dissociation.
[H+] × [OH-] = 1 × 10^-14 (at 25°C)
[OH-] = 1 × 10^-14 / [H+]
[OH-] ≈ 1 × 10^-14 / (8.94 × 10^-3)
[OH-] ≈ 1.12 × 10^-12 M
Step 3: Calculate [H3AsO4] and [H2AsO4-] using [H+] and Ka1 and Ka2.
[H3AsO4] = [H+] × [H2AsO4-] / Ka1
[H3AsO4] ≈ (8.94 × 10^-3 × 8.94 × 10^-3) / (5 × 10^-3)
[H3AsO4] ≈ 1.60 × 10^-5 M
[H2AsO4-] = [H+]^2 / Ka1
[H2AsO4-] ≈ (8.94 × 10^-3)^2 / (5 × 10^-3)
[H2AsO4-] ≈ 1.58 × 10^-2 M
Step 4: Calculate [HAsO4^2-] and [AsO4^3-] using [H+] and Ka2 and Ka3.
[HAsO4^2-] = [H+] × [AsO4^3-] / Ka2
[HAsO4^2-] ≈ (8.94 × 10^-3 × 1.58 × 10^-2) / (8 × 10^-8)
[HAsO4^2-] ≈ 1.79 × 10^-2 M
[AsO4^3-] = [H+]^2 / Ka2
[AsO4^3-] ≈ (8.94 × 10^-3)^2 / (8 × 10^-8)
[AsO4^3-] ≈ 1 × 10^-7 M
Therefore, in a 0.16 M arsenic acid solution, the concentrations of each species are:
[H+] ≈ 8.94 × 10^-3 M
[OH-] ≈ 1.12 × 10^-12 M
[H3AsO4] ≈ 1.60 × 10^-5 M
[H2AsO4-] ≈ 1.58 × 10^-2 M
[HAsO4^2-] ≈ 1.79 × 10^-2 M
[AsO4^3-] ≈ 1 × 10^-7 M