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 0.53 M arsenic acid solution.
To solve this problem, we will use the principles of acid-base equilibrium and consider the stepwise ionization of arsenic acid.
Step 1: Determine the concentration of [H3AsO4] using the given molarity of the arsenic acid solution.
[H3AsO4] = 0.53 M
Step 2: Calculate the concentration of [H+] using the first ionization constant (Ka1).
Ka1 = [H+][H2AsO4-] / [H3AsO4]
[H+] = sqrt(Ka1 * [H3AsO4])
[H+] = sqrt(5 * 10^-3 * 0.53)
[H+] = 1.28 * 10^-2 M
Step 3: Calculate the concentration of [H2AsO4-] using the equation for second ionization constant (Ka2).
Ka2 = [H+][HAsO4^2-] / [H2AsO4-]
[H2AsO4-] = Ka1 * [H3AsO4] / [H+]
[H2AsO4-] = 5 * 10^-3 * 0.53 / (1.28 * 10^-2)
[H2AsO4-] = 2.08 * 10^-3 M
Step 4: Calculate the concentration of [HAsO4^2-] using the equation for the third ionization constant (Ka3).
Ka3 = [H+][AsO4^3-] / [HAsO4^2-]
[HAsO4^2-] = Ka2 * [H2AsO4-] / [H+]
[HAsO4^2-] = 8 * 10^-8 * 2.08 * 10^-3 / (1.28 * 10^-2)
[HAsO4^2-] = 1.31 * 10^-10 M
Step 5: Calculate the concentration of [AsO4^3-] using the equation for the third ionization constant (Ka3).
[AsO4^3-] = Ka3 * [HAsO4^2-] / [H+]
[AsO4^3-] = 6 * 10^-10 * 1.31 * 10^-10 / (1.28 * 10^-2)
[AsO4^3-] = 6.09 * 10^-21 M
Step 6: Calculate the concentration of [OH-] using the equation for water ionization constant (Kw).
Kw = [H+][OH-]
[OH-] = Kw / [H+]
Kw at 25°C is 1×10^−14
[OH-] = 1 * 10^-14 / (1.28 * 10^-2)
[OH-] = 7.81 * 10^-13 M
Therefore, in a 0.53 M arsenic acid solution, the concentrations are as follows:
[H+] = 1.28 * 10^-2 M
[OH-] = 7.81 * 10^-13 M
[H3AsO4] = 0.53 M
[H2AsO4-] = 2.08 * 10^-3 M
[HAsO4^2-] = 1.31 * 10^-10 M
[AsO4^3-] = 6.09 * 10^-21 M