The triprotic acid H3A has ionization constants of Ka1 = 2.5× 10–4, Ka2 = 3.4× 10–8, and Ka3 = 5.4× 10–13. Calculate the following values for a 0.0760 M solution of NaH2A
To calculate the values for a 0.0760 M solution of NaH2A, we need to determine the concentrations of the various species present in solution.
1. Calculate the concentration of NaH2A:
Since the solution concentration is given as 0.0760 M, the concentration of NaH2A is also 0.0760 M.
2. Calculate the concentration of HA-:
NaH2A dissociates to form Na+ and H2A-. Since NaH2A is a monoprotic acid, it fully dissociates, meaning the concentration of H2A- is equal to the concentration of NaH2A. Therefore, [HA-] = 0.0760 M.
3. Calculate the concentration of H2A:
H2A- can further dissociate to form H+ and HA2-. To calculate the concentration of HA2-, we need to consider the dissociation constant Ka1.
Ka1 = [H+][A-]/[HA-]
Since [H+] = [HA2-] (initially), and [A-] = [HA-] (initially), we can rewrite the equation as:
Ka1 = [H+]^2 /[HA-]
Rearranging the equation, we have:
[H2A] = sqrt(Ka1 * [HA-])
Plugging in the values, we get:
[H2A] = sqrt(2.5×10^(-4) * 0.0760) ≈ 0.00349 M
4. Calculate the concentration of HA2-:
We can use the same approach as in step 3, but now we consider Ka2.
Ka2 = [H+][HA-]/[HA2-]
Since [H+] = [HA3-] (initially), and [HA-] = [H2A] (initially), we can rewrite the equation as:
Ka2 = [H+]^2 /[HA-]
Rearranging the equation, we have:
[HA2-] = sqrt(Ka2 * [HA-])
Plugging in the values, we get:
[HA2-] = sqrt(3.4×10^(-8) * 0.00349) ≈ 4.74x10^(-5) M
5. Calculate the concentration of A3-:
A3- can further dissociate to form H+ and A2-. To calculate the concentration of A3-, we need to consider the dissociation constant Ka3.
Ka3 = [H+][HA2-]/[A3-]
Since [H+] = [A-] (initially), and [HA2-] = [HA3-] (initially), we can rewrite the equation as:
Ka3 = [H+]^2 /[A-]
Rearranging the equation, we have:
[A3-] = sqrt(Ka3 * [A-])
Plugging in the values, we get:
[A3-] = sqrt(5.4×10^(-13) * 0.0760) ≈ 6.20x10^(-8) M
These are the concentrations of the different species present in the 0.0760 M solution of NaH2A:
[NaH2A] = 0.0760 M
[HA-] = 0.0760 M
[H2A] ≈ 0.00349 M
[HA2-] ≈ 4.74x10^(-5) M
[A3-] ≈ 6.20x10^(-8) M