Calculate the pH of a solution prepared by mixing 200.0 mL of 0.100 M NaOH with 150.0 mL of
0.200 M oxalic acid (H2A). Ka1 = 5.62 × 10−2
, Ka2 = 7.24 × 10−5
To solve this problem, we need to determine the concentration of H+ ions in the solution.
First, let's calculate the moles of NaOH:
moles of NaOH = volume of NaOH (in L) x concentration of NaOH
moles of NaOH = 0.200 L x 0.100 M
moles of NaOH = 0.020 mol
Next, let's calculate the moles of H2A:
moles of H2A = volume of H2A (in L) x concentration of H2A
moles of H2A = 0.150 L x 0.200 M
moles of H2A = 0.030 mol
Since oxalic acid is a diprotic acid, it can donate two protons (H+) in a reaction. The first hydrogen ion has Ka1 = 5.62 x 10^-2, and the second hydrogen ion has Ka2 = 7.24 x 10^-5.
Let's start by calculating the concentration of the first dissociation product (H+ and H2A-) using the equation for Ka1:
Ka1 = [H+][H2A-] / [H2A]
We assume that the concentration of [H2A] is equal to the initial concentration of H2A, since the dissociation is relatively small.
5.62 x 10^-2 = [H+][H2A-] / 0.030
[H+][H2A-] = 5.62 x 10^-2 * 0.030
[H+][H2A-] = 1.686 x 10^-3
The concentration of [H2A-] is equal to the concentration of [H+], so we can rewrite the equation as:
[H+]^2 = 1.686 x 10^-3
[H+] = √(1.686 x 10^-3)
[H+] ≈ 4.1 x 10^-2
Now, let's calculate the concentration of the second dissociation product (H+ and A2-) using the equation for Ka2:
Ka2 = [H+][A2-] / [HA-]
Since the concentration of [HA-] is equal to the initial concentration of H2A, we can substitute it in the equation:
7.24 x 10^-5 = [H+][A2-] / 0.030
[H+][A2-] = 7.24 x 10^-5 * 0.030
[H+][A2-] = 2.172 x 10^-6
The concentration of [H+], which is equal to the concentration of [A2-], is:
[H+] = √(2.172 x 10^-6)
[H+] ≈ 4.7 x 10^-3
Since both Ka values are relatively small, we can assume that the concentration of [H+] is equal to the concentration of [H2A-] in the final solution.
Now, let's calculate the total volume of the solution:
Total volume = volume of NaOH + volume of H2A
Total volume = 0.200 L + 0.150 L
Total volume = 0.350 L
To find the concentration of [H+], we divide the moles of H+ by the total volume:
[H+] concentration = (moles of H+) / (total volume)
[H+] concentration = (0.020 mol + 4.7 x 10^-3 mol) / 0.350 L
[H+] concentration = 5.7 x 10^-2 M
Finally, let's calculate the pH of the solution:
pH = -log[H+]
pH = -log(5.7 x 10^-2)
pH ≈ 1.24
Therefore, the pH of the solution is approximately 1.24.