A titration of 100.0 mL of 1.00 M malonic acid (H₂A) was done with 1.00 M NaOH. For malonic acid, K_a1 = 1.49 × 10^-2, K_a2 = 2.03 × 10^-6.
-Calculate the [H⁺] after 50.00 mL of 1.00 M NaOH has been added.

Question

Quizplus · Accepted Answer

Malonic acid is a diprotic acid, meaning it can donate two protons (H+) in solution. The balanced chemical equation for the reaction between malonic acid and NaOH is:

HS1U1P1+S1S1P0(aq) + 2NaOH(aq) → Na2S1U1P1+S1S1P0(aq) + 2H2O(l)

This equation shows that for every mole of NaOH added, two moles of H+ are neutralized. Therefore, to calculate the concentration of HS1U1P1+S1S1P0 after 50.00 mL of 1.00 M NaOH has been added, we need to first calculate how many moles of NaOH were added:

1.00 mol/L × 50.00 mL = 0.0500 mol NaOH

Since the reaction is 1:1 between NaOH and H+, this means that 0.0500 moles of H+ were neutralized. However, we don't know how many of these moles came from the first proton of malonic acid (HS1U1P1+S1S1P0H) and how many came from the second proton (S1S1P0S1U1B1a1S1U1B0). To find out, we need to know the pKa values for each proton.

The pKa values for malonic acid are given by:

pKa1 = 2.83
pKa2 = 5.69

These values tell us that the first proton is much more acidic than the second, so we can assume that most of the H+ that were neutralized came from HS1U1P1+S1S1P0H.

To calculate the concentration of HS1U1P1+S1S1P0 after the addition of NaOH, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

where [A-] is the concentration of the conjugate base (in this case, S1S1P0S1U1B1a1S1U1B0), and [HA] is the concentration of the acid (in this case, HS1U1P1+S1S1P0H). We can rearrange this equation to solve for [A-]:

[A-]/[HA] = 10^(pH - pKa)

At the halfway point of the titration, the volume of NaOH added is 50.00 mL, which corresponds to a total volume of 150.00 mL (since we started with 100.0 mL of malonic acid). At this point, we can assume that most of the H+ have been neutralized by NaOH, so the pH is close to the pKa value of the second proton (pKa2 = 5.69). We can use this value to calculate [A-]/[HA]:

[A-]/[HA] = 10^(7.00 - 5.69) = 3.87

Since the total concentration of malonic acid is 1.00 M, we can use this ratio to calculate the concentration of S1S1P0S1U1B1a1S1U1B0:

[A-] = 3.87/[3.87 + 1] × 1.00 M ≈ 0.795 M

Therefore, the concentration of HS1U1P1+S1S1P0 after 50.00 mL of 1.00 M NaOH has been added is:

[HS1U1P1+S1S1P0] = [HA] = (1.00 - 0.795) M = 0.205 M

This corresponds to a pH of about 2.57, which is consistent with most of the H+ being neutralized at this point in the titration. The closest answer choice is C) 1.49 × 10^-2 M, which is the pKa value of the first proton multiplied by 10^-2. This value is not particularly useful for calculating the concentration of HS1U1P1+S1S1P0 at the halfway point of the titration because most of the H+ have been neutralized by NaOH by this point. Therefore, we should use the Henderson-Hasselbalch equation and the pKa values to calculate the concentration of [HS1U1P1+S1S1P0].

A Titration of 100