A titration of 100.0 mL of 1.00 M malonic acid (H 2 A) was done with 1.00 M NaOH. For malonic acid, K a1 = 1.49 × 10 -2 , K a2 = 2.03 × 10 -6 . -Calculate [H⁺] after 150.0 mL of 1.00 M NaOH has been added. A) 2.03 × 10^-6 M B) 1.41 × 10^-10 M C) 1.49 × 10^-2 M D) 1.00 × 10^-7 M E) none of these

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The correct answer is A.The titration of malonic acid with NaOH is a diprotic acid-base reaction. The first equivalence point is reached when 1 mole of NaOH has been added for every mole of malonic acid. At this point, all of the malonic acid has been converted to the hydrogen malonate ion (HS1U1B12S1U1B0A-). The second equivalence point is reached when 2 moles of NaOH have been added for every mole of malonic acid. At this point, all of the hydrogen malonate ion has been converted to the malonate ion (S1U1B12S1U1B0A2-).The pH of the solution at the first equivalence point can be calculated using the Henderson-Hasselbalch equation:```pH = pKa + log([A-]/[HA])```where pKa is the dissociation constant of the acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.For malonic acid, the first dissociation constant is KS1U1B1a1S1U1B0 = 1.49 × 10S1U1P1-2S1S1P0. At the first equivalence point, the concentration of the hydrogen malonate ion is equal to the initial concentration of malonic acid, which is 1.00 M. The concentration of the malonate ion is 0 M. Therefore, the pH at the first equivalence point is:```pH = pKa + log([A-]/[HA])pH = -log(1.49 × 10S1U1P1-2S1S1P0) + log(0)pH = 1.83```After 150.0 mL of 1.00 M NaOH has been added, the solution is past the first equivalence point and is in the buffer region. The pH of the solution can be calculated using the Henderson-Hasselbalch equation:```pH = pKa + log([A-]/[HA])```where pKa is the dissociation constant of the acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.For malonic acid, the second dissociation constant is KS1U1B1a2S1U1B0 = 2.03 × 10S1U1P1-6S1S1P0. At this point, the concentration of the hydrogen malonate ion is equal to the initial concentration of malonic acid minus the amount of NaOH that has been added, which is 1.00 M - 0.150 M = 0.850 M. The concentration of the malonate ion is equal to the amount of NaOH that has been added, which is 0.150 M. Therefore, the pH at this point is:```pH = pKa + log([A-]/[HA])pH = -log(2.03 × 10S1U1P1-6S1S1P0) + log(0.150/0.850)pH = 5.69```Therefore, the [H+] concentration after 150.0 mL of 1.00 M NaOH has been added is:```[H+] = 10^-pH[H+] = 10^-5.69[H+] = 2.03 × 10^-6 M```Therefore, the correct answer is A.

A Titration of 100