Question

Problem 2 - 12 pts You want to prepare a buffer around a pH of 3.90, so you choose lactic acid (C3H.Os) and its conjugate base. The K. of lactic acid is 1.4 x 10 A. What is the formula of the conjugate base of lactic acid? (2 pts) B. In order to prepare your buffer at the intended pH of 3.90, you start with a solution of 1.25 moles of lactic acid. How many moles of the conjugate base do you need to add to reach a pH of 3.90? (5 pts) C. If you add 10.0 mL of a 0.85 M solution of HCl to your buffered solution, what will the resulting pH be? (5 pts)

Answer 1

A.

Formula of lactic acid = C₃H₆O₃

After removal of a proton whatever remains of the acid is the conjugate base of that acid. The conjugate base of lactic acid is the lactate anion.

Hence, the formula of the conjugate base = C₃H₅O₃^-

B.

Let us say that the moles of conjugate base need to be added = Z mol

Let us say that the volume of the solution = X L

Moles of lactic acid = 1.25 mol

Thus, [C₃H₆O₃] = 1.25 mol/ X L = 1.25/X M

[C₃H₅O₃^-] = Z mol/X L = Z/X M

Hence,

[C₃H₅O₃^-]/[C₃H₆O₃] = Z/1.25

From Henderson-Hasselbalch equation,

pH = pK_a + log([base]/[acid])

or, pH = - logK_a + log([C₃H₅O₃^-]/[C₃H₆O₃])

or, 3.90 = - log(1.4 x 10^-4) + log(Z/1.25)

or, 3.90 = 3.85 + log(Z/1.25)

or, log(Z/1.25) = 0.05

or, Z/1.25 = 10^0.05

or, Z = 1.40

Hence, the moles of conjugate base needed = 1.40 mol

C.

Moles of HCl added = 10 mL x 0.85 M = 8.5 mmol = 0.0085 mol

0.0085 mol of HCl reacts with 0.0085 mol of lactate to form 0.0085 mol of lactic acid.

Moles of lactic acid = (1.25 + 0.0085) mol = 1.2585 mol

Moles of lactate = (1.40 - 0.0085) mol = 1.3915 mol

Let us say that the volume of the solution = Y L

Thus,

[C₃H₆O₃] = 1.2585 mol/ Y L = 1.2585/Y M

[C₃H₅O₃^-] = 1.3915 mol/Y L = 1.3915/Y M

Hence,

[C₃H₅O₃^-]/[C₃H₆O₃] = 1.3915/1.2585

From Henderson-Hasselbalch equation,

pH = pK_a + log([base]/[acid])

or, pH = - logK_a + log([C₃H₅O₃^-]/[C₃H₆O₃])

or, pH = - log(1.4 x 10^-4) + log(1.3915/1.2585)

or, pH = 3.85 + 0.044

or, pH = 3.89

Hence, the pH of the solution after the addition of HCl = 3.89