1. Lactic acid is a carboxylic acid with a Ka = 8.3 x 10-4. Calculate the pH of

Question

1.

Lactic acid is a carboxylic acid with a Ka = 8.3 x 10-4.

Calculate the pH of a 1.000 L solution of 0.200 M lactic acid.

2.

From the Lactic acid question above:

How would the pH change if it were a 2.000 L solution of 0.200 M lactic acid? (This is not a dilution of the previous solution)

3. Continuing further with the Lactic acid question: If the 0.2 M lactic acid solution were heated from room 298 K to 323 K, how would the pH of the solution change?

a.) It would increase slightly

b.) It would decrease slightly

c.) It would stay the same

d.) There is not enough information provided to answer this question

4. Continuing further with the Lactic acid question:

If the 0.2 M lactic acid solution were heated from room 298 K to 323 K, how would the pH of the solution change?

There is not enough information provided to answer this question

5.

Continuing even further with lactic acid: If the 0.2 M lactic acid solution had the pH decrease with the temperature increase described above, which of the following would be true.

Explanation / Answer

Ans 1

The formula for Ka is

Ka = [H+][B-]/[HB]

where
[H+] = concentration of H+ ions
[B-] = concentration of conjugate base ions
[HB] = concentration of undissociated acid molecules
for a reaction HB ? H+ + B-

Lactic acid(C3H6O3) dissociates one H+ ion for every C3H5O3- ion, so [H+] = [C3H5O3-].

Let x represent the concentration of H+ that dissociates from HB, then [HB] = C - x where C is the initial concentration.

Enter these values into the Ka equation

Ka = x · x / (C -x)
Ka = x²/(C - x)
(C - x)Ka = x²
x² = CKa - xKa
x² + Kax - CKa = 0

Solve for x using the quadratic equation

x = [-b ± (b² - 4ac)½]/2a

x = [-Ka + (Ka² + 4CKa)½]/2

Note: there are two solutions for x. Since x represents a concentration of ions in solution, the value for x cannot be negative.

Enter values for Ka and C

Ka = 8.3 x 10-4
C = 0.2 M

x = {-8.3 x 10-4 + [(8.3 x 10-4)² + 4(0.2)(8.3 x 10-4)]½}/2
x = (-8.3 x 10-4 + 1510 x 10-4)/2
x = (1501.7 x 10-4)/2
x = 0.0750

Find pH

pH = -log[H+]

pH = -log(x)
pH = -log(0.0750)
pH = -(-1.124)
pH =1.12

The same procedure can be followed for other questions with some modifications.

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