-51. Lactic acid is produced in our muscle cells when we work too strenuously to

Question

-51. Lactic acid is produced in our muscle cells when we work too strenuously to maintain aerobic respiration. Lactic acid buildup lowers the pH in the muscle tissues from about 7.0 to 6.5, at which point enzymes required for an anaerobic energy supply no longer function and the muscles do not work adequately. What concentra- tion of lactic acid must be present in water to create a pH of 6.5 at 25°C? 52· Th e amount of ascorbic acid (vitamin C) in clinical samples is to be determined by a colorimetric assay (a) To prepare for this analysis, a laboratory worker pre- pares a standard solution that contains 10 mg of ascor- bic acid that is dissolved in 100.00 mL of water. What is the expected pH of this standard at 25°C? (b) A second standard is prepared that contains 10 mg of ascorbate (the conjugate base of ascorbic acid) in 100.00 mL of water. What is the expected pH of thi second standard at 25°C?

Explanation / Answer

51. The desired pH of lactic acid solution = 6.5

pH = -log[H+] = 6.5

[H+] = 3.16 x 10^-7 M

Ka for lactic acid = 1.38 x 10^-4

let HA represent lactic acid then,

Ka = [H+][A-]/[HA]

[H+] = [A-]

or,

1.38 x 10^-4 = (3.16 x 10^-7)^2/[HA]

[HA] = 7.23 x 10^-10 M

So the concentration of lactic acid needed to maintain pH 6.5 is 7.23 x 10^-10 M

52. Ascorbic acid

(a) molarity of ascorbic acid solution = mols/L of solution

                                                          = 0.01 g/176.12 g/mol x 0.1 L

                                                          = 5.7 x 10^-4 M

Ka of ascorbic acid = 8 x 10^-5

let HA represent ascorbic acid

HA + H2O <==> H3O+ + A-

let x amount has dissociated

Ka = [H3O+][A-]/[HA]

8 x 10^-5 = x^2/5.7 x 10^-4

x = [H3O+] = 2.13 x 10^-4 M

pH = -log[H3O+] = -log(2.13 x 10^-4) = 3.67

So pH of solution is 3.67

(b) let A- represent ascorbate, then

[A-] molarity = 0.01 g/175.12 g/mol x 0.1 L = 5.7 x 10^-4 M

A- hydrolyzes in water

A- + H2O <==> HA + OH-

let x amount has hydrolyzed

Kb = 1 x 10^-14/8 x 10^-5 = x^2/5.7 x 10^-4

x = [OH-] = 2.67 x 10^-7 M

pOH = -log[OH-] = -log(2.67 x 10^-7) = 6.57

pH = 14 - pOH = 7.43

So pH of solution would be 7.43

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