Question

A, What are the elemental chemical reactions (forward and reverse) that exist when you put lactic acid (CH3COHCOOH) into water. Draw all structures.

B. Write the equilibrium constant for acetic acid as a function of the species concentrations. Is water included? Why or why not.

C. Given that pH = -log[h+] and the definition of pKa, derive the Henderson-Hasselbalch equation that relates the pH to the pK and the equlibirium concentrations among reacting species. Show work.

D. The equilibrium constant , Ka, for lactic acid is 1.4 *10^-4. What's the pKa, show work.

E. What fraction of the carboxylic acid is protonated when the ph is 4?

F. What fraction of the carboxylic acid is protonated when the pH is 2?

G. What fraction of the carboxylic acid is protonated when the pH is 7?

Answer 1

a)

Lactic acid is a weak acid and thus form an equilibrium in water.

Forward reaction is the dissociation of acid to give H+ ions

CH3COHCOOH +H2O -> CH3COHCOO- + H3O+

Backward reaction will be when the dissociated ion CH3COHCOO- attracts H3O+ from water to form CH3COHCOOH again -

CH3COHCOO- + H3O+ -> CH3COHCOOH + H2O

b) Acetic acid is CH3COOH

The equilibrium with water will be as follows -

CH3COOH(aq) + H2O(l) <-> CH3COO-(aq) + H3O+(aq)

Equilibrium constant Ka will be [CH3COO-][H3O+]/[CH3COOH]

H2O is not included because its in pure liquid form and thus we do not take it into account as it doesn't affect our equilibrium constant.

c)

For an acid HA if it dissociates as H+ and A- in water we have -

HA <-> H+ + A-

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

taking log on both the sides we have

log(Ka) = log([H+][A-]/[HA])

so, log(Ka) = log([H+]) + log([A-]/[HA])

so, -log([H+]) = -log(Ka) + + log([A-]/[HA])

so, pH = pKa + log([A-]/[HA]) where pH = -log([H+]), pKa = -log(Ka)

d)

Given that Ka for lactic acid is 1.4*10^-4 so, pKa = -log(Ka) = -log(1.4*10^-4) = 3.854