Question

Muscle physiologists study the accumulation of lactic acid [ch3ch(oh)cooh] during exercise. food chemists study its occurrence in sour milk, beer, wine, and fruit. industrial microbiologists study its formation by various bacterial species from carbohydrates. a biochemist prepares a lactic acid-lactate buffer by mixing 225 ml of 0.85 m lactic acid (ka = 1.38 × 10−4) with 435 ml of 0.68 m sodium lactate. what is the buffer ph?

Answer 1

The provided information are:
volume of 0.85 M lactic acid = 225 ml
volume of 0.68 M sodium lactate = 435 ml
Ka of the lactate buffer = 1.38 x 10⁻⁴
 The equation for dissociation of lactic acid is:
CH₃CH(OH)COOH(aq) + H₂O ⇄ CH₃CH(OH)COO⁻(aq) + H₃O⁺(aq)
The pH of buffer is calculated from Henderson-Hasselbalch equation, which is:
pH = pKa + log $\frac{[conjugated base]}{[Acid]}$
pKa = - log Ka = - log (1.38 x 10⁻⁴) = 3.86 
The number of moles of lactic acid and lactate are as follows:
n (Lactic acid) = 225 ml x 0.85 mmol/ml = 191.25 mmol
n (Lactate) = 435 ml x 0.68 mmol/ml = 295.8 mmol
The number of moles of lactic acid and lactate in total volume of the solution:
[CH₃CH(OH)COOH] = n (lactic acid) / 660 ml = 191.25 mmol / 660 ml = 0.29 M
[CH₃CH(OH)COO⁻] = n (lactate) / 660 ml = 0.45 M
pH = 3.86 + log $\frac{0.45 M}{0.29 M}$ = 3.86 + 0.191 = 4.05
So the pH of given solution is 4.05