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?

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 
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  = 3.86 + 0.191 = 4.05
So the pH of given solution is 4.05

The pH of the buffer solution is 4.1.

We have to obtain the number of moles of each solution as follows;

Number of moles of acid = 225/1000 × 0.85 M = 0.191 moles

Number of moles of lactate = 435/1000 × 0.68 M = 0.296 moles

Total volume of solution = 225 ml + 435 ml = 660 ml or 0.66 L

Next we obtain the final molarity of each solution

Final molarity of acid = 0.191 moles/0.66 L  = 0.289 M

Final molarity of lactate = 0.296 moles/0.66 L  = 0.448 M

Using the Henderson-Hasselbalch equation;

pH = pKa +log [conjugate base]/[acid]

pka = - log( 1.38 × 10−4) = 3.86

pH = 3.86 + log [0.448 M]/[0.289 M]

pH = 4.1

Hence, the pH of the buffer solution is 4.1.

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The answer to the question is d