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.

Learn more about buffer solution: link

Answer : option c) heat is released during the process.explanation :

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.

Sorry i dont understand the question can you rephase it?