Lactic acid (CH3CHOHCO2H) is found in sour milk, in saurkraut, and in muscles after activity. Ka = 1.4 * 10^-4

Question

a. If 2.75g of NaCH3CHOHCO2, sodium lactate, is added to 5 * 10^2 ml of .100 M lactic acid, what is the pH of the resulting buffer solution?

b. Is the final pH lower or higher than the pH of the lactic acid solution?

Answer 1

OK...You're going to use the Henderson-Haselbalch equation to solve this, but you first need to convert the mass of NaLac into moles, and then calculate the molar concentration of that ion when you dissolve it in .050 L (The volume of the solution you have).

Then, the pH of the final solution is calculated by:

pH = pKa + log [lactate ion]/[lactic acid]

where pKa is the -log(Ka). The other terms are the molar concentrations of each thing.

You'll find that the pH is higher than just the lactic acid solution. Why is that?

Answer 2

Use the Henderson-Hasselbalch equation:

pH = pKa + log[A-]/[HA]

where [A-] is [CH3CHOHCO2-]
and [HA] is [CH3CHOHCO2H]