Ask Question
27 November, 05:38

A buffer contains 0.020 mol of lactic acid (pKa = 3.86) and 0.100 mol sodium lactate per liter of aqueous solution.

a. Calculate the pH of this buffer.

b. Calculate the pH after 8.0 mL of 1.00 M NaOH is added to 1 liter of the buffer (assume the total volume will be 1008 mL).

+2
Answers (1)
  1. 27 November, 05:57
    0
    pH = 4.8

    Explanation:

    We will use the Henderson-Hasselbach equation to calculate the pH of the buffer:

    pH = pKₐ + log [A⁻]/[HA]

    From the information given:

    pKₐ = 3.86

    [A⁻] = 0.100 M

    [HA] = 0.020 M

    Plugging our values:

    pH = 3.86 + log (0.100/0.020) = 4.6

    For part b the same equation is utilized.

    However we have to realize that the concentrations of the acid and its conjugate base have changed according to the neutralization reaction:

    NaOH + lactic acid ⇒ sodium lactate + H₂O

    # mol NaOH reacted = (8.0 mL x 1 L / 1000 mL) x 1.00 M

    = 8.0 x 10⁻³ mol

    mol sodium lactate produced = 8.0 x 10⁻³ mol (1:1)

    number of moles mol lactic acid originally = 1 L x 0.020 mol/L = 0.020 mol

    new mol lactic acid after reaction = 0.020 - 8.0 x 10⁻³ = 0.012 mol

    new mol sodium lactate after reaction = 0.100 mol/L x 1 L + 8.0 x 10⁻³ = 0.108

    Here we do not need to calculate the new concentrations since molarity is mol/V, and the volumes cancel each other in the Henderson-Hasselbach equation because they are in a ratio.

    Now we are in position to determine the pH.

    pH = 3.86 + log (0.108/0.012) = 4.8

    This the usefulness of buffers, we are adding a 1.00 M strong base NaOH, and the pH did not change that much (a long as they are small additions within reason)
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A buffer contains 0.020 mol of lactic acid (pKa = 3.86) and 0.100 mol sodium lactate per liter of aqueous solution. a. Calculate the pH of ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers