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6 February, 08:25

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).

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  1. 6 February, 08:50
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    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)
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