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4 December, 13:01

You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (p K a = 4.20) and 0.220 M sodium benzoate. How many milliliters of each solution should be mixed to prepare this buffer?

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Answers (2)
  1. 4 December, 14:08
    0
    Volume benzoic acid = 77.74 mL

    Volume sodium benzoate: = 22.26 mL

    Explanation:

    Step 1: Data given

    Volume = 100.0 mL

    pH = 4.00

    Molarity benzoic acid = 0.100 M

    pKa benzoic acid = 4.20

    Molarity sodium benzoate = 0.220 M

    Step 2

    pH = pKa + log ([base]/[acid])

    4 = 4.2 + log ([base]/[acid])

    -0.2 = log ([base]/[acid])

    10^-0.2 = [base]/[acid]

    0.63 = [base]/[acid]

    Step 3

    start with 100 mL 0.1 M benzoic acid

    0.100 L * 0.1M = 0.01 moles.

    Since base/acid = 0.63, add 0.63*0.01 moles base = 0.0063 moles base.

    And 0.0063 moles of a 0.22 M salt solution are x mL:

    0.22 moles = 1000 mL

    1 mole = 1000/0.22 mL

    0.0063 moles = 1000*0.0063/0.22 mL = 28.63 mL.

    So a mixture of 100 mL 0.1 M bencoic acid + 28.63 mL of 0.22 benzoate has pH = 4.00

    Step 4: For a total volume of 100 mL we'll have:

    Volume benzoic acid: 100*100 / (128.63) = 77.74 mL

    Volume sodium benzoate: 28.63*100 / (128.63) = 22.26 mL
  2. 4 December, 16:59
    0
    34.5 mL

    Explanation:

    Given, we ned to prepared buffer solution with pH = 4.00, volume = 100.0 mL.

    [C6H5COOH] = 0.100 M, [C6H5COO-] = 0.120 M, pKa = 4.20

    First we need to calculate the ratio of the conjugate base and acid

    We know, Henderson Hasselbalch equation

    pH = pKa + log [C6H5COO-] / [C6H5COOH]

    4.00 = 4.20 + log [C6H5COO-] / [C6H5COOH]

    log [C6H5COO-] / [C6H5COOH] = 4.00-4.20

    = - 0.20

    Antilog from both side

    [C6H5COO-] / [C6H5COOH] = 0.631

    Now we are given the molarity of each acid and its conjugate base and we need to calculate for volume

    Sum of volume = 100 mL = 0.100 L

    volume of benzoic acid = x

    volume of benzoate = 0.10 - x,

    so Volume of acid + volume of conjugate base = 0.100 L

    [C6H5COO-] / [C6H5COOH] = 0.631

    [C6H5COO-] = 0.631 * [C6H5COOH]

    0.120 (0.1-x) = 0.631 * 0.100x

    So, x = 0.065

    So, volume of benzoic acid = x = 0.0655 L

    = 65.5 mL

    So, volume of sodium benzoate = 0.1 - x

    = 0.1-0.0655

    = 0.0345 L

    = 34.5 mL

    So we need to mix 65.5 mL of 0.100 M benzoic acid and 34.5 mL of 0.120 M sodium benzoate form 100.0 mL of a pH=4.00 buffer solution.
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