Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid is titrated

Question

Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid is titrated to the equivalence point with 45.4 mL of 0.020 M sodium hydroxide solution. What is the pH of the resulting solution at the equivalence point

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Verity 1 week 2021-07-22T04:43:52+00:00 1 Answers 0 views 0

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    2021-07-22T04:45:04+00:00

    Answer:

    pH of resulting solution = 7.98

    Explanation:

    The balanced equation  

    HA + NaOH - Na+ + A- + H2O

    Number of moles of A = Number of moles of HA  = Number of moles of NaOH

    = 35.8/1000 * 0.020 = 0.000716 mol

    Initial concentration of A = 0.000716/0.0608 = 0.01178 M

    pKb = 14 – pKa = 14 -3.9 = 10.1

    Kb = 10^{-Kb} = 10^{-10.1} = 7.943 * 10^-11

    Kb = [HA][OH-]/[A-]

    Kb = a^2/(0.01178 -a) = 7.943 * 10^-11

    a^2 + 7.943 * 10^-11 a – 9.357 * 10^-13 = 0

    a = 9.673 * 10^-7

    OH- = a = 9.673 * 10^-7 M

    pOH = -log [OH-] = -log (9.673 * 10^-7) = 6.02

    pH = 14-6.02 = 7.98

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