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    What is the PH at equivalence point when 25ml of 0.100M HClO is titrated with 0.200M KOH ..ka HClO=2.9*10^-8

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    What is the PH at equivalence point when 25ml of 0.100M HClO is titrated with 0.200M KOH ..ka HClO=2.9*10^-8

    © BrainMass Inc. brainmass.com December 24, 2021, 9:50 pm ad1c9bdddf
    https://brainmass.com/chemistry/acids-and-bases/solving-ph-titration-question-416504

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    SOLUTION:
    At the equivalence point, the number of moles of HClO and KOH are the same. Hence,
    HClO + OH- ↔ ClO- + H2O
    Initial mole 0.0025 0.0025 ----
    Change -0.0025 -0.0025 +0.0025
    Equilibrium mole ------------ ---------- +0.0025
    Equilibrium concentration ------------ ---------- 0.0025 mol/0.0375 L
    = 0.0667 M
    NOTE:
    Computing for the volume of KOH required to reach the equivalence point:
    Volume = 0.0025 mol / 0.2 M
    Volume = 0.0125 L = 12.5 mL
    Hence,
    Total volume = 25 mL + 12.5 mL = 37.5 mL = 0.0375 L

    At the equivalence point, the pH of the solution is dependent on the hydrolysis of ClO-:
    ClO- + H2O ↔ HClO + OH-
    Initial 0.0667 ----- ----
    Change -x +x +x
    Equilibrium 0.0667 - x x x

    Kb = [HClO][OH-] / [ClO-]
    Kb= Kw / Ka = 1 x 10-14 / 2.9 x 10-8 = x2 / 0.0667 - x
    3.45 x 10-7 = x2 / 0.0667 - x
    Assume x as negligible:
    3.45 x 10-7 = x2 / 0.0667
    x2 = 2.3 x 10-8
    x = 1.517 x 10-4 = [OH-]

    pOH = -log [OH-]
    pOH = - log 1.517 x 10-4
    pOH = 3.819

    pH = 14 - pOH
    pH = 14 - 3.819
    pH = 10.181

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 9:50 pm ad1c9bdddf>
    https://brainmass.com/chemistry/acids-and-bases/solving-ph-titration-question-416504

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