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4 May, 11:30

Calculate the pH for each case in the titration of 50.0 mL of 0.220 M HClO (aq) with 0.220 M KOH (aq). Use the ionization constant for HClO. What is the pH before addition of any KOH?

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  1. 4 May, 14:38
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    Before adding any KOH, the pH is 4.03

    Explanation:

    Step 1: Data given

    Volume of a 0.220 M HClO = 50.0 mL = 0.050 L

    Molarity of KOH = 0.220 M

    The ionization constant for HClO is 4.0*10^-8

    Step 2: The balanced equation

    HClO + KOH → KClO + H2O

    Step 3: pH before any addition of KOH

    When no KOH is added, we only have HClO, a weak acid.

    To calculate the pH of a weak acid, we need the Ka

    Ka = [H+] / [acid]

    4.0*10^-8 = [H+]² / 0.220

    [H+]² = (4.0*10^-8) * 0.220

    [H+]² = 8.8*10^-9

    [H+] = √ (8.8*10^-9)

    [H+] = 9.38*10^-5 M

    pH = - log [H+]

    pH = - log (9.38*10^-5)

    pH = 4.03

    Before adding any KOH, the pH is 4.03
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