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15 September, 05:29

Calculate the pH for each case in the titration of 50.0 mL of 0.110 M HClO (aq) with 0.110 M KOH (aq). Use the ionization constant for HClO. What is the pH before addition of any KOH? pH = What is the pH after addition of 25.0 mL KOH? pH = What is the pH after addition of 35.0 mL KOH? pH = What is the pH after addition of 50.0 mL KOH? pH = What is the pH after addition of 60.0 mL KOH? pH =

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  1. 15 September, 07:42
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    Before KOH is added: pH = 3.93

    After adding 25.0 mL KOH: pH = 7.52

    After ading 35.0 mL KOH: pH = 7.88

    After adding 50.0 mL KOH: pH = 10.13

    After adding 60.0 mL KOH: pH = 12

    Explanation:

    Step 1: Data given

    Volume of HClO = 50.0 mL = 0.050 L

    Molarity of HClO = 0.110 M

    Molarity of KOH = 0.110 M

    Ka of HClO = 3.0 * 10^-8

    Kb = Kw / Ka = 10^-14 / 3.0 * 10^-8 = 3.33 * 10^-7

    Before KOH is added

    HClO + H2O → ClO - + H3O+

    Calculate the initial concentrations

    [HClO] = 0.110 M * 0.050 L = 0.455 M

    [ClO-] = 0M

    [H3O+] = 0M

    The concentration at the equilibrium

    [HClO] = 0.455 - XM

    [ClO-] = XM

    [H3O+] = XM

    Ka = [ClO-][H3O+] / [HClO]

    3.0 * 10^-8 = X² / 0.455 - X

    3.0 * 10^-8 = X² / 0.455

    X² = 1.365*10^-8

    X=0.000117 = [H3O+]

    pH = - log (0.000117)

    pH = 3.93

    pH after addition of 25.0 mL of KOH

    Moles HClO = 0.050 L * 0.11 M = 0.00550 moles

    Moles KOH = 0.025 L * 0.11 M = 0.00275 moles

    The limiting reactant is KOH. It will completely be consumed (0.00275 moles). HClO is in excess. There will react 0.00275 moles. There will remain 0.00550 - 0.00275 = 0.00275 moles.

    Moles KClO (ClO-) produced = 0.00275 moles

    pH = pKa + log[ClO-]/[HClO]

    pH = 7.52 + 0

    pH = 7.52

    pH after addition of 35.0 mL of KOH

    Moles HClO = 0.050 L * 0.11 M = 0.00550 moles

    Moles KOH = 0.035 L * 0.11 M = 0.00385 moles

    The limiting reactant is KOH. It will completely be consumed (0.00385 moles). HClO is in excess. There will react 0.00385 moles. There will remain 0.00550 - 0.00385 = 0.00165 moles

    Moles KClO (ClO-) produced = 0.00385 moles

    [HClO] = 0.00165 moles / 0.085 L = 0.0194 M

    [ClO-] = 0.00385 moles / 0.085 L = 0.0453 M

    pH = pKa + log[ClO-]/[HClO]

    pH = 7.52 + log (0.0453/0.0194)

    pH = 7.88

    pH after addition of 50.0 mL of KOH

    ClO - + H2O ⇄ HClO + OH-

    Moles HClO = 0.050 L * 0.11 M = 0.00550 moles

    Moles KOH = 0.050 L * 0.11 M = 0.00550 moles

    Both reactants will be consumed, there are no limiting reactants. No HClO or KOH will be left.

    There will be 0.00550 moles ClO - produced

    [ClO-] = 0.00550 moles / 0.100 L = 0.055 M

    The initial concentrations:

    [ClO-] = 0.055 M

    [HClO] = 0 M

    [OH-] = 0M

    Cocnentration at the equilibrium

    [ClO-] = 0.055 - X M

    [HClO] = X M

    [OH-] = XM

    Kb = [HClO][OH-] / [ClO-]

    3.33 * 10^-7 = X² / 0.055 - X

    3.33 * 10^-7 = X² / 0.055

    X² = 1.83*10^-8

    X = 0.0001354 = [OH-]

    pOH = - log (0.0001354)

    pOH = 3.87

    pH = 14 - 3.87 = 10.13

    pH after addition of 60.0 mL of KOH

    Moles HClO = 0.050 L * 0.11 M = 0.00550 moles

    Moles KOH = 0.060 L * 0.11 M = 0.00660 moles

    We have 0.0011 moles of KOH remaining. KOH is a strong base

    [KOH] = 0.0011 moles / 0.11 L

    [KOH] = 0.01 M

    pOH = - log (0.01)

    pOH = 2

    pH = 14 - 2 = 12
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Home » Chemistry » Calculate the pH for each case in the titration of 50.0 mL of 0.110 M HClO (aq) with 0.110 M KOH (aq). Use the ionization constant for HClO. What is the pH before addition of any KOH? pH = What is the pH after addition of 25.
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