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11 March, 04:07

A volume of 25.0 mL of 0.100 M HCl is titrated against a 0.100 M CH3NH2 solution added to it from a buret. (a) Calculate the pH value of the solution after 10.0 mL of CH3NH2 solution has been added. (b) Calculate the pH value of the solution after 25.0 mL of CH3NH2 solution have been added. (c) Calculate the pH value of the solution after 35.0 mL of CH3NH2 solution have been added.

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  1. 11 March, 07:28
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    a) pH = 1.37

    b) pH = 5.76

    c) pH = 10.24

    Explanation:

    Step 1: Data given

    Volume of a 0.100 M HCl = 25.0 mL = 0.025 L

    Molarity of a CH3NH2 solution = 0.100 M

    Step 2: The balanced equation

    2CH3NH2 + 2HCl → 2CH3NH3 + Cl2

    Calculate the pH value of the solution after 10.0 mL of CH3NH2 solution has been added.

    Step 3: Calculate moles

    Moles HCl = molarity * volume

    Moles HCl = 0.100 M * 0.025 L

    Moles HCl = 0.0025 moles

    Moles CH3NH2 = 0.100 M * 0.010 L

    Moles CH3NH2 = 0.001 moles

    Step 4: Calculate the limiting reactant

    For 2 moles CH3NH2 we need 2 moles HCl to produce 2 moles CH3NH3 and 1 mol Cl2

    CH3NH2 is the limiting reactant. There will react 0.001 moles. HCl is in excess. There will react 0.001 moles. There will remain 0.0025 - 0.001 = 0.0015 moles HCl

    Step 5: Calculate molarity HCl

    Molarity = moles / volume

    Molarity = 0.0015 moles / 0.035 L

    Molarity = 0.043 M

    Step 6: Calculate pH

    pH HCl = - log[H+]

    pH HCl = - log (0.043)

    pH HCl = 1.37

    Calculate the pH value of the solution after 25.0 mL of CH3NH2 solution have been added.

    Step 3: Calculate moles

    Moles HCl = molarity * volume

    Moles HCl = 0.100 M * 0.025 L

    Moles HCl = 0.0025 moles

    Moles CH3NH2 = 0.100 M * 0.025 L

    Moles CH3NH2 = 0.0025 moles

    Step 4: Calculate the limiting reactant

    For 2 moles CH3NH2 we need 2 moles HCl to produce 2 moles CH3NH3 and 1 mol Cl2

    Both reactants will completely be consumed.

    Ka = Kw/Kb = 1.0 * 10^-14 / 4.2 * 10^-4 = 2.38 * 10^-11 = x² / 0.125-x

    x = [H+] = 1.73 * 10^-6 M

    pH = - log[H+]

    pH = - log (1.73 * 10^-6)

    pH = 5.76

    Calculate the pH value of the solution after 35.0 mL of CH3NH2 solution have been added.

    Step 3: Calculate moles

    Moles HCl = molarity * volume

    Moles HCl = 0.100 M * 0.025 L

    Moles HCl = 0.0025 moles

    Moles CH3NH2 = 0.100 M * 0.035 L

    Moles CH3NH2 = 0.0035 moles

    Step 4: Calculate the limiting reactant

    For 2 moles CH3NH2 we need 2 moles HCl to produce 2 moles CH3NH3 and 1 mol Cl2

    There will remain 0.0010 moles of CH3NH2

    There will be 0.0025 mole of CH3NH3 be produced

    Step 5: Calculate the molarity

    [CH3NH2] = 0.0010 moles / 0.060 L

    [CH3NH2] = 0.0167 M

    [CH3NH3] = 0.0025 moles / 0.060 L

    [CH3NH3] = 0.0417 M

    Step 5: Calculate pOH

    pOH = pKb + log ([B+]/[BOH]

    ⇒[B+] = the concentration of CH3NH3 = 0.0417 M

    ⇒[BOH] = the concentration of CH3NH2 = 0.0167 M

    ⇒pKb = 3.36

    pOH = 3.36 + log (0.0417 / 0.0167)

    pOH = 3.36 + 0.40

    pOH = 3.76

    pH = 14 - 3.76

    pH = 10.24
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