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8 February, 23:17

A 25.0 ml sample of a 0.2900 m solution of aqueous trimethylamine is titrated with a 0.3625 m solution of hcl. calculate the ph of the solution after 10.0, 20.0, and 30.0 ml of acid have been added; pkb of (ch3) 3n = 4.19 at 25°c.

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  1. 9 February, 02:50
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    a) After adding 10 mL of HCl

    first, we need to get moles of (CH3) 3N = molarity * volume

    = 0.29 m * 0.025 L

    = 0.00725M moles

    then, we need to get moles of HCl = molarity * volume

    = 0.3625 m * 0.01L

    = 0.003625 moles

    so moles of (CH3) 3N remaining = moles of (CH3) 3N - moles of HCl

    = 0.00725 - 0.003625

    = 0.003625 moles

    and when the total volume = 0.01 L + 0.025L = 0.035 L

    ∴ [ (CH3) 3N] = moles remaining / total volume

    = 0.003625 moles / 0.035L

    = 0.104 M

    when we have Pkb so we can get Kb:

    pKb = - ㏒Kb

    4.19 = - ㏒Kb

    ∴Kb = 6.5 x 10^-5

    when Kb = [ (CH3) 3NH+] [OH-] / [ (CH3) 3N]

    and by using ICE table we assume we have:

    [ (CH3) 3NH+] = X & [OH] = X

    and [ (CH3) 3N] = 0.104 - X

    by substitution:

    ∴ 6.5 x 10^-5 = X^2 / (0.104-X) by solving for X

    ∴X = 0.00257 M

    ∴[OH-] = X = 0.00257 M

    ∴POH = - ㏒[OH]

    = - ㏒0.00257

    = 2.5

    ∴ PH = 14 - POH

    = 14 - 2.5

    = 11.5

    b) after adding 20ML of HCL:

    moles of HCl = molarity * volume

    = 0.3625 m * 0.02 L

    = 0.00725 moles

    the complete neutralizes of (CH3) 3N we make 0.003625 moles of (CH3) 3NH + So, now we need the Ka value of (CH3) 3NH+:

    and when the total volume = 0.02L + 0.025 = 0.045L

    ∴ [ (CH3) 3NH+] = moles / total volume

    = 0.003625 / 0.045L

    = 0.08 M

    when Ka = Kw / Kb

    and we have Kb = 6.5 x 10^-5 & we know that Kw = 1 x 10^-14

    so, by substitution:

    Ka = (1 x 10^-14) / (6.5 x 10^-5)

    = 1.5 x 10^-10

    when Ka expression = [ (CH3) 3N][H+] / [ (CH3) 3NH+]

    by substitution:

    ∴ 1.5 x 10^-10 = X^2 / (0.08 - X) by solving for X

    ∴X = 3.5 x 10^-6 M

    ∴ [H+] = X = 3.5 x 10^-6 M

    ∴PH = - ㏒[H+]

    = - ㏒ (3.5 x 10^-6)

    = 5.5

    C) after adding 30ML of HCl:

    moles of HCl = molarity * volume

    = 0.3625m * 0.03L

    = 0.011 moles

    and when moles of (CH3) 3N neutralized = 0.003625 moles

    ∴ moles of HCl remaining = moles HCl - moles (CH3) 3N neutralized

    = 0.011moles - 0.003625moles

    = 0.007375 moles

    when total volume = 0.025L + 0.03L

    = 0.055L

    ∴[H+] = moles / total volume

    = 0.007375 mol / 0.055L

    = 0.134 M

    ∴PH = - ㏒[H+]

    = - ㏒ 0.134

    = 0.87
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