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9 November, 09:02

A 20.0 ml sample of 0.150 m ethylamine is titrated with 0.050 m hcl. you may want to reference (page 805) section 17.4 while completing this problem. part a what is the ph after the addition of 5.0 ml of hcl? for ethylamine, pkb = 3.25.

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  1. 9 November, 10:47
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    First, we need to get moles of OH-:

    moles of ethylamine = molarity * volume

    = 0.15 M * 0.02 L

    = 0.003 mol

    moles of H + = molarity * volume

    = 0.05 M * 0.005 L

    = 0.00025 mol

    when the total volume = 0.02 L + 0.005L

    = 0.025 L

    [H+] = moles / total volume

    = 0.00025 / 0.025 = 0.01 M

    [ethaylamine] = moles / total volume

    = 0.003 / 0.025 = 0.12 M

    when Pkb = 3.25 so we can get Pka from this formula:

    Pka = 14 - Pkb

    = 14 - 3.25

    = 10.75

    by using H-H equation:

    PH = Pka + ㏒[salt]/[acid]

    = 10.75 + ㏒[0.12/0.01]

    = 11.8
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