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22 April, 07:56

A 25.0 ml sample of a 0.1700 m solution of aqueous trimethylamine is titrated with a 0.2125 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. 22 April, 08:25
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    Volume required for neutralization V will be:

    V * 0.2125 M HCl = 25 mL * 0.17 M

    V = 20 ml

    First part:

    When 10 mL is added we can apply Henderson equation to get the result, so:

    The pH will be of basic buffer

    pOH = pKb + log (salt/base)

    or pOH = 4.19 + log (0.2125*10 / 25*0.17 - 10*0.2125)

    pOH = 4.19 and pH = 14 - 4.19 = 9.81

    Second part:

    When 20 ml is added, there is only salt formed

    The pH will be salt of strong acid and weak base

    So pH = 7 - 0.5 pKb - 0.5 log C

    where C is the concentration of the salt formed so:

    pH = 7 - (0.5*4.19) - (0.5 log (25*0.17) / (25+20))

    = 5.42

    Third part:

    When 30 ml of the acid has been added,

    The pH will be of the remaining strong acid

    pH = - log (0.2125*10 / 25 + 30)

    = 1.326
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