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23 August, 19:06

What is the ph of a 0.010 m triethanolammonium chloride, (hoc2h2) 3nhcl, solution?

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  1. 23 August, 19:30
    0
    The answer is 4.89

    The explanation:

    when the Kb of ((HOC2H2) 3N) = 5.9 x 10^-7

    1 - we can get the value of Ka from Kb:

    when Ka = Kw / Kb

    = 1 x 10^-14 / 5.9 x 10^-7

    = 1.69 x 10^-8

    2 - now we will calculate the [H+] value:

    when [H+] = √ (Ka*[HA])

    =√ ((1.69E-8) (0.010))

    = 1.3 x 10^-5 M

    3 - the final step we will calculate the PH value from the value of [H+]:

    PH = - ㏒ [H+]

    = - ㏒ 1.3 x 10^-5

    = 4.89
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