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

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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