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

1 Answer

6 votes

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

User Lyborko
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