23 August, 19:06
What is the ph of a 0.010 m triethanolammonium chloride, (hoc2h2) 3nhcl, solution?
23 August, 19:30
The answer is 4.89
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
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