What is the pH of a 0.050 M triethylamine, (C2H5)3N, solution? Kb for triethylamine is 5.3 ´ 10-4. Question options: 1) 11.69 2)
8.68 3) 5.32 4) 2.31 5) < 2.0
1 answer:
<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine,
![(C_{2}H_{5})_{3}N](https://tex.z-dn.net/?f=%28C_%7B2%7DH_%7B5%7D%29_%7B3%7DN)
, the reaction will be
![(C_{2}H_{5})_{3}N + H_{2}O ---\ \textgreater \ ( C_{2}H_{5})_{3}NH^{+} + OH^{-}](https://tex.z-dn.net/?f=%28C_%7B2%7DH_%7B5%7D%29_%7B3%7DN%20%2B%20H_%7B2%7DO%20---%5C%20%5Ctextgreater%20%5C%20%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DNH%5E%7B%2B%7D%20%2B%20OH%5E%7B-%7D%20)
and we know, pH = -log[H+] and pOH = -log[OH-]
Also, pOH + pH = 14
Now, the Kb value = 5.3 x 10^-4
And
![kb = \frac{( [( C_{2}H_{5})_{3}NH^{+} ]* OH^{-} )}{[( C_{2}H_{5})_{3}N]}](https://tex.z-dn.net/?f=kb%20%3D%20%20%5Cfrac%7B%28%20%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DNH%5E%7B%2B%7D%20%5D%2A%20%20OH%5E%7B-%7D%20%29%7D%7B%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DN%5D%7D%20)
thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050
=0.00516 M
Thus, pOH = 2.30
pH = 14 - pOH = 11.7
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