<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine,
, the reaction will be
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
Rows of elements are called periods. The period number<span> of an element signifies the highest unexcited energy level for an electron in that element.
</span>Columns of elements help define element groups<span>. </span>Elements within a group share<span> several common properties. Groups are elements have the same outer electron arrangement.</span>
The question is incomplete.
You need two additional data:
1) the original volume
2) what solution you added to change the volume.
This is a molarity problem, so remember molarity definition and formula:
M = n / V in liters: number of moles per liter of solution
To give you the key to answer this kind of questions, supppose the original volumen was 1 ml and that you added only water (solvent).
The original solution was:
V= 1 ml
M = 0.2 M
Using the formula for molarity, M = n / V
n = M×V = 0.2 M × (1 / 10000)l = 0.0002 moles
For the final solution:
n = 0.0002 moles
M = 0.04
From M = n / V ⇒ V = n / M = 0.002 moles / 0.04 M = 0.05 l
Change to ml ⇒ 0.05 l × 1000 ml / l = 50 ml. This would be the answer for the hypothetical problem that I assumed for you.
I hope this gives you all the cues you need to answer similar problems about molarity.
