Chemical equilibrium, getting molarity given pH and Kb

In summary, the molarity of Trimethylamine in an aquatic dissolution with a pH of 11.2 can be calculated using the equilibrium constant Kb and the initial concentration of trimethylamine. Using ICE tables, the initial concentration of trimethylamine is found to be approximately 0.04M. This can be confirmed by calculating the pH of the solution using the initial concentration. It is recommended to use ICE tables for equilibrium calculations as they are more general and can be applied to most typical equilibrium problems.
  • #1
Telemachus
835
30
What should be the molarity in an aquatic dissolution for Trimethylamine, ##(CH_3)_3N## if ##pH=11,2##?

##(CH_3)_3N+H_2O \rightleftharpoons (CH_3)_3NH^++OH^-##, ##K_b=6.3 \times 10^{-5}##

I know that ##pH+pOH=14 \rightarrow pOH=2.8##, then ##[OH^-]=1.58 \times 10^{-3}##

And ##K_b=\displaystyle\frac{ [ (CH_3)_3NH^+] [OH^-]}{[(CH_3)_3N]}## I used the approximation that water remains constant.
I think I should use some other approximation to get the concentration for ##(CH_3)_3N##, but I'm not sure.
 
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  • #2
Many ways to skin that cat. Do you know ICE tables?
 
  • #3
Not really. We see all this really fast, we had only one class for this.

Now that I googled it and I know what I mean I think I saw some of these tables in examples in books. But the exercise don't give any information, I would expect to know the initial concentration or something.
 
  • #4
In this case original concentration is your unknown. Pretend you use ICE table to calculate pH and use it to derive formula for the concentration of H+ (or OH- as it should make calculations easier) as a function of the concentration of trimethylamine. Then solve this formula for the concentration of trimethylamine - it will be the only unknown.

It is possible to solve the problem using approximate equations, but then you will have no idea what and why you are doing. ICE tables are much more general and can be applied to most typical equilibrium calculations.
 
  • #5
Alright, I would like to try both ways. Where can I find an ICE table? an what approximation should I use?
 
  • #6
Telemachus said:
Where can I find an ICE table?

ICE stands for Initial/Change/Equilibrium. Ask uncle google or aunt wikipedia for help.
 
  • #7
Alright. I was thinking. Would it work just to suppose some arbitrary initial concentration for each species? let's say k, l, m. I thought of doing that using stoichiometry.
 
  • #8
Why arbitrary? Assume concentration of the trimethylamine to be C, and other concentrations to be 0.
 
  • #9
Nice!
 
  • #10
Ok, this is what I did. I know the concentration for ##[OH^-]## at equilibrium.
So I used that I have a concentration ##C_0## for trimethylamine that I want to find.
The changes given by the stoichiometry are -X for trimethylamine, +X for ##(CH)_3NH^+## and +X for ##[OH^-]##

Then, if initially there is only trimethylamine, the initial concentration ##[OH^-]_0=[(CH)_3NH^+]=0## and then in equilibrium the thrimethylamine concentration is given by ##[(CH)_3N]=C_0-X## and for the products ##[(CH)_3NH^+]=0+X## and ##0+X=[OH^-]\rightarrow X=1.58 \times 10^{-3}##

So, using the equilibrium constant:

##K_b=\displaystyle\frac{[(CH)_3NH^+][OH^-]}{[(CH)_3N]}=\frac{X1.58\times10^{-3}}{C_0-X}=6.3\times10^{-5}##

Then ##C_0=25.12M ## and at equilibrium ##[(CH)_3N]\approx C_0##

Is this fine?
 
  • #11
Check your math. In general you are on the right track, but for some reason you mixed X and 1.58x10-3 in the final equation (even if you know X=1.58x10-3) and your answer is wrong.

Whenever you see calculated concentration being higher than 2 or 3 M it should raise red flag for you.
 
  • #12
I've made the numbers again and it gave: ##C_0=1.04## I think I forgot to square ##1.58\times10^{-3}## in the last equality for Kb.

Thanks.
 
  • #13
This is still wrong.

As a final check you should try to calculate pH of the solution.
 
  • #14
This equation is fine?

##K_b=\displaystyle\frac{[(CH)_3NH^+][OH^-]}{[(CH)_3N]}=\frac{X1.58\times10^{-3}}{C_0-X}=6.3\times10^{-5}##

Alright, Iv'e tried again and get to ##C_0=0,04## I've forgot another product last time. Your purpose was that I should calculate pH from the initial concentration as a check? I'm not sure on how to do that, I should look for it in a book.

Thank you Borek.

BTW, can you help me with this one? https://www.physicsforums.com/showthread.php?t=660731
 
Last edited:
  • #15
  • #16
Alright. Thank you Borek.
 

1. What is chemical equilibrium?

Chemical equilibrium refers to a state in which the forward and reverse reactions of a chemical reaction are occurring at equal rates, resulting in a constant concentration of reactants and products. This means that the overall concentration of the substances involved in the reaction does not change over time.

2. How is equilibrium affected by changes in concentration, temperature, and pressure?

Changes in concentration, temperature, and pressure can shift the equilibrium of a chemical reaction. If the concentration of one of the reactants or products is increased, the equilibrium will shift towards the opposite side to reduce the concentration. Similarly, an increase in temperature will favor the endothermic reaction, while an increase in pressure will favor the reaction with fewer moles of gas.

3. How do you calculate the molarity given pH and Kb?

To calculate the molarity of a solution given the pH and Kb, you can use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where [A-] and [HA] represent the concentrations of the conjugate base and acid, respectively. Then, you can solve for [A-] and [HA] using the Kb value and the equation Kb = [A-][H+]/[HA]. Finally, you can use the molar concentration formula to calculate the molarity of the solution.

4. How does the addition of a catalyst affect equilibrium?

The addition of a catalyst does not affect the equilibrium position of a chemical reaction. However, it can increase the rate at which the equilibrium is reached by lowering the activation energy required for the reaction to occur. This results in a faster approach to equilibrium, but the actual equilibrium concentrations of the reactants and products remain the same.

5. What is the relationship between molarity and pH?

Molarity and pH are not directly related. Molarity is a measure of concentration, specifically the number of moles of solute per liter of solution. pH, on the other hand, is a measure of the acidity or basicity of a solution and is determined by the concentration of hydrogen ions (H+) in the solution. However, the pH of a solution can affect the molarity of certain substances, such as weak acids or bases, through the Henderson-Hasselbalch equation.

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