Titration of Trimethylamine with HCl (g)

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    Hcl Titration
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Discussion Overview

The discussion revolves around the titration of trimethylamine with gaseous HCl, focusing on how to calculate the amount of HCl needed to achieve a specific pH of 10.30 in a solution of trimethylamine. Participants explore the implications of using gaseous HCl and the relevant equilibrium expressions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses uncertainty about how gaseous HCl affects pH and where to start with the problem.
  • Another participant suggests that the problem is simpler than it appears and recommends calculating the number of moles of HCl needed using the ideal gas law (pV=nRT).
  • A participant questions how to set up the expression for Ka, proposing a formula based on the concentrations of trimethylamine and its conjugate acid.
  • There is a calculation presented for Ka derived from Kb, with a value of 1.6 x 10^-10, and a corresponding hydronium ion concentration calculated from the given pH.
  • One participant attempts to set up an equilibrium expression but expresses doubt about its correctness.
  • Another participant suggests using the pH equation involving pKa and the ratio of base to acid concentrations, noting a minor detail regarding the form of the concentrations in the log term.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to solve the problem, with differing opinions on whether to use Ka or Kb and how to set up the equilibrium expressions. Uncertainty remains regarding the correct application of the equations.

Contextual Notes

Participants highlight potential limitations in their understanding of how to apply the ideal gas law in this context and the appropriate equilibrium expressions for the reaction.

plexus0208
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Homework Statement


Trimethylamine has a Kb of 6.3 x 10-5. How many liters of HCl gas, measured at 1.20 bar and 298 K, must be added to 250.0 mL of 0.500 M trimethylamine to give a pH of 10.30?

Homework Equations



The Attempt at a Solution


I don't know where to start because I don't know how the gaseous HCl affects the pH.
 
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It is gaseous just to make you think it is a difficult question. Just solve for number of moles of HCl that have to be added to get given pH, then it is simple pV=nRT.

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How do I set up the expression for Ka for this reaction?

Is Ka = [(CH3)3N][H3O+] / [(CH3)3NH]?

Since Kb = 6.3*10^-5, I found Ka by dividing Kw by Kb, and that turned out to be 1.6*10^-10
Also, since pH = 10.3, [H3O+] = 5*10^-11

I'm not sure what to do from that point on...

I tried to set an equation up:
Ka = (x)(5*10^-11 + x) / (0.500 - x)
But this can't be right...

Am I supposed to use the equation pH = pKa + log [A-]/[HA]?
 
Last edited:
The form of your equation looks reasonable, but you might be much more comfortable using the Kb value and any other corresponding changes.
 
plexus0208 said:
Am I supposed to use the equation pH = pKa + log [A-]/[HA]?

That sounds like the simplest approach, although part under the log would take

[tex]\frac {<b>} {[BH^+]}</b>[/tex]

form. But that's just a minor detail.

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