Solving Trimethylacetic Acid Concentration Questions

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SUMMARY

The discussion focuses on calculating the pKa of Trimethylacetic Acid (C4H11COOH), a weak monoprotic acid. Given that 0.0010 mol of Trimethylacetic acid is dissolved in 100 mL of water, the concentration of trimethylacetic ion is determined to be 3.0 x 10-4 mol/L. The participants discuss the formula for pKa as pKa = -log[Ka] and the relationship between the concentrations of H+ and A- ions in the solution. Additionally, they explore the calculation of the volume of 0.2M KOH required for neutralization.

PREREQUISITES
  • Understanding of weak acids and their dissociation
  • Knowledge of pKa and Ka calculations
  • Familiarity with molarity and concentration calculations
  • Basic acid-base neutralization concepts
NEXT STEPS
  • Learn how to calculate pKa from concentration data
  • Study the Henderson-Hasselbalch equation for buffer solutions
  • Explore titration techniques for weak acids with strong bases
  • Investigate the properties and applications of Trimethylacetic Acid in organic chemistry
USEFUL FOR

Chemistry students, organic chemists, and anyone involved in acid-base chemistry or titration experiments will benefit from this discussion.

dcl
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Heya's
Having some problems with the following question:

Trimethylacetic acid [tex]C_4 H_{11} COOH[/tex]is a weak monoprotic acid. When 0.0010 mol of Trimethylacetic acid was dissolved in 100mL of water, the concentration of trimethylacetic ion was found to be [tex]3.0 \times 10^{ - 4} mol.L^{ - 1}[/tex]

a) What is monoprotic acid?
An Acid that donates only one proton.

b) Use the above data to work out [tex]pK_a[/tex] for Trimethylacetic acid.
now pKa = -log[Ka] think? How would I go about working Ka out?

c) What volume of 0.2M KOH would be required to neutralise this solution of trimethylacetic acid?
For this I would just work out how many moles are in the 100mL solution, and then use that to work out how how much volume of KOH I would need to add.
However I'm unsure how to work out how many mols are in the 100mL solution, It's an easy units conversion problem I am sure, but I suck at it :(
 
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Look at the concentration of the ion.

If there's that much in the solution that exists as an ion, an equal amount must exist as H+.
 
[tex]Ka= \frac {[H^+][A^-]} {[HA]}[/tex]

and

[tex][H^+] = 10^{-7} + [A^-][/tex]
 

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