Chemistry - Concentration from known pH and Ka.

Click For Summary
SUMMARY

The discussion focuses on calculating the concentration of acetic acid in vinegar with a pH of 2.90 and a dissociation constant (Ka) of 1.8 x 10-5. The user attempted to derive the concentration using the equations pH = -log[H+] and Ka = [Products]/[Reactants]. However, the calculation led to an incorrect concentration of 8.8 x 10-12 M, while the correct concentration is 0.089 M. The error lies in the interpretation of the equilibrium expression and the relationship between [H+] and the concentration of acetic acid.

PREREQUISITES
  • Understanding of pH and its calculation from hydrogen ion concentration
  • Familiarity with acid dissociation constants (Ka) and their significance
  • Knowledge of equilibrium expressions in chemical reactions
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Review the concept of acid-base equilibrium and how to apply the ICE table method
  • Learn how to derive concentrations from pH using logarithmic relationships
  • Study the relationship between Ka and the concentrations of reactants and products in weak acid solutions
  • Practice similar problems involving weak acids and their dissociation in aqueous solutions
USEFUL FOR

Chemistry students, particularly those studying acid-base equilibria, and educators looking for examples of pH calculations and concentration determinations in weak acid solutions.

Mholnic-
Messages
43
Reaction score
0

Homework Statement


A particular sample of vinegar has a pH of 2.90. If acetic acid is the only acid that vinegar contains (Ka = 1.8 x 10-5), calculate the concentration of acetic acid in the vinegar.

Homework Equations


Ka = [Products]/[Reactants]
pH = -log[H+]




The Attempt at a Solution


I arranged the problem in my usual lazy way:
Acid + Water ---> Conjugate Base + Hydrogen Ions

Assuming that [H+] is equal to [Conjugate Base] I calculated the concentration of the conjugate base and hydrogen ions.

pH = -log[H+] = 2.90
[H+] = 10-2.90 = [Conjugate Base]

Then I plugged those numbers and the given Ka value into:

Ka = [products]/[reactants]

1.8 x 10-5 = [10-2.90]2/[Acid]

[10-2.90]2/1.8 x 10-5 = [Acid]

I get 8.8 x 10-12. The book says the answer is .089 M. Not quite sure where I'm going wrong.. maybe I am not using the correct equations or perhaps it's an overlooked error in algebra. Help!
 
Physics news on Phys.org
Mholnic- said:
[10-2.90]2/1.8 x 10-5 = [Acid]

Equation is correct, but it doesn't yield 8.8x10-12.

Note that concentration of the acid that you can calculate in this step is not the final answer yet.
 

Similar threads

Chemistry How to calculate pH at equivalence point using this alternative?
  • · Replies 7 ·
Replies
7
Views
2K
Chemistry Understanding how to calculate pH of a buffer solution
  • · Replies 2 ·
Replies
2
Views
2K
Chemistry How to take into account autoprotolysis in weak base solution?
  • · Replies 8 ·
Replies
8
Views
2K
Chemistry Comparing methods of computing pH at equivalence point in titration
  • · Replies 4 ·
Replies
4
Views
2K
Titration of acetic acid w/ NaOH problem
  • · Replies 2 ·
Replies
2
Views
3K
Chemistry Moles of NaOH to Create Buffer Solution pH=6 in 0.42M Ethanoic Acid
  • · Replies 3 ·
Replies
3
Views
3K
How Do You Calculate NaCH3COO Concentration and Predict pH at Equivalence Point?
  • · Replies 4 ·
Replies
4
Views
4K
Chemistry Solving pH Changes: Adding Acid to a Buffer or Water
  • · Replies 16 ·
Replies
16
Views
3K
Chemistry Moles of a conjugate base in a buffer solution
  • · Replies 7 ·
Replies
7
Views
4K
Calculating the pH of a Buffer Solution
  • · Replies 7 ·
Replies
7
Views
4K