Why Is the ICE Table Set Up This Way for CH3COOH and NaCH3COO Solution?

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SUMMARY

The discussion centers on the pH calculation of a solution containing 0.100 M acetic acid (CH3COOH) and 0.100 M sodium acetate (NaCH3COO), with a dissociation constant (Ka) of 1.8x10-5. Participants clarify that the ICE table setup is unnecessary since both the acid and its conjugate base are introduced at equal concentrations. The sodium ion (Na+) is identified as a spectator ion, and the molarity of sodium acetate is equated to that of acetate ions (CH3COO-) due to dissociation stoichiometry. The discussion also references the Henderson-Hasselbalch equation for further context.

PREREQUISITES
  • Understanding of weak acid dissociation and equilibrium constants (Ka).
  • Familiarity with the Henderson-Hasselbalch equation.
  • Basic knowledge of ICE tables in chemical equilibrium.
  • Concept of spectator ions in chemical reactions.
NEXT STEPS
  • Study the Henderson-Hasselbalch equation in detail.
  • Learn about weak acid and conjugate base equilibria.
  • Explore the concept of spectator ions in various chemical reactions.
  • Investigate the implications of dissociation stoichiometry in buffer solutions.
USEFUL FOR

Chemistry students, educators, and anyone involved in acid-base chemistry or buffer solution preparation will benefit from this discussion.

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Homework Statement



What will be the pH of a solution that contains 0.100 M CH3COOH and 0.100 M NaCH3COO?Ka = 1.8x10-5
IMPORTANT: THE REACTION IS NOT BETWEEN CH3COOH AND CH3COO-

- My question here is why the ice table was setup as so by the instructor? see below...



Homework Equations





The Attempt at a Solution



CH3COOH + H20 ---> CH3COO- + H3O
.100M-x .100m+x +x

Where did the Na go? Why did the Molarity of NaCH3COO = Molarity of CH3COO- ?
 
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Raul3140 said:
My question here is why the ice table was setup as so by the instructor? see below...

No idea. No need for an ICE table. You can safely assume concentrations of both acid and conjugate base are these introduced into solution.

Where did the Na go?

Doesn't matter, Na+ is just a spectator.

Why did the Molarity of NaCH3COO = Molarity of CH3COO- ?

Dissociation stoichiometry, plus the assumption explained above.

Compare Henderson-Hasselbalch equation.
 

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