Calculate pH of CH3COOH & NaOH Solution | Ka=1.8x10^-5

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SUMMARY

The discussion focuses on calculating the pH of a solution created by mixing 200 mL of 0.2 M acetic acid (CH3COOH) with 100 mL of 0.1 M sodium hydroxide (NaOH). The relevant equilibrium reaction is CH3COOH + NaOH ⇔ NaCH3OO + H2O, with the acid dissociation constant (Ka) given as 1.8 x 10^-5. Participants emphasize the importance of applying the Henderson-Hasselbalch equation and understanding the equilibrium concentrations to determine the final pH accurately.

PREREQUISITES
  • Understanding of acid-base reactions and equilibrium
  • Familiarity with the Henderson-Hasselbalch equation
  • Knowledge of calculating concentrations in mixed solutions
  • Basic grasp of pH and pKa concepts
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  • Study the Henderson-Hasselbalch equation in detail
  • Learn how to calculate equilibrium concentrations in buffer solutions
  • Explore the concept of acid dissociation constants (Ka) and their applications
  • Practice similar pH calculation problems involving weak acids and strong bases
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Chemistry students, educators, and anyone preparing for exams involving acid-base chemistry and pH calculations.

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Homework Statement


A solution is prepared by mixing 200 mL of 0.2 M CH3COOH with 100 mL of 0.1 M of NaOH solution.Calculate the pH of the solution.(Ka=1.8x10^-5)


I really don't know how to start this, so please help me.Its going to be a similar one on my exam tomorrow.

Thanks.
 
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Some of what you are missing is to fill in THIS information more precisely:

Homework Equations

and rules[/B]

CH3OOH + NaOH ⇔ NaCH3OO + H2O

Ka = \frac{[H][CH3OO]}{[CH3OOH]}
 
Buffer and Henderson-Hasselbalch equation.
 

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