Calculate the molarities of CH3COOH and CH3COO-

  • Thread starter surahyot
  • Start date
In summary, the conversation is about calculating the molarities of CH3COOH and CH3COO- in a buffer solution. After adding NaOH, the equilibrium of acetic acid will be altered by the reaction with OH-. To find the new concentrations, the initial concentrations of both OH- and acetic acid must be determined and used in the equilibrium equation. It is important to add the concentration of OH- to the conjugate base concentration of acetic acid.
  • #1
surahyot
5
0
Can someone help me how to calculate the molarities of CH3COOH and CH3COO- in a buffer solution containing 40.00 ml of 0.100M acetic acid, 10.00 ml of distilled water and 30.00ml of 0.100M NaOH?
 
Chemistry news on Phys.org
  • #2
What will happen after sodium hydroxide, which dissociates completely in water, is added? The hydroxide will alter the equilibrium of the acetic acid by reacting with _______?

You need to find the new concentrations after the equilibrium.
 
Last edited:
  • #3
Still don't get it. Chemistry is a nightmare for me.Can someone explain in detail.:frown:
 
  • #4
Thus the first thing to do is to find the initial concentrations of both OH and acetic acid.

The base, OH- (which dissociates completely), will react with the acetic acid directly. So subtract the original concentration of acetic acid by the concentration of hydroxide anion. Using this concentration substitute into the equilibrium equation and find the rest of the concentrations.

Hope this answers your question.
 
  • #5
don't forget to add the concentration value of OH- to the conjugate base concentration of the acetic acid.
 
  • #6
thanks a lot
 
Back
Top