Calculate the molarities of CH3COOH and CH3COO-

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This discussion focuses on calculating the molarities of acetic acid (CH3COOH) and its conjugate base acetate (CH3COO-) in a buffer solution comprising 40.00 ml of 0.100M acetic acid, 10.00 ml of distilled water, and 30.00 ml of 0.100M sodium hydroxide (NaOH). Upon the addition of NaOH, which dissociates completely, the hydroxide ions (OH-) react with acetic acid, altering the equilibrium. The new concentrations of CH3COOH and CH3COO- can be determined by adjusting the initial concentrations based on the reaction between OH- and acetic acid.

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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?
 
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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:
Still don't get it. Chemistry is a nightmare for me.Can someone explain in detail.:frown:
 
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.
 
don't forget to add the concentration value of OH- to the conjugate base concentration of the acetic acid.
 
thanks a lot
 

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