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# Titration of a weak acid with a strong base

Calculating the pH for titration of acetic acid with strong base NaOH before adding any base and at half-equivalence point.  Created by Jay.

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• Hi,
At where he states that we started with 0.01 moles of CH3COOH, didnt we actually start with 0.01 - x moles of CH3COOH, where x is the number of moles that dissociated into H+ and CH3COO- ions?
• Yes, technically, but at about he explains that x is so small that in comparison with the initial concentration/amount that it hardly makes a difference. In other words, 0.01-x is almost equal to 0.01, to the point where subtracting x unnecessary, especially given the low precision of the data. The % ionization of acetic acid, by the calculations in this problem, is around 1%, so only a tiny amount of the acetic acid is dissociated at any given time. Thus, it makes more sense to write the amount of acetic acid in solution as 0.01 moles– it is 99% accurate and it simplifies the problem immensely. However, if you are dealing with the equivalence point, where all of the available acetic acid has been neutralized, the amount of conjugate base (CH3COO-) becomes significant, since it is the only species left in solution other than water, and you will want to calculate that x to find out the pH of the solution at the equivalence point, or what have you. I hope this helps!
• how do you know this a buffer solution?
• because it's a solution of a weak acid and it's conjugate base
• Why did you not account for the reaction of acetate and water to calculate the pH in part b like you did in part c (next video)?
• for the 1st section on initial pH of the solution before adding any base, why is it incorrect to estimate the pH by using the concentration of acetic acid and plugging it into the pH formula?
• Because pH = -log [H3O+]. The initial concentration of CH3COOH is not equal to [H3O+]. You need to do ICE in order to figure out how much H3O+ has been formed by the reaction of CH3COOH with water.
• For problem b, could we just have looked at solely the H3O+ and OH- and figured out how much H30+ was left after adding NAOH, then using that to find the pH of the solution? We can find the amount of moles of H30+ from the calculations we made in a).
• Can the moles of h30+ be used to determine the ph instead of moles of weak acid. It's seems to be more direct.
• You need to know the moles of the weak acid and the pKa to determine how much of the acid dissociates and forms H3O+, which is ultimately used to determine pH. You need to go through this route because acid will (rarely) dissociate completely, and therefore 1M of [acid] will not always give you 1M of H3O+.
• why did we not account for the dissociation of acetic acid in water to form some deprotonated acetic acid. With initial concentration of 0.019 similar to the concentration of the hydronium ion
• At it talks about the Henderson-Hasselbach equation. What other equation could be used to find the pH of a buffer solution?
• There are alternates, but this is the most useful. It won't be any easier than with this form of the HH equation.
(1 vote)
• neutralisation of ch3cooh is reversible reaction ,isn't?
(1 vote)
• Yes, CH3COOH (Acetic Acid) is a weak acid and does not dissociate completely in a solution.
• why it is erong to check the PH of the acid after we have the concentration of it? I know it's different number, but i'm looking for the logic behind it