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# Titration of a weak acid with a strong base (continued)

Calculating the pH for titration of acetic acid with strong base NaOH at equivalence point and past the equivalence point.  Created by Jay.

## Want to join the conversation?

• For the last part, how come you didn't have to solve for whatever OH- concentration the acetate then turned the water into like in part c?
• You're right. Hydrolysis of A⁻ does generate some OH⁻.
A⁻ + H₂O ⇌ HA + OH⁻
But the amount is so small compared to the excess NaOH that you have added that it is negligible.
It's like saying that x is negligible in all your other calculations.
• At why did we further react acetate with H2O instead of just using the Molar concentration of 0.04 of acetate to calculate the pH?
• Since CH3COO- is a weak basic salt, it would not stay disassociated like Cl- salt. So some CH3COO- will be protonated to form CH3COOH. This reassociation to form the acid is important here since there is only the acetate present because as the acetate converts back to acetic acid, OH- is formed. And there is no CH3COOH present in the original solution to neutralize the OH- to form water, and will result in a pH change. Basically, the 0.04 M acetate solution is not a buffer solution, so protonation of the acetate will result in a pH change.
• How can you tell if X is going to be a very small number to count as negligible ?
• If the initial concentration is greater than 400×Ka or 400× Kb, x is negligible.
• If I have a solution which contains is affected by common ion effect (lets say 0.5 mol HF and 0.9 mol NaF) and then I apply titration using 2g of NaOH, how do i take account of both common ion effect equilibrium and the weak acid titration equilibrium?
• You are starting with 0.50 mol of HF and 0.90 mol of F⁻.
You are adding 0.05 mol of OH⁻.
This will react with 0.05 mol of HF and form 0.05 mol of F⁻.
You will then have 0.45 mol of HF and 0.95 mol of F⁻.
These are the numbers you use to calculate the new pH.
• For part D, why didnt you use the same steps for part C, given a second ice table?
• In part D, the acetic acid had all been neutralized, so all you had was excess NaOH.
There was effectively no equilibrium to calculate, so there was no need for an ICE table.
• I understand this completely, and this process. Will we have to memorize Ka values for acids and bases on the MCAT? I can't do these calculations without a calculator.
• No you will not have to memorize the Ka values. On the MCAT, the values they give you are easily doable without the calculator so it won't be a big deal.
• Why can't we use the Henderson-Hasselbach Equation for part c?
• The CH3COOH is all used up in the reaction with OH from NaOH. That means that it is no longer a buffer solution, and Henderson-Hasselbach is only for buffers.
• At , are we ignoring Hydrolysis of CH3COO- due to common ion effect of OH- ?
• at 13.48, why can we calculate the POH from left-over OH- concentration directly instead of putting both CH3COO- and OH- together and re-calculate the [OH-]. though the increase amount is too small and the final result is the same. but, why we can do it? under what situation we can and cannot? Tks.

When dealing with a strong base and a weak acid/weak base buffer system, once you pass the equivalence point the contribution from the weak acid/weak base becomes trivial. This can easily be checked using the known excess of OH¯, the concentration of the weak conjugate base (CH₃OO¯) and the Kb.

We first assume that the OH¯ only comes from NaOH and calculate what happens to the acid-base equilibrium at this point.
Plug those values into the equation for Kb:
Kb = [OH¯] • [CH₃OOH] / [CH₃OO¯]
5.6e-10 = 0.014 M • [CH₃OOH] / [CH₃OO¯]
[CH₃OOH] / [CH₃OO¯] = 5.6e-10 / 0.014 M = 4.0e-8

So, for every CH₃OO¯ that reacts with water to form CH₃OOH and OH¯, there will be 25 million CH₃OO¯ that don't react.
• The reactions vary a lot in different stages. How to know which one to use at which point?
• Trust the Henderson Hasselbach equation and know precisely what it is you are trying to do. If you want pKa or pH and have the other, use the Hend-Hass equation. If you want to know how much dissociate, use the other formula.
(1 vote)