Simple Neutralization Question

[sanguinee]

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To neutralize 25mL of 0.10M acetic acid, how many mLs of 0.10M of KOH would be necessary?
TBR says 25mL, but I thought a weak acid only needs half the volume of a strong acid to be neutralized - so 12.5 mLs would be needed. Can someone tell me where I'm going wrong?

 

[Eaglefan]

As you neutralize the H+ that the weak acid has dissociated, it will shift to make more. this continues until the whole eq of acid is used. Revisit this, I'm not sure where you got that you only need half the volume!

 

[amy_k]

This is a common misconception - weak acids still need the same amount of OH- to be neutralized as strong acids.

 

[NextStepTutor_3]

Just to add - this misconception usually stems from confusion involving buffers. Let's say you do have 25 mL of 0.10 M acetic acid. After you've added 12.5 mL, you've neutralized exactly half of the original acid, leaving you at the half-equivalence point. Where people get confused is that this is the point where an optimal buffer is formed - in other words, [HAc] = [Ac-]. The moles of acetic acid remaining will equal the moles of its conjugate base.

To actually neutralize the original 25 mL, as said above, you'd need 25 mL of 0.10 M base. At this point, the moles of original acid will equal the moles of base added, and all of your original HAc will be converted to Ac-.

 

[shefv]

You can't generalize this based on the strength of the acid or base - you have to think about the moles of H+ and OH- that each one can contribute in solution.

Use the equation - n1C1V1 = n2C2V2

In this case, n values are the same