titration question.

[dorjiako]

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Although this question seems easy on the face value, I have been getting different answers on it depending on whether I used PH=PKA + Log(A-/HA) or I used the Ka constant to find it. So, I need somebody who is good with this kind of questions to explain to me step by step the right way to solve this question. Thanks a lot in advance.

Question: What is the PH of a 100ml solution of 0.1 M acetic acid to which 50ml of 0.1M NaoH has been added (ka of acetic acid is 1.8 x 10^-5)?

A.5, B.7, C.3, D.10.

Again, I am truly thankful for helping me understand this question once and for all.

 

[tendiw]

Hi,

Acetic acid is a monoprotic acid, meaning one mole of acetic acid will react with on mole of NaOH.

The amount of NaOH added would mean that half of the acetic acid has been titrated. So basically 50% of the solution is made of acetic acid and 50% is made of the conjugate base of acetic acid. At this point you have a buffer solution.

The pH of the buffer solution is the pKa of acetic acid. So all you need to do is use the ka to find the pKa and ou answer should be pH 5.

If you use the henderson hasselbach equation you provided, you should realize that the log portion of the equation is equal to zero as log(1)=0. So pH=pKa like I stated. If you dont know where the "1" came from then simply remember that because we have a buffer [A-]=[HA], when you divide, you get 1.

If you look at buffers again you will see thats all there is to. Remember that usually the MCAT asks questions about half equivalence points (where pH=pKa). If 100mL of NaOH was used, all the acid would have been tirated (equivalence point) and the pH would be above 7 which happens when a weak acid is titrated with a strong base.

To find the equivalence pH( which the MCAT is unlikely to ask as it is more time consuming) just use the pKb of the conjugate base to find the [OH-] and find -log[OH-] to get a pOH then convert it to pH (pH=14-pOH) if your pH is less than 7 then its wrong. The equivalenc point hsould be above 7...but not extremely high

 

[dorjiako]

Hi,

Acetic acid is a monoprotic acid, meaning one mole of acetic acid will react with on mole of NaOH.

The amount of NaOH added would mean that half of the acetic acid has been titrated. So basically 50% of the solution is made of acetic acid and 50% is made of the conjugate base of acetic acid. At this point you have a buffer solution.

The pH of the buffer solution is the pKa of acetic acid. So all you need to do is use the ka to find the pKa and ou answer should be pH 5.

If you use the henderson hasselbach equation you provided, you should realize that the log portion of the equation is equal to zero as log(1)=0. So pH=pKa like I stated. If you dont know where the "1" came from then simply remember that because we have a buffer [A-]=[HA], when you divide, you get 1.

If you look at buffers again you will see thats all there is to. Remember that usually the MCAT asks questions about half equivalence points (where pH=pKa). If 100mL of NaOH was used, all the acid would have been tirated (equivalence point) and the pH would be above 7 which happens when a weak acid is titrated with a strong base.

To find the equivalence pH( which the MCAT is unlikely to ask as it is more time consuming) just use the pKb of the conjugate base to find the [OH-] and find -log[OH-] to get a pOH then convert it to pH (pH=14-pOH) if your pH is less than 7 then its wrong. The equivalenc point hsould be above 7...but not extremely high

I think you have helped set me on the path of MCAT success. Thanks again.