Kaplan Acid/Base Buffer Question

[arsenal14]

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What is the approximate pH when .1 mol NaOH has been added to .2 mol HCN? (Ka for HCN 6x10^-10)

A. 10
B. 9.2
C. 7
D. 3.7

Answer is B.


At first glance, I thought that I would be able to get the answer without doing any calculations and just reasoning through the problem. My thought process was that there was a weak acid being titrated with a strong base. There is .1 mol of NaOH reacting with .2 mol HCN which means that the there will be .1 mol HCN left unreacted and .1 mol of NaCN and H20 produced correct? Wouldn't this buffer solution of HCN and NaCN be able to resist the .1 mol of NaOH, and thus keep the pH of the solution slightly acidic? Wouldn't it take .2 mol of NaOH, i.e., the equivalence point to completely react the HCN and thereby raise the pH significantly? Maybe I'm completely wrong, but to me it seems counter intuitive. The answer suggests using the HH equation to find the pH.

 

[ezsanche]

What is the approximate pH when .1 mol NaOH has been added to .2 mol HCN? (Ka for HCN 6x10^-10)

A. 10
B. 9.2
C. 7
D. 3.7

Answer is B.


At first glance, I thought that I would be able to get the answer without doing any calculations and just reasoning through the problem. My thought process was that there was a weak acid being titrated with a strong base. There is .1 mol of NaOH reacting with .2 mol HCN which means that the there will be .1 mol HCN left unreacted and .1 mol of NaCN and H20 produced correct? Wouldn't this buffer solution of HCN and NaCN be able to resist the .1 mol of NaOH, and thus keep the pH of the solution slightly acidic? Wouldn't it take .2 mol of NaOH, i.e., the equivalence point to completely react the HCN and thereby raise the pH significantly? Maybe I'm completely wrong, but to me it seems counter intuitive. The answer suggests using the HH equation to find the pH.

This was a PKA question. All you had to do was take the negative log of the KA.

 

[Element]

Plus NAOH is a strong base, whereas HCN is a very weak acid.

 

[arsenal14]

I guess what I'm trying to get at has a lot to do with my understanding of the equivalence point.

Hypothetically, if .2 mol of NaOH is added, then wouldn't that make the HCN fully neutralized? But there is only .1 mol of NaOH which means that there is still .1 HCN left in the solution, which also means that HCN hasn't been fully neutralized, correct? So all equations aside, can we not deduce that the pH would still be slightly acidic and below the equivalence point. Would this not be the half equivalence point of the reaction, since .1 mol of HCN has been reacted and .1 mol of CN- has been made? I really need to nail this concept down. Thanks.

 

[wait4me]

how do we know when we can just get the pH from the Ka rather than doing calculations and factoring in the base?

 

[ezsanche]

I guess what I'm trying to get at has a lot to do with my understanding of the equivalence point.

Hypothetically, if .2 mol of NaOH is added, then wouldn't that make the HCN fully neutralized? But there is only .1 mol of NaOH which means that there is still .1 HCN left in the solution, which also means that HCN hasn't been fully neutralized, correct? So all equations aside, can we not deduce that the pH would still be slightly acidic and below the equivalence point. Would this not be the half equivalence point of the reaction, since .1 mol of HCN has been reacted and .1 mol of CN- has been made? I really need to nail this concept down. Thanks.

Yes you are right!! And because there are equal concentration of HCN and CN the solution is at its PKA. The PH of the solution is then the negative log of the Ka. Different compounds have different PKAs so we can not deduce if the PH will be acidic or not at it's PKA until we look at the Ka value.

 

[minutemen11]

ok questions like these are what i hate the most in mcat... i struggle with these, but i think i can help..

the fact is that HCN is not a strong acid as you said, so it will not completely dissociate as given by the Ka values. If you had a strong acid, HCl, THEN your arguement is correct because you need exactly the same Molarity of strong base to neutralize that acid. So since you have a really weak acid, adding a strong base will make the solution more basic, ie the equivalence point will be higher.

i'm acutally wondering how to solve this problem so here goes:
the pKa for HCN will be around 9.1ish- you get this value because -log(6*10^-10) will not be 10, it will be less than 10 and since 6 is pretty close to the number 10 you can assume that the pka will be much closer to 9 than to ten, this is how you get around 9.1-9.2ish (if it was 3*10^-10, 3 is much closer to number 1 so the value will be much closer to 10 like 9.7ish i think)

now you use the HHE equation: pH=pKa+ log(A/HA)
this is where im totally stuck.. i need help from here on out because the OP only gives moles not the Molarity.. i do not know how the heck to go from moles to Molarity if the volume of the solution is not given... is there a trick to this? help..

 

[ezsanche]

ok questions like these are what i hate the most in mcat... i struggle with these, but i think i can help..

the fact is that HCN is not a strong acid as you said, so it will not completely dissociate as given by the Ka values. If you had a strong acid, HCl, THEN your arguement is correct because you need exactly the same Molarity of strong base to neutralize that acid. So since you have a really weak acid, adding a strong base will make the solution more basic, ie the equivalence point will be higher.

i'm acutally wondering how to solve this problem so here goes:
the pKa for HCN will be around 9.1ish- you get this value because -log(6*10^-10) will not be 10, it will be less than 10 and since 6 is pretty close to the number 10 you can assume that the pka will be much closer to 9 than to ten, this is how you get around 9.1-9.2ish (if it was 3*10^-10, 3 is much closer to number 1 so the value will be much closer to 10 like 9.7ish i think)

now you use the HHE equation: pH=pKa+ log(A/HA)
this is where im totally stuck.. i need help from here on out because the OP only gives moles not the Molarity.. i do not know how the heck to go from moles to Molarity if the volume of the solution is not given... is there a trick to this? help..

you are done! PKA=PH when the concentration of the conjugate acid and base are equal. And in this case, these concentrations are equal!

 

[minutemen11]

you are done! PKA=PH when the concentration of the conjugate acid and base are equal. And in this case, these concentrations are equal!

whatt.. umm explain that to me as if i were a pre-schooler.

how are the concentrations equal, you dont know what the concentrations are.. it only gives you moles not Molarity..

 

[ezsanche]

whatt.. umm explain that to me as if i were a pre-schooler.

how are the concentrations equal, you dont know what the concentrations are.. it only gives you moles not Molarity..

You are right we dont know the concentration but that is not required.

molarity = Moles of solute/ Volume of solution

We know moles..... But we dont know volume.

However we do know that the conjugate acid (HCN) and the conjugate base (CN) are in the same solution. Hence the volume of solution has to be equal for both (even though we dont know the exact volume is). and since the conjugate acid and base have the same amount of moles (.1 moles) they have to have the same amount of concentration (although we dont know the exact concentration of the solution is). I hope that helps. This concept is quite difficult to explain over text.

 

[minutemen11]

would this mean that the amount of NaOH you add is irrelevant? like if i dumped 2 moles of NaOH the pH will not still be 9.2, right??
im just confused because the volume of NaOH you add does matter and will make a difference..

pka is 9.2 for HCN (that itself i dont understand because it should be less than 7 because its a weak acid..)

 

[arsenal14]

Oh wow now I get it! Thanks ezsanche!!!


Ok minutemen11, pKa is not the same as the pH, usually a strong acid will have a low pKa and a weak one will have a high pka. 9.2 is the pKa of the NaOH in the HCN.

pKa= -log (Ka)

sooo... if the Ka is big like HCl it will have a small pKa. This means that it will dissociate more so than a weaker acid. Also, I believe pKa's (just like Ka's) are set values.


If you dumped 2 moles of NaOH only .2 moles of the NaOH would react with the .2 moles of the HCN. After that there is just 1.8 moles of NaOH floating around in solution because there is no more HCN for it to react with. At this time, the amount of OH (1.8mol because 2-.2=1.8) will control the pH all of the HCN is gone. In order to calculate the pH (assuming there is 1L soln), we take the
-log [1.8]= pOH-----> 14-pOH = pH
If anything is wrong plz correct me.
And again thanks ezsanche! 6 more days till doomsday! F*** my life...

 

[ezsanche]

Oh wow now I get it! Thanks ezsanche!!!


Ok minutemen11, pKa is not the same as the pH, usually a strong acid will have a low pKa and a weak one will have a high pka. 9.2 is the pKa of the NaOH in the HCN.

pKa= -log (Ka)

sooo... if the Ka is big like HCl it will have a small pKa. This means that it will dissociate more so than a weaker acid. Also, I believe pKa's (just like Ka's) are set values.


If you dumped 2 moles of NaOH only .2 moles of the NaOH would react with the .2 moles of the HCN. After that there is just 1.8 moles of NaOH floating around in solution because there is no more HCN for it to react with. At this time, the amount of OH (1.8mol because 2-.2=1.8) will control the pH all of the HCN is gone. In order to calculate the pH (assuming there is 1L soln), we take the
-log [1.8]= pOH-----> 14-pOH = pH
If anything is wrong plz correct me.
And again thanks ezsanche! 6 more days till doomsday! F*** my life...

that is right! GL with your Test!