[cornfused]

Hi There,

I was given this question on an exam (i don't know if I got it right or wrong yet) and am hoping someone can help me understand this

I was given 4 options - asked which was NOT a buffer.
1)100 mL of 0.1 M Na2CO3 and 50 mL of 0.1 M HCl
2)100 mL of 0.1 M NaHCO3 and 25 mL of 0.2 M HCl
3)100 mL of 0.1 M Na2CO3 and 75 mL of 0.2 M HCl
4)100 mL of 0.2 M Na2CO3 and 100 mL of 0.4 M HCl

I wrote out the reaction and made ice tables for all of them
I was able to eliminate the first 2 options because it left .005 moles of Na2CO3 and HCO3 - weak base/conj acid.

For the first one:
Na2CO3 + HCl <--> HCO3 + NaCl
.01         .005            0      0
-.005     -.005        +.005   +.005
--------------------------------------
 .005                      .005
equal amouts of a weak base and its conjugate acid = Buffer

The second one is the same
Na2CO3 + HCl <--> HCO3 + NaCl
.01         .005            0      0
-.005     -.005        +.005   +.005
--------------------------------------
 .005                      .005
  = Buffer

the third one and last one confused me...

Na2CO3 + HCl <--> HCO3 + NaCl
.01        .015           0          0
-.01      -.01           +.01    +.01
--------------------------------------…
0         .005          .01         .01
not a buffer b/c no Na2CO3 left, and HCl is a strong acid

the last I had as
Na2CO3 + HCl <--> HCO3 + NaCl
.02          .04            0        0
-.02        -.02         +.02    +.02
--------------------------------------…
0             .02           .02      .02
also not a buffer???

I'm confused about the 3rd and 4th option... both seem to leave no Na2CO3, and only HCl which couldn't make a buffer because it's a strong acid.  The exam had specifically stated there was ONLY 1 CORRECT ANSWER. 

I then thought that maybe we have to do a second dissociation for the HCO3 - but still not sure how to tell which one is not the buffer.

Can someone please help me with this and explain?  I thought I was understanding these types of problems until I came across this one... I chose the 3rd option on the exam...

Any help and explanation would be appreciated!!!

[Borek]

In all cases try simple stoichiometry - assume protonation reaction went to completion. What is left in the solution in each case?

[cornfused]

Borek - thanks for responding

Do you mean that for option 3, the .005 would react with the .01 HCO3 and then you would have .005 HCO3 left that goes into the second dissociation, leaving you with HCO3 and CO3?

And the last one would leave you with neither HCl orHCO3 since it's .02 of each?  

So the correct answer should have been option 4 ?

Option 3 would make a buffer

[AWK]

4 - completely wrong calculations. Take into account this reaction:
Na₂CO₃ + 2HCl = 2NaCl + H₂CO₃

[Borek]

Do you mean that for option 3, the .005 would react with the .01 HCO3 and then you would have .005 HCO3 left that goes into the second dissociation, leaving you with HCO3 and CO3?

No idea what you are talking about. Not only you are ignoring charges which makes following what you write difficult, you are also most likely wrong, as the final solution will not (well, almost not) contain CO₃^2-.