TBR Buffer Problem

[jsmith1]

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Q: Which mixture does NOT produce a buffer?

Answer: A) H2CO3 with 1.5 equiv KOH
B) H3CNH2 with 1.5 equiv HCl
I eliminated choices C and D.

The answer is B. And the book says that the answer is NOT A because H2CO3 is diprotic so the 1.5 equiv. of KOH completely removes the first H+ to make HCO3- and then pulls off half of the protons from bicarbonate ions which results in a solution with equal parts HCO3- and CO3(2-)

How does that give equal parts HCO3- and CO3(2-)? Shouldnt it result in .5 HCO3- and 1.5 CO3(2-)??

 

[Vurday]

Let's break down answer choice A into two steps:

1eq H2CO3 + 1eq NaOH --> 1eq HCO3-
At this point, you've deprotonated ALL of your H2CO3. Since we started with 1.5eq NaOH, after 1 eq is used up, we'll still have 0.5eq NaOH left.

Then, we'll use that 1eq of HCO3-:
1eq HCO3- + 1/2 eq NaOH --> 1/2 eq CO3(2-) + 1/2 eq HCO3-
In this step, that 0.5 eq NaOH can only deprotonate half of the 1 equivalent of HCO3- leaving you with 0.5eq HCO3- and 0.5eq fully deprotonated CO3(2-)

 

[jsmith1]

Let's break down answer choice A into two steps:

1eq H2CO3 + 1eq NaOH --> 1eq HCO3-
At this point, you've deprotonated ALL of your H2CO3. Since we started with 1.5eq NaOH, after 1 eq is used up, we'll still have 0.5eq NaOH left.

Then, we'll use that 1eq of HCO3-:
1eq HCO3- + 1/2 eq NaOH --> 1/2 eq CO3(2-) + 1/2 eq HCO3-
In this step, that 0.5 eq NaOH can only deprotonate half of the 1 equivalent of HCO3- leaving you with 0.5eq HCO3- and 0.5eq fully deprotonated CO3(2-)

Ohh man dumb mistake on my part. Thanks