# Number 44 in C/P Section Bank

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#### deleted647690

I was confused on this question, but I think I figured out my confusion, could someone tell me if I'm thinking about this correctly?

At first, I was confused as to why you would just calculate the number of moles of HEW. I knew that it was a 1:1 molar ratio at the equivalence point, so moles HEW = moles NAG3, and I didn't see why you couldn't just use moles of NAG3.

But then I realized that 4 equivalents of NAG3 were needed to neutralize 1 equivalent of HEW:

1*10^-3 L * .1*10^-3 mol/L HEW = 1 *10^7 mol HEW

10*10^-6 L * 2.5 * 10^-3 mol/ L NAG3 = 25 * 10 ^-9 mol NAG3

0.1 * 10^-6 mol HEW / 25 * 10 ^ -9 mol NAG 3 = 4

So the moles of NAG3 must be multiplied by a factor of 4 to get an equivalent of HEW, thus giving you the same number of moles of HEW needed for the equivalence point.

Not sure if I'm using the term "equivalence" correctly for that factor of 4 that I found above

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#### theonlytycrane

##### Full Member
7+ Year Member
Yeah, that's correct. The shortcut is knowing that the mol of NAG3 must equal mol of HEW at the equivalence pt, so you could just figure out the # of mol of HEW.

Each injection adds 2.5 * 10^-8 mol of NAG3 into the titration. After 4 injections the curve levels off at the equivalence pt.

D

#### deleted647690

Yeah, that's correct. The shortcut is knowing that the mol of NAG3 must equal mol of HEW at the equivalence pt, so you could just figure out the # of mol of HEW.

Each injection adds 2.5 * 10^-8 mol of NAG3 into the titration. After 4 injections the curve levels off at the equivalence pt.
Thanks!

#### traveller12#12#

##### New Member
2+ Year Member
How did you know 4 equivalents of NAG3 were needed to neutralize 1 equivalent of HEW?
Thanks!

#### traveller12#12#

##### New Member
2+ Year Member
Yeah, that's correct. The shortcut is knowing that the mol of NAG3 must equal mol of HEW at the equivalence pt, so you could just figure out the # of mol of HEW.

Each injection adds 2.5 * 10^-8 mol of NAG3 into the titration. After 4 injections the curve levels off at the equivalence pt.

Why 4 injections? How did you know 4 equivalents of NAG3 were needed to neutralize 1 equivalent of HEW?
Thanks!

#### Ro300!

##### New Member
Why 4 injections? How did you know 4 equivalents of NAG3 were needed to neutralize 1 equivalent of HEW?
Thanks!

acetylmandarin found the molar ratio of HEW to NAG3 in the first thread above.

"0.1 * 10^-6 mol HEW / 25 * 10 ^ -9 mol NAG 3 = 4"

#### Flufpot12

##### Full Member
It says that the “graphs both imply that 1:1 ratio of NAG3 was added at the equivalence point”...how does the second graph show that? I don’t even understand what’s happening in the second though. At 0 moles of NAG3, less heat is released. That changes the equivalence point right?

#### Ro300!

##### New Member
It says that the “graphs both imply that 1:1 ratio of NAG3 was added at the equivalence point”...how does the second graph show that? I don’t even understand what’s happening in the second though. At 0 moles of NAG3, less heat is released. That changes the equivalence point right?

From what I can gather (someone please correct me if I'm wrong), the decrease in energy released is as a result of the addition of a weakly competitive substrate (NAG) in the 2nd graph to the right. The higher the affinity between the substrate (NAG3) and the enzyme (HEW), the greater the heat released. Due to the presence of NAG is the graph to the right, the affinity of NAG3 for HEW is slightly weakened. In the 2nd graph to the right, NAG3 serves as the displacement titrant when NAG is weakly bound to HEW b/c the interaction btw NAG3 & HEW is easier to detect than interactions btw NAG & HEW (as stated in the passage).

With regard to the ratio, I reasoned that the 1:1 ratio of NAG3/HEW would be independent of the heat released (think of the reaction N2 + 3H2 --> 2NH3 for example). The stoichiometry (similar to moles at equivalent point) remains the same and is independent of the yield of the reaction (or in this case, the affinity as indicated by the decrease in heat released) which may change.

Honestly, I'm also in the process of fully understanding this question and any clarification/elaboration will be greatly appreciated!