# GenChem Acid-Base Titration Question

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#### Pre-Med Oso

##### New Member
Hi everyone, I am struggling to understand how the question creators know to use the value 1.5 N H2CO3 in the explanation for this problem. If there is anyone who could explain to me how they got this value it would be much appreciated or if you have a different method of solving these problems please share I would like to know if there is another way to solve the problem.

A.83 mL

B.166.5 mL

C.333 mL

D.1000 mL

Answer is A: Point A is the first half-equivalence point, meaning that we need to add exactly half of the volume required to neutralize the first proton, or one-quarter of the amount needed to fully neutralize H2CO3. We can use N1V1 = N2V2: (1.5 N H2CO3)(500 mL) = (2.25 M NaOH)(x mL). Solving for x, we get a volume of 333 mL. However, remember that this equation gives us the volume of base needed to fully neutralize the acid! To answer this question, we need to divide by four, giving us 83.3 mL.

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It looks like a typo to me, they just doubled the 0.75M H2CO3 and then divided the volume of base by 4. It works out to the same answer if you follow their steps and divide the volume of base needed to fully neutralize the base by 2.

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Hi everyone, I am struggling to understand how the question creators know to use the value 1.5 N H2CO3 in the explanation for this problem. If there is anyone who could explain to me how they got this value it would be much appreciated or if you have a different method of solving these problems please share I would like to know if there is another way to solve the problem.
View attachment 351843

A.83 mL

B.166.5 mL

C.333 mL

D.1000 mL

Answer is A: Point A is the first half-equivalence point, meaning that we need to add exactly half of the volume required to neutralize the first proton, or one-quarter of the amount needed to fully neutralize H2CO3. We can use N1V1 = N2V2: (1.5 N H2CO3)(500 mL) = (2.25 M NaOH)(x mL). Solving for x, we get a volume of 333 mL. However, remember that this equation gives us the volume of base needed to fully neutralize the acid! To answer this question, we need to divide by four, giving us 83.3 mL.
N is normality. Normality is molarity times the number of protons (or OH-) in one mole of acid (or base). Because H2CO3 yields two moles of protons for every one mole of H2CO3, the normality is 0.75 M x 2 = 1.50 N.

The final equivalence point for all titrations can be solved by using Normality instead of Molarity; hence, N1V1 = N2V2. However, like the solution mentions, point A is half-way to the first equivalence point; divide V2 by 4 to get 83.3 mL.

For this question, you need to focus on the fact that a is halfway to the first equivalence, b is the first equivalence point, and c is halfway between the first and second equivalence points.

You can ignore normality for this, because point a is before the equivalence point, so you are not dealing with the second proton.

To reach the first equivalence point, you know that MHA x VHA = MOH- x VOH-, as a general definition. We can ignore the second proton of carbonic acid, because we are in the region of the diprotic curve associated with the first proton. To reach point b, we get:
(0.75 M) x (500 mL) = (2.25 M) x (Vfirst eq) SO Vfirst eq = (0.75 x 500)/2.25 = 500/3 = 166.7 mL.

Point a is halfway to point b, so we need half of 166.7 mL to reach point a, the halfway point. That is 83.33 mL, choice A. I hope this helps.

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