<p>Ok, this has to do with dilution/titration.
I'm working out of the Kaplan prep book. Here is an example of a questions:</p>

<p>What volume of water would be needed to dilute 50mL of 3M HCl solution to 1M?</p>

<p>so, MV=MV: 50x3=1xV, V=150. However they say the answer is 100, cuz you already have 50 ml of water (100+50=150). Is this what the question means? It kinda sounds like they would have to say how much water would have to be added for an answer of 100 to be correct. But I guess not?? Is it cuz of the ''to dilute phrases?''</p>

<p>Also, I remember learning when you have a strong Polyprotic acid like H2PO4, and u are going to do a titration/dilution, then you have to factor in the the amount of H's in the MV=MV equation? Is this right, if not, when?</p>

<p>Here is an example question from Kaplan:</p>

<p>What volume of water would be needed to dilute 50ml of 3M H2SO4 TO .75 ml?</p>

<p>They do it just by: 50 x 3 = .75 x V, and u get V=200. However the FINAL answer they have is 150, cuz 150 is added to 50 to get 200.</p>

<p>I'm in AP Chem, and don't remember this kind of methodoolgy. Help, please, haha.</p>

<p>im not sure if their answer is correct, ask your ap chem teacher. as far as the polyprotic acid, i think that only comes into play with titration since titration involves titrating acid/acid, base/base, acid/base, whereas dilution only deals with water and an acid or base. so, the method they have for the volume of water needed for diluting the H2S04 seems correct, but again check with your teacher about the 200 vs 150 thing for final answer.</p>

<p>i dont think my teacher would know lol. but thanks!</p>

<p>anyone else have any insight?</p>

<p>The answer is correct. The correct formula for dilution/titration is n1M1V1=n2M2V2, where n=number of H+ (which is where polyprotic acids come in), M=molarity, and V=volume. The n is not important for the HCl (one H). Therefore, you have: </p>

(3 M)(50 mL)=(1 M)(X mL)
X=150 mL </p>

<p>This is the total amount of AQUEOUS SOLUTION, not water added. To actually dilute the acid, you would pour the 50 mL of 3 M HCl into a volumetric flask and add water until the total volume is 150 mL. In this case, the volume of water added would be 100 mL.</p>

<p>Ok. Thanks a lot. Duh . Here is a practice AP Exam question we took in class:</p>

<p>How many milliliters of 11.6M HCl must be diluted to obtain 1.0 L of 3M HCl?
So: 11.6V= 1 x 3, V = about .25= 250 ml. Now, that is the final answer for this question. </p>

<p>So, what is the difference here? Is it because the question asks just ''How many millilters,'' and does not specify water? In other words, is it asking here for the plain total volume?</p>

<p>And for the second question I had originally put, how come you do not take into account the 2 H's in H2SO4?</p>

<p>Well for the practice problem, its asking for the total volume you need to add at first to dilute the HCl, so there's no need to add the volume you origninally had. </p>

<p>For your second question, the number formula only comes into play when you have different numbers of H that you're diluting/titrating. So if you were titrating the H2SO4 with NaOH, youd have to multiply the H2SO4 side by two to account for the two H's.</p>

<p>Hope that helps.</p>

<p>Ok, thanks. But don't you multiply the two on the opposite side? I think I remember that. And it makes sense because the point of titration is to get both species in equal stoichiometric proportions. So to make up for the extra H's, wouldn't you ''help out'' the other side by multiplying it by two?</p>

<p>And also, for the practce problem (How many milliliters of 11.6M HCl must be diluted to obtain 1.0 L of 3M HCl?), how come the final answer you get is the actual final answer, as opposed to the subtraction required for the previous problems??</p>

<p>I am obviously not clear on the concept of these, haha. I understand the concept now of the first 2 problems. But what is the difference with this one? I might understand it, if the key is that it is the same situation as the others, but that it's asking for total volume, not the amount of water added.</p>

<p>I have never seen that forumla. Ever. I think plain old dimensional analysis is easiest way to go. Lol.</p>

<p>wow, your teacher wouldn't know how to answer that question? I'm a little rusty on my chem but the question should be very easy for a teacher to answer.</p>