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<p class="MsoNormal"><span style="font-family:"Courier New"">I often get questions about two very similar equations you’ve<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Courier New"">seen in class, M2*V2=M1*V2 and Ma*Va=Mb*V. They look similar<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Courier New"">but are different and that difference is important.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Courier New""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Courier New"">In Chem 1210 (or the equivalent) you learned about dilution and
<br>
using the eqn. for dilution, M2*V2 = M1* V1. <br>
<br>
Last week we discussed acid-base titrations and the fact at the <br>
equivalence pt. of a titration the moles of acid equals the moles <br>
of base. Technically, this is only true for monoprotic acids and <br>
monobasic bases (contribute 1 OH-). More generally, it's the<br>
moles of H+ equals the moles of OH- at the eq. pt. (which will be <br>
the same as moles of acid = moles of base for a monoprotic <br>
acid reacting with a monobasic base or when titrating a <br>
polyprotic acid to the first eq. pt. with a monobasic base).<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Courier New"">We saw that based on this we have an eqn that looks similar to<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Courier New"">the one above, <br>
<br>
Ma*Va = Mb*Vb. <br>
<br>
While the two eqns look the same they are <b>NOT</b>. There are <br>
different meanings for the M and V. <br>
<br>
For dilution, M2*V2 = M1* V1, <br>
<br>
M1 = initial molarity of solute in solution <br>
V1 = initial volume of the solution <br>
M2 = final molarity of solute after dilution <br>
V2 = final TOTAL volume of the new solution <br>
<br>
Remember, it's not dilution only when water is added. If one <br>
solution is added to another (like the BAR exp) the molarity of <br>
everything changes (the final molarities of everything are decreased). <br>
<br>
At the eq. pt. in an acid-base titration,<br>
<br>
moles acid = moles base<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Courier New""><br>
Ma*Va = Mb*Vb, <br>
<br>
Ma = molarity of the acid (technically the H+) <br>
Va = volume of the acid solution <br>
Mb = molarity of the base (technically the OH-) <br>
Vb = volume of the base solution <br>
<br>
Note, neither volume is total volume of the solution resulting <br>
from mixing the acid and base solutions during the titration. <br>
This eqn can be used to determine the volume of base needed <br>
to reach the eq. pt. when an acid (strong or weak) is titrated <br>
by a base (the titrant). In this case Ma & Va are for the acid<br>
being titrated and the Mb & Vb are for the base titrant. It can<br>
also be used to determine the volume of acid needed to reach<br>
the eq. pt. when a base (strong or weak) is titrated by an acid<br>
(the titrant). In this case Mb & Vb are for the base being<br>
titrated and the Ma & Va are for the acid titrant. <br>
<br>
Understanding the difference is important. If you do, you know<br>
when to use which (i.e. is it a dilution problem or acid-base<br>
neutralization problem). A lot of people don't like to use the<br>
Ma*Va = Mb*Vb equation for neutralization problems and titrations.<br>
If you understand it, how to use it and when to use it you should<br>
be fine. <br>
<br>
Also, for a polyprotic acid you have to be careful with this eqn. It <br>
can be used to determine the volume of titrant required to reach each <br>
eq. pt. The volume of titrant required to reach the second eq. pt. will <br>
be the same as the volume to reach the first eq. pt. This is true for <br>
each eq. pt. If you want to determine the volume of base to completely <br>
neutralize a polyprotic acid (react with all the acidic protons) you can <br>
think of what I just said and use that. Alternatively, you could keep in <br>
mind for complete neutralization of an acid the Ma is the molarity of the <br>
H+. The Mb is the molarity of the OH-. This would allow one to <br>
use the Ma*Va=Mb*Vb eqn for complete neutralization. For example, <br>
to completely neutralize H2SO4 with NaOH it will take 2 moles of NaOH <br>
for every one mole of H2SO4 because there are 2 protons in the H2SO4 <br>
which have to react. The neutralization reaction is, <br>
<br>
H2SO4 + 2 NaOH ---> 2 Na<sup>+</sup> + SO4<sup>^2- </sup> + 2 H2O
<br>
<br>
You can use this eqn to determine the volume of NaOH it would take to <br>
get to the 2nd eq. pt. by doing a stoichiometry problem or you could use <br>
the MaVa = MbVb eqn with making sure the Ma is the conc. of H+. <br>
For example, lets say you are titrating 30 mL of 0.10 M H2SO4 with 0.15 M <br>
NaOH to the 2nd eq. pt. (complete neutralization). You need to find the <br>
volume of NaOH it will take to get to the 2nd eq. pt. Lets do it both ways <br>
as described above. <br>
<br>
Using the neutralization eqn and stoichiometry: <br>
<br>
0.10 mol H2SO4 2 mol NaOH 1 L NaOH soln <br>
? L NaOH = 0.030 L H2SO4 x ----------------- x ------------- x --------------- <br>
1 L H2SO4 soln 1 mol H2SO4 0.15 mol NaOH<br>
<br>
= 0.040 L NaOH soln (40 mL) <br>
<br>
(it takes 20 mL to get to the first eq. pt. and another 20 mL to <br>
get to the 2nd eq. pt.) <br>
<br>
<br>
Using the Ma*Va=Mb*Vb eqn for Ma you need to account for the fact <br>
there are 2 H+ ions in H2SO4 which will ultimately react, <br>
<br>
Ma = 2(0.10 M) = 0.20 M in H+ ions <br>
<br>
even though in solution they don't both come off completely, in the <br>
titration they both react completely with the strong base to get to<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Courier New"">the 2nd eq. pt. <br>
<br>
Ma * Va = Mb * Vb <br>
<br>
Vb = (Ma * Va)/Mb = {(0.20 M H+)(30 mL)}/(0.15 M OH-) = 40 mL <br>
<br>
You get the same answer. <br>
<br>
If only going to the first eq. pt. you would use Ma = 0.10 M <br>
because only the first proton is reacting and the Vb would be <br>
20 mL (and it takes another 20 mL to get to the 2nd eq. pt., i.e. <br>
react with the 2nd proton). <br>
<br>
When using this Ma * Va = Mb * Vb eqn for a strong or weak acid<br>
being titrated by a strong base the Ma and Va are for the acid being<br>
titrated and the Mb and Vb are for the strong base used as the titrant.<br>
Sometimes people think the Mb and Vb are for the conjugate base of<br>
the acid (particularly when it's a weak acid) that's being titrated.<br>
<br>
For a strong or weak base being titrated by a strong acid the Mb and<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Courier New"">Vb would be for the base being titrated and the Ma and Va are for the<br>
strong acid used as the titrant. <br>
<br>
For titrations we are generally using this eqn to determine the <br>
volume of titrant needed to reach the eq. pt. When you know where <br>
the eq. pt. is (in terms of the volume of titrant required to reach <br>
it) determining other things (pH at various points) becomes much <br>
easier. For instance if you know it takes 30 mL to reach the <br>
eq. pt. in a WA-SB titration then if you want to know the pH at <br>
15 mL you know this is halfway to the eq. pt. and thus pH = pKa. <br>
You also know at the eq. pt. what reaction is taking place (hydrolysis,<br>
a salt soln problem). In the case of a WA-SB titration it would be the <br>
hydrolysis rxn for the conj. base of the WA and thus the soln should <br>
be basic and you need the total volume of solution in order to <br>
calculate the concentration of this weak base (i.e. you need the <br>
volume of SB required to reach the eq. pt. so you can add it to the <br>
volume of WA you started with to get the total volume). In the case <br>
of a WB-SA titration it would be the hydrolysis rxn for the conj. acid <br>
of the WB and thus the soln should be acidic and you need the total <br>
volume of solution in order to calculate the concentration of this <br>
weak acid at the eq. pt. <br>
<br>
Hopefully this makes sense, particularly if you've been in lecture. <br>
I've gone over titrations more than once in lecture at this point so <br>
with a little work things shouldn't be too bad. <br>
<br>
Dr. Zellmer <o:p></o:p></span></p>
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