Chemistry 103A; Sections 5, 6, 7, 8; Lecture 11, 15 Sep 00

Redox reactions, continued

One more redox equation:

The following reaction takes place in acid solution

Br2(l) ® Br- (aq) + BrO3- (aq).

(Water and H+ ions are not shown.)


This is called a "disproportionation" reaction.
 

Concentration Units

There are many occasions when it is very important to know how much solute is present in a given amount of solvent. The amount of solute in a given amount of solvent is called the concentration.
 

There are lots of different units of concentration which are useful for different situations. Ones that we will learn about at the appropriate times are wt/wt percent, wt/vol percent, vol/vol percent, ppm, ppb, mole fraction, etc.
 
 

The unit that is most important for the stoichiometry of solutions is called molarity. The molarity is defined as

.
In short, we write
,
where it is understood that the V is the volume of the solution in liters. (Not volume of the solvent added.)
 
 

We know all we need to know to be able to calculate the molarity of a solution.

Example.

What is the molarity of a solution obtained by diluting 10.0 g of NaOH to a final volume of 500 mL?

FW for NaOH is 40.00 g/mol

Recall that

So,
Lets try a different version of the problem. How many moles and how many grams of HCl are contained in 26.7 mL of 0.435 M HCl?
26.7 mL = 0.0267 L
Rearrange the definition of molarity to get

mol = V M.

Then

0.0267 L ´ 0.435 mol/L = 0.0116 mol HCl,
and the weight of HCl is
wt = mol ´ FW = 0.0116 mol HCl ´ 36.46 g/mol = 0.423 g HCl
Molarity is useful in chemical analysis because volumes of solutions are much easier to control or measure out than weights.

Now we can do stoichiometry with solutions. All we have to do is add some new paths to our stoichiometry roadmap.

Recall that our old roadmap for stoichiometry was

.


Our expanded roadmap is


This new roadmap offers all sorts of possibilities. We can start with either a weight or a volume and end up with either a weight or a volume (or even a new molarity).

Stoichiometry is used a lot in quantitative analyses.

Example.

A sample is said to contain some sodium carbonate. If it takes 32.6 mL of 0.435 M HCl to react with all the sodium carbonate in the sample, how many grams of Na2CO3 are in the sample?

We have seen this reaction before,
 

Na2CO3 + 2 HCl ® 2 NaCl + CO2 + H2O.


(FW of Na2CO3 is 106.0 g/mol)

We must use the path: vol ® mol ® new mol ® new wt

Find mol,

convert to new mol,

convert to new wt (weight of Na2CO3).

 One can make known molarity solutions of redox reagents.

Let's say we have a 0.228 M solution of KMnO4. We want to use this to analyze for the concentration of iron in another solution.

Take a 20.0 mL sample of the solution with the unknown iron concentration.

(Make sure that all of the iron is in the Fe2+ oxidation state by mixing the sample with Zn metal and then filtering out the Zn metal remaining. The reaction for this reduction is

2 Fe3+(aq) + Zn(s) ® 2 Fe2+(aq) + Zn2+(aq).)

 Now our Fe2+ can be oxidized by the permanganate ion back to Fe3+ according to the reaction
Fe2+ + MnO4- + H+ ® Fe3+ + Mn2+ + H2O
Oops, we better balance this one.

Do the experiment and find that it takes 36.8 mL of the permanganate solution to react with the iron (II) ions.

Our road map is vol ® mol ® new mol ® Oops again!
If we knew the molarity of the new solution we could get the new volume. But this time we know the volume of the new solution and we know the number of moles so we can get a new molarity from
.

So, convert the volume of the 0.228 M permanganate solution into moles of permanganate ion.

Use the reaction fraction to convert mol of permanganate to mol of Fe2+.

Use the mole of Fe2+ and the volume of the iron solution sample to find the molarity of the iron in the sample.