Chapter 12 Chemical Calculations for solutions
In the last chapter you learned how to do chemical calculations with chemicals where
you could weigh out each component on a balance. In this chapter we add another
layer of difficulty. What happens if the chemical is dissolved in water?
Well you first response might be....no big deal, I will put it in a beaker an then weigh it.
But you have missed on important point. Once you dissolve something in water, the
bulk of what you are dealing with is water, not the chemical, so even if you weigh your
solution in a beaker, what you are weighing is the water not the chemical, so all your
calculations are going to be way off. (So to speak)
12-1 Solutions
Going back to chapter 2, a solution is a mixture that is homogeneous at the
molecular level. That means all the components are uniformly dispersed. While
a solution can be a solid, liquid or gas, in this chapter I will focus primarily on
aqueous solutions
First some terminology
Key Definitions:
Solute - a solid that is to be dissolved in a liquid
Solvent - the liquid the solute is to be dissolved in
Solution - The mixture of solvent and solute
The best chemical equation to describe NaCl dissolving in water is:
This equation explicitly shows that water is the solvent. Most of the time I will
write this simply as NaCl(s)6Na+(aq) + Cl-(aq) so you will have to assume that
water is the solvent. If I get really lazy I may write NaCl(s)6NaCl(aq)
and make you remember that NaCl(aq) is actually composed of the two
individual aqueous ions. You should remember that these ions are hydrated ions
and look more like Figure 12.2
When you first put some salt in water, all the salt dissolves, leaving no solit in the
bottom of the beaker. However, if you keep adding more and more salt, you get
to the point where no more can go into solution, and you have solid left on the
bottom
Key Definition:
A saturated solution is one in which the maximum amount of a solute has been
dissolved in a solute
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12-2 Molarity
Molarity
Over the next year you will be shown several different ways to define the
concentration of solute in a solution. Lets start with the most common measure
of concentration, Molarity
Key definition/equation:
Molarity (M) is defined as moles of solute/liters of solution. A solution that is 1.0
M contains 1 mole of solute for each liter of solution.
Example calculation 1:
When I make Koolaid for my kids, I use 1 cup of sugar in ½ gallon of water. If
we assume that 1 cup of sugar is about 190 grams and ½ gallon is about 2 liters,
what is the molarity of Kool-aid?
Sugar (sucrose has a molecular formula of C12H22O11 so its mass is
12.01(12) + 1.01x22 + 16(11 = 342.34
190 g = 190x1 mole/342.34 g = .555 Moles
.555 mole/2 l = 0.277 M
Example calculation 2:
Now let’s try this in a different direction. How many moles of sugar are in 1 cup
of Koolaid? How many grams of sugar?
Let’s say that one cup is about .25 l
Then all we need is to use our molarity equation, plugging in for volume and
molarity:
And
.06925 mole x 342.34 g/mole = 2.93g
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It is important to remember that what we say is in solution may not be the actual
chemicals species that are in solution
POINT 1
For instance I might tell you that I have a 1M NaCl solution. But really, if you are
chemically exact there is NO NaCl in the solution. Instead, what is there? 1M
Na+ (aq) and 1M Cl-(aq). Chemically speaking NaCl will always ionize in
aqueous solution so NaCl(s) does not exist in this solution.
POINT 2
This gets a little tricky for ionic compound with more than 2 ions. What if I had a
solution 1M CaCl2? Again we put in 1 mole of CaCl2 in 1L of water to make our
solution. But as we just discussed, CaCl2 complete ionizes, so what we actually
have is a solution of 1M Ca+2 and 2M Cl- ions. This gets even messier in a poor
electrolyte like Acetic Acids (HC2H3O2) because then you end up with three
species in solution, H+, C2H3O2-, and HC2H3O2 and without lot’s of math, you
can’t figure out who is where. Solution concentrations are often given in terms of
solute before it dissolves, and you, as a chemist must figure the details out for
yourself.
POINT 3
If you are being a chemist rather than a Kool-aid maker, how would you actually
make the above 0.277 M solution of sugar. Well, first you weigh out the sugar.
Next you dissolve it in water. But how much water? 2L?
NO. Remember the definition of molarity of moles of solute per liters of solution?
When you add ,277 moles of sugar to 2L of water, the resulting volume will be
something bigger that 2L, so that won’t work
Next time you are in the lab look for the volumetric flasks that are in the drawer.
This flask has a line that etched in the next so when you fill it to that line it
contains exact 1 volume of solution So the actual procedure is to put the sugar
into the volumetric flask, then add water and mix and fill it exactly to the line so
you have the correct volume of the solution you need.
Mass %
On page 397 the book mentions mass percentage as a measure of
concentration. We won’t use that until chapter 16 so let’s skip it for now
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Dilution
Another way you will frequently see molarity used is when take one solution and
dilute it by mixing it with water or another solution.
For instance, let’s go back to the 1M NaCl solution. Say I was working in the lab
and I needed a 0.1M NaCl solution, How would I make it up?
If you are quick you can see that my concentration is 10 times lower, so it seems
quite reasonable that you just dilute the solution by a factor of 10 (put 1 ml in a
total of 10)
Let’s make it a little more difficult. Let’s say I wanted to make just 15 mls of a
.75M NaCl solution. How much 1M of my 1M NaCl stock solution would I need
now?
The trick to this problem is to realize that when you make this solution you don’t
change the number of moles of material. You do add more water to dilute it but
you don’t add or remove any of the material. Thus, we have:
Moles of material in concentrated solution = Moles of material in dilute solution
Now how would you calculate moles of material in a solution?
Molarity = Moles/liters;
Molarity x Liters = moles
So we can change our equation to:
Moles x Liters (of concentrated solution) = Moles x Liters (of dilute solution)
Key equation:
McVc = MdVd
-or- M1V1 = M2V2
Example:
MC = 1M
VC = what we are trying to find, or X
MD = .75M
VD = 15 ml
We need to have the same units of volume on both sides of the equation so:
15 ml X (1l/1000ml) = .015 L
So now we have:
1(X) = .75(.015)
(.75M x .015l) / 1M = X
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X= .01125L or 11.25 ml
So to make this solution you would take 11.25 ml of the stock solution and dilute
it to a total of 15 ml by adding 3.75 ml of additional water. (15-11.25=3.75)
Other problems to try:
1.Our Kool-aid is 0.277 M sucrose. And we want to deliver to out
experiment exactly 0.01 mol of sucrose. How much volume do we need?
From the above equation you can see that we want
Liter = mol of substance /molarity
? = .01/.277 = .0361 l = 36.1 ml
2. A typical blood sample is about 0.14 M NaCl. How much blood would it
take to give me 1 mg of sodium? This is similar to the last problem, by I
want to deliver a set number of grams. How do we handle this? Well this
is the same kind of problem, only we have to covert our grams of NaCl to
moles of NaCl.
First, how many moles of sodium is 1 mg?
1 mg x 1 g/1000 mg x 1 mole/58.45 g = 1.7x10-5 mol NaCl
Here we want a volume so we take Molarity = mols /volume and rearrange
to
vol = mols/molarity
now fill in the blanks
X= 1.7x10-5 mol/ 0.14 mol/L = 1.2x10-4 L = 0.12 mL
3. When you buy acids you frequently will purchase a stock bottle with a
very high concentration, than make dilutions. HCl, for instance, comes as
a 12.1 M solution. How much of this would you use if you wanted to make
500 mls of a 1.0 M HCl.
12.1(X)=1.0(.5)
1.0(.5)/12.1 = .04132 L or 41.32 ml.
12-3 Electrolytes
Back in Chapter 6 I told you that when ionic substances dissolve they produce
ions that carry current and conduct electricity. On the other hand a covalent
compound like sucrose does not make any ions, so it cannot conduct electricity
Demo of strong, weak and medium electrolytes
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Key Definition
1.) Ionic compounds that dissolve to solutions that conduct electricity are called
electrolytes
2.) Covalent compounds that dissolve, but do not conduct electricity are called
nonelectrolytes.
3.) Compounds that dissolve to form a limited amount of ions so they make
solutions that weakly conduct electricity are called weak electrolytes
Strong electrolytes
There are three classes of compounds that are strong electrolytes
Strong Acids
Strong Bases
Soluble salts
The reason they are strong electrolytes is that they dissociate completely
(100%) into their component ions
Nonelectrolytes
These are covalent compounds that are polar so they dissolve in water,
but they do not ionize, so they can’t produce the ions to carry current.
Good examples are ethanol, and sugar, sand ethylene glycol (antifreeze)
Weak electrolytes
Weak electrolytes are substances that exhibit only a little ionization in
water. They produce relatively few ions when dissolved in water. Acetic
acid, the acid component of vinegar is a good example of a weakly acid.
The chemical formula of acetic acid is HC2H3O2. Notice that in this formula
1 H is separated from the rest. This is to help us distinguish the 1 H that
can ionize in the molecule
That is
HC2H3O2. º H+ + C2 H3O2This molecule is very different from the strong acids because the degree
of ionization is much lower. In the strong acids for every 100 molecules of
acid you put in solution you got 200 ions (100 + and 100 -) In acetic acid
for every 100 molecules you put in solution you get about 2 ions (1 + and
1 -!)
Weak bases can also be weak electrolytes. The base ammonia is a fine
example NH3 + H2O º NH4+ + OH-. Again only about 1 molecule in 100
of NH3 ionized to NH4+
Beside weak acids and weak bases there are some ionic compounds that
dissolve, but only partially dissociate into ions. For instance HgCl2
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The uppermost form is a complex that is an aqueous compound, but it is not
ionized. And then there are the other ions, but very few ions are produced.
Since we don’t do molar conductance experiments in the lab I will skip the rest of
this section.
12-4 Using Molarities in Solution calculations
Now that you can use volumes and molarities to determine moles, let’s try a
calculation that has a little of both
What volume of 8.50M HBr is required to completely react with 3.05 grams of
MnO2 in the reaction
MnO2(s) + 4HBr(aq) 6 MnBr2(aq) + Br2(l) + 2H2O(l)
Problem analysis
Starting point
3.05 g MnO2
Ending point
Volume of MnO2
Steps
g MnO2 6moles MnO2 6Moles HBr 6Volume HBr
Set up of unit to do the above steps
g MnO2 x Moles MnO2 x Moles HBr
g MnO2
Moles MnO2
From Molar mass
1 mole Ô 86.94g
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x Volume HBr = Volume HBr
Moles HBr
From Molarity
1 L Ô 8.50 mole
From reaction stiochiometry
4 moles HBr Ô 1 mole MnO2
Putting in the #’s
3.05g MnO2 x 1 Moles MnO2 x 4 Moles HBr x 1L Volume HBr = .0165 L HBr
86.94g MnO2 1 Moles MnO2 8.50 Moles HBr
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12-5 Precipitation Reactions
Now lets apply your molarity calculations to precipitation type reactions
Example 1 How much reagent do we need
How much 0.350M KI will react with 30.0 mL of 0.205M Hg2(NO3)2 in the
reaction
Hg2(NO3)2 + 2KI 6 2KNO3(aq) + Hg2I2(s)
Problem analysis
Start with moles of Hg2(NO3)2 end with mLs of KI
Steps
Volume Hg2(NO3)2 6moles Hg2(NO3)2 6moles KI6volume KI
Set up for above steps
Vol Hg2(NO3)2 x Moles Hg2(NO3)2 x Moles KI
X Volume KI = Volume KI
Vol Hg2(NO3)2
Mol Hg2(NO3)2
Mole KI
From Molarity
From Molariy
.205 mol Ô 1L
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.350M Ô 1L
From Equation
1 mol Hg2(NO3)2Ô2 Mole KI
The equation
.0300L Hg2(NO3)2 x .205Moles Hg2(NO3)2 x 2 Moles KI
X 1L KI = .0351L KI
1LHg2(NO3)2
1Mol Hg2(NO3)2 .350 mol KI
Example 2 limiting reagent
If I mix 50 mL of .150M AgNO3 with 45 mL .250 M Na2CrO4, how many grams of
precipitate will I get in the reaction
2AgNO3(aq) + Na2CrO4(aq) 6 2NaNO3(aq) + AgCrO4(s)
Problems analysis
Need to find g AgCrO4(s) if AgNO3 is limiting reactant
Volume AgNO3 6 moles AgNO3 6moles AgCrO4(s) 6g AgCrO4(s)
Then find g AgCrO4(s) if Na2CrO4 is limiting reactant
Vol Na2CrO4 6 moles Na2CrO4 6moles AgCrO4(s) 6 g AgCrO4(s)
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Since this is the second example I will cut straight to the equations:
So the AgNO3 yields the smalles amount of product ans so is my limiting reactant
and my final theoretical yield is 1.24 g of Ag2CrO4
12-6 Acid-Base Titrations
In the last chapter we talked a little about how acids and bases neutralize each
other. This leads to a very common lab technique called an acid-base titration
that is used to determine the amount of acid or base in any solution
Bring in Demo Buret, and run demo titration
So if you were stating with an unknown amount of acid in the beaker, you would
add base from the buret until the solution was exactly neutral. Because at that
neutralization point the reaction that has occurred is:
H+ + OH- 6H2O
and moles acid Ô mole of base and we can use that in stoichiometric calculations
that I will show you in a minute
How do you know when you are neutral? So moles acid Ô moles base?
Either use a pH meter or an indicator to say you are at pH 7
Calculations
Say your buret initially had 0.12 mL of .1561M HCl in it
and you had to deliver 26.78 ml of acid to neutralize 50 mL of base.
What is the molarity of the base?
Problem analysis
volume of acid6 moles of acid 6moles of base 6molarity of base
Volume of acid = 26.78-0.12 = 26.66 mL = .02666 L
Volume of base = 50 mL = .05L
Set up as two calculation
Volume of acid x Moles Acid
x mol base = moles of base
Volume of acid Mol of acid
And moles of base = Molarity of base
Volume of base
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Calculation
.02666L HCl x .1561 mol Hcl x 1 mole NaOH = 4.159x10-3 moles base
1 L Hcl
1 mole acid
4.559x10-3 moles base / .05L base = .08318 moles/L = .08318 M Base
12-7 Formula mass from Titration data
The above titration data can be taken one step further to determine formula
mass
Example
Say in the above titration I had started out with .1664 g of an unknown base
instead of a 50 mL solution What is the formula mass of the base?
Set up
Volume of acid 6 moles of acid 6moles of base 6 Formula mass of base
Again this sets up in two calculations:
Volume of acid x Moles Acid
x mol base = moles of base
Volume of acid Mol of acid
And Formula mass = mass
Mole
The first equation is exactly the same as before:
.02666L HCl x .1561 mol Hcl x 1 mole NaOH = 4.159x10-3 moles base
1 L Hcl
1 mole acid
But for the second equation you have:
.1664 g Base
= 40.0 g/mol
-3
4.159x10 moles