Chemical Calculations for
Solutions (Ch 12)
Dr. Harris
Lecture 12
Suggested HW: Ch 12: 1, 10, 15, 21, 53, 67, 81
Solutions
• As we learned in a previous chapter, solutions are homogenous
mixtures, meaning that the components comprising the solution are
uniformly dispersed
• The most common type of solution is a solid dissolved in a liquid.
The dissolved solid is the solute, the liquid is the solvent.
• Solutes and solvents do not react, merely co-exist, as is the case with
an aqueous solution like NaCl(aq)
NaCl (s) -----> NaCl(aq)
H2O (L)
Solubility
• When NaCl is dissolved, the ions
are surrounded by water
molecules, and dipole interactions
disperse causes the ions to
separate and disperse
• As more and more NaCl is added, a
point is reached where further
dissociation ceases, the salt simply
drops to the bottom of the beaker
• The solution is now saturated
• The quantity of NaCl dissolved at
this point is its solubility.
Figure Above: Dissolution of
NaCl and uniform distribution
of solute and solvent.
Electrolytes
• Reminder: Strong electrolytes fully dissociate in water. All strong
electrolytes are ionic compounds. These include:
• All salts with group 1 cations
• All salts with ammonium cations
• All salts with nitrate, perchlorate, and acetate anions
• Hydroxides of Ca, Sr, and Ba (plus group 1 cations and
ammonium)
• All sulfates except Ca and Ba
Strong vs. Weak Electrolytes
100%
CaCl2(s) ----------->
H2O(l)
Ca2+ (aq)
+
2Cl-(aq)
Full dissociation of
strong electrolyte
99.8%
--------> HgCl2(aq)
0.18%
HgCl2(s) ------------->
--------> HgCl+(aq) + Cl-(aq)
H2O(l)
0.02%
--------> Hg2+(aq) + 2Cl-(aq)
Minimal dissociation
of weak electrolyte
Strong Acids and Strong Bases Are Strong
Electrolytes
• Strong ACIDS
•
•
•
•
•
•
HCl
HBr
HI
HNO3
HClO4
H2SO4
• Strong BASES
• Hydroxides of group 1
metals
• Hydroxides of Ca, Ba,
and Sr
pH Scale
• At this point, we will not go into full detail of pH. However, it is important
to know how acids and bases are distinguished. The pH scale allows us to
do this.
Acids
Bases
WATER
Concentration (Molarity)
• The concentration of a solute describes the number of solute
ions/molecules in a certain volume of solvent
• Concentration is most commonly expressed using MOLARITY,
represented by the letter M. Molarity is defined as the moles of
solute per liter of solution.
𝐌=
𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆
𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏(𝑳)
Examples
• 30 g of NaCl are dissolved in 450 mL of H2O. What is the
concentration of NaCl?
𝑚𝑜𝑙 𝑁𝑎𝐶𝑙
𝑚𝑜𝑙𝑒𝑠: 30 𝑔 𝑁𝑎𝐶𝑙 𝑥
= .513 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙
58.45 𝑔 𝑁𝑎𝐶𝑙
10−3 𝐿
𝑣𝑜𝑙𝑢𝑚𝑒: 450 𝑚𝐿 𝐻2 𝑂 x
= 0.450 𝐿 𝐻2 𝑂
𝑚𝐿
.513 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙
𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑁𝑎𝐶𝑙 =
= 1.14 𝐌
.450 𝐿 𝐻2 𝑂
• How many moles of NaCl are there in 500 mL of this solution?
.500 𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑥
1.14 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙
= 0.57 𝑚𝑜𝑙
1 𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Example
• 15 g of Aluminum nitrate, Al(NO3)3, is dissolved in 200 mL of H2O. What
is the concentration of nitrate in the solution?
• Aluminum nitrate will dissociate into aluminum and nitrate ions, as
according to the chemical formula:
Al(NO3)3 ------> Al3+(aq) + 3NO3-(aq)
H2O(L)
• Therefore, every mole of aluminum nitrate yields 3 moles of nitrate
𝑚𝑜𝑙 𝐴𝑙(𝑁𝑂3 )3
𝟑 𝒎𝒐𝒍 𝑵𝑶−
𝟑
15 𝑔 𝐴𝑙(𝑁𝑂3 )3 𝑥
𝑥
= .542 𝑚𝑜𝑙 𝑁𝑂3−
83 𝑔 𝐴𝑙(𝑁𝑂3 )3 𝒎𝒐𝒍 𝑨𝒍(𝑵𝑶𝟑 )𝟑
𝑁𝑖𝑡𝑟𝑎𝑡𝑒 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 =
.542 𝑚𝑜𝑙
= 𝟐. 𝟕𝟏 𝑴
.200 𝐿
Dilution
• In many instances (especially in lab), you may need to
prepare a solution of some desired concentration
from a pre-existing stock solution.
• An example of this would be a water enhancer, like
Mio. You wouldn’t drink the Mio directly because it is
extremely sweet.
• Instead, you add a small amount (aliquot) to your
water, until you’ve attained the desired level of taste
and sweetness.
• This is dilution. The flavored water is our diluted
aqueous solution, and the bottle of Mio is the stock
solution
Dilution
• Keep in mind that dilution does not
change the total moles of solute, only
the molarity.
• We know that the moles (n) of solute in V
liters of a solution with molarity M is:
n = MV
• Therefore, we know the concentration of
a solution before and after dilution:
𝑴𝟏 𝑽𝟏 = 𝑴𝟐 𝑽𝟐
V1
V2
How to perform a Series Dilution
Take an aliquot of the stock
solution, add it to a new
container
Dilute with
solvent to
desired
volume, V2
High concentration
stock solution of
concentration M1
Aliquot of stock solution
with volume V1 and
concentration M1.
After complete mixing, we have a
dilute solution with volume V2
and concentration M2
Example
• A concentrated stock solution of NaOH is 19.1 M. How would you
prepare 500 mL of a 3.0 M solution?
We are given:
• initial concentration of the NaOH stock (M1 = 19.1 M),
• the desired final concentration of NaOH (M2 = 3.0 M),
• and the final volume of the solution (V2 = 0.5 L).
 We need to find the volume of the aliquot (V1)
𝑴𝟏 𝑽𝟏 = 𝑴𝟐 𝑽𝟐
19.1 𝑀 𝑉1 = 3.0 𝑀 (0.500 𝐿)
𝑉1
3.0 𝑀 (0.500 𝐿)
=
= .0785 𝐿 = 78.5 𝑚𝐿
(19.1 𝑀)
• A 78.5 mL aliquot of the stock solution is added to 421.5 mL of
water to make a 3.0 M solution
Example
• a.) Explain how would you make a 500 mL stock solution that is 1.0
M NaBr (aq) ? (molar mass NaBr: 102.9 g/mol)
• b.) From this stock solution, you decide to make 100 mL of a 0.10M
solution. Explain how you would do this?
Dilution
Applying Molarity to Stoichiometric Calculations
• For reactions of solutions, we can use molarity to calculate product yields
• Example:
MnO2(s) + 4HBr(aq) -----> MnBr2(aq) + Br2(L) + 2H2O(L)
What volume of an 8.84 M HBr solution is needed to completely react with 3.62
g of MnO2?
Convert MnO2 to
moles
.0416 𝑚𝑜𝑙 𝑀𝑛𝑂2
Determine the
required moles of HBr
.0416 𝑚𝑜𝑙 𝑀𝑛𝑂2 𝑥
Calculate volume
𝑉=
4 𝑚𝑜𝑙 𝐻𝐵𝑟
= .167 𝑚𝑜𝑙 𝐻𝐵𝑟
𝑚𝑜𝑙 𝑀𝑛𝑂2
𝑛
.167 𝑚𝑜𝑙
=
= .0188𝐿 = 18.8 𝑚𝐿
𝑀 8.84 𝑚𝑜𝑙 𝐿−1
The Reaction of Strong Acids and Strong Bases is A DoubleReplacement Reaction Known as a Neutralization Reaction
• When acids and bases react, they neutralize each other, and the product
is salt and water
HCl (aq) + NaOH(aq)
H2O(L) + NaCl(aq)
• This is a double replacement reaction. The net ionic equation is:
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)
H+(aq) + OH-(aq)
H2O(L) + Na+(aq) + Cl-(aq)
H2O(L)
Titrations
• Knowing that acids and bases neutralize each other, lets imagine
that we have an acid/base of unknown concentration.
• How can we find the concentration?
• Perform an acid/base titration
Titrations
• In a titration, an indicator is added to the basic
solution.
• In the example to the right, as long as the pH is
above 7 (basic) the indicator will make the
solution pink.
• An exact volume of an acid solution with a known
concentration is added to a buret.
• The acid solution is added drop-by-drop until the
solution just turns clear (neutralized, pH =7 ).
Titrations
• Say we have 100 mL of a basic NaOH
solution of an unknown concentration.
Before
titration
After
titration
• We titrate with 5 mL of 1.0 M HCl, and
the solution just turns clear.
NaOH(aq) + HCl(aq)  H2O(L) + NaCl(aq)
• We know that the acid and base are completely neutralized, and
none is left in solution.
Moles of acid added = Stoichiometric equivalent of base
.005 𝐿 𝐻𝐶𝑙 𝑥
1.0 𝑚𝑜𝑙 𝐻𝐶𝑙 1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
𝑥
= .005 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
𝐿 𝐻𝐶𝑙
1 𝑚𝑜𝑙 𝐻𝐶𝑙
Concentration of base solution =
.005 𝑚𝑜𝑙
= .05 𝑀 𝑁𝑎𝑂𝐻
.100 𝐿