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Molarity (M), Equation 5.1
>> Parts of this equation/concept include:
| molarity |
= moles of solute/liters of solution |
| = mmol solute/mL solution |
Tips: Memorize the correct formula with units. Most mistakes
are caused by incorrect formulas. Use appropriately. Molarity is
used in stoichiometric calculations and osmotic pressure problems.
It is not used for boiling point or freezing point problems.
| A. Determining Molarity
from Amount of Solute and Solution Volume |
>> Example 1
What is the molarity of a solution made by dissolving 0.479 g
of Li2CO3 to make 250.0 mL of solution?
Solution:
M = mol solute/L solution
0.479 g of Li2CO3 = solute
250.0 mL = solution
Grams of Li2CO3 must be changed to moles,
the gram-mole relationship is molar mass. The molar mass of lithium
carbonate is
2(6.941) + 12.011 + 3(15.999) = 73.890 g/mol
| 0.479 g Li2CO3 |
 |
|
= |
0.00648 mol |
mL of solution must be changed to liters; the mL-L relationship
is the metric prefixes
1000 mL = 1 L
| 250.0 mL |
|
|
= |
0.2500 L |
Then the molarity formula can be used
The answer will have only three significant figures; it was limited
by the mass of lithium carbonate.
>> Example 2
What is the molarity of a solution made by adding water to 0.33
mol of CaBr2 until there is 500.0 mL of solution?
Solution:
M = mol solute/L solution
solute = 0.33 mol CaBr2
solution = 500.0 mL aqueous
The solute is already in the correct units (avoid the temptation
to change it!) The solution volume must be changed from mL to
L. (This is typical, but not always the case.)
| 500.0 mL |
|
|
= |
0.5000 L (don't lose your significant figures) |
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| B. Determining Amount
Solute from Molarity and Solution Volume |
Tips: you can use one of two equally correct methods
to solve these problems.
Method 1:
Plug the values into the formula and solve algebraically.
Method 2:
Use molarity as a dimensional analysis relationship. An advantage
of method 2 is that it is very similar to the way molarity is used
in stoichiometry problems.
>> Example 3
How many grams of AlBr3 are required to make 500.0
mL of 0.20 M AlBr3(aq)?
Solution:
Method 1
M = mol solute/L solution
solute = AlBr3
solution = 500.0 mL
| 500.0 mL |
|
|
= |
0.5000 L |
|
|
|
0.5000 L |
= |
mol AlBr3 |
0.10 = mol AlBr3
molar mass of AlBr3 = 26.982 + 3(79.904) = 266.694
g/mol
| 0.10 mol AlBr3 |
|
|
= |
27 g AlBr3 |
Method 2
0.20 M can be interpreted as 0.20 mol AlBr3
= 1 L solution
| 500.0 mL solution |
|
|
|
|
|
|
= |
27 g AlBr3 |
>> Example 4
How many mg of silver nitrate are required to make 1.0 mL of
0.042 M solution?
Solution:
For either method you will need to know the formula for silver
nitrate. Recall naming rules from Chapter 4. As
a metal and a nonmetal it would be an ionic compound. Silver ion
always has the same charge: Ag+. Nitrate is one of
the polyatomic ions you should have memorized: NO3–.
Since both ions have the same charge, the formula is AgNO3.
You will also need its molar mass: 107.87 + 14.007 + 3(15.999)
= 169.87 g/mol.
Method 2
Using molarity as 0.042 mol AgNO3 = 1 L solution
Method 1
Using the formula M = mol/L
| 1.0 mL |
|
|
= |
0.0010 L |
| 4.210–5
mol AgNO3 |
|
|
= |
7.110–3
g
|
| 7.110–3
g |
|
|
= |
7.1 mg |
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| C. Using Molarity in
Stochiometry Problems |
Stoichiometry problems require you to relate amounts of different
reactants and products. These problems normally involve the use
of a chemical reaction. Specific to this chapter are titrations,
where quantities of reactants are related and the reactants are
in solution form. Because they are solutions (mixtures) you must
use concentration (molarity) to relate volume of solution to moles
of solute (pure substance). However, the basic steps for solving
these types of problems are the same as in Chapter 4.
Steps for solving stoichiometry problems:
- Balance the chemical reaction
- Change unit given to moles
- atoms or molecules to moles: use Avogadro's number
- grams to moles: use molar mass
- (new) volume of solution to moles: use molarity
- Change moles of given substance to moles of desired substance
- stoichiometric coefficients of chemical reaction
- Change moles of desired substance to units desired
(the reverse of 2a, b, and c)
>> Example 5
What is the molarity of a solution of hydrochloric acid, if 0.1511
g of sodium carbonate in 100.0 mL of water required 27.31 mL of
the acid solution to reach the equivalence point?
HCl + Na2CO3 NaCl
+ H2O + CO2
Solution:
Before solving any stoichiometry problem, make sure the reaction
is balanced. In this example, it is not. Balancing the reaction
give you
2 HCl + Na2CO32
NaCl + H2O + CO2
Because this example asks you to find molarity (a formula or
relationship) as your final answer, it is useful write the formula
and see what you have and what you need to find.
Since it is moles of hydrochloric acid solution, HCl is your
solvent. Looking closely at the question, you will also notice
that the volume of HCl solution is given as 27.31 mL. Since you
need volume in liters for use in the formula
| 27.31 mL |
|
|
= |
0.2731 L |
Now all you need is the moles of HCl. No further information
is given directly about HCl, however information (grams) is given
about sodium carbonate and that can be related to HCl from the
chemical reaction
| 0.1511 g Na2CO3 |
|
|
|
|
= |
2.85110–3
mol HCl |
Note that step 4 was not needed, because the desired unit was moles!
Now there is sufficient information to use the formula
| M |
= |
|
= |
| 2.85110–3 mol |
|
| 0.2731 L |
|
= |
0.1044 M |
Note that the volume of water was not used in these calculations.
It is not a reactant, nor does it affect the moles of sodium carbonate,
nor was it part of the HCl solution which the question asked about.
It is not usual to have extra information in problems of this
type.
>> Example 6
Magnesium hydroxide was titrated with standard 0.1478 M
HCl(aq). It required 39.11 mL of the hydrochloric acid
solution to reach the equivalence point. How many grams of magnesium
hydroxide were titrated?
Solution:
The reaction is not given, but titrations relate reactants. Therefore,
the question tells you that Mg(OH)2 and HCl are your
reactants. For tips on predicting the products, see the later
section on acid-base reactions. The balanced chemical reaction
will be
Mg(OH)2 + 2 HClH2O
+ MgCl2
In picking out what is "given," it helps to remember
that molarity is a ratio rather than a specific value. Therefore
it is not your "given." The ratio is between volume of solution
and moles. In this example: 0.1487 mol HCl = 1 L solution.
| 39.11 mL HCl |
|
|
|
|
|
|
|
|
|
| = 0.1696 g Mg(OH)2 |
>> Example 7
15.27 mL of 0.113 M NaOH was used to neutralize 25.00
g of vinegar. What is the percent of acetic acid (CH3COOH)
in the vinegar?
NaOH + CH3COOHNaCH3CO2
+ H2O
Solution
Note that the reaction is balanced. The question asks for percent
as the final answer. Recall that percent is a formula and the
formula is
| % |
= |
|
|
100 |
where part and whole must have the same units
In this example the question asks for the percent of acetic acid
(part) in vinegar (whole). Note that the total amount of vinegar
is given in the problem.
whole = grams of vinegar = 25.00 g
To complete the formula, grams of acetic acid is needed, but
information is given about sodium hydroxide instead. Therefore,
stoichiometry will be used.
| 15.27 mL NaOH |
|
|
|
|
|
|
|
|
|
| = 0.104 g CH3COOH |
Then to use in the formula
part = grams of acetic acid = 0.104 g CH3COOH
| % |
= |
|
|
100 |
= |
0.414% |
Other examples: You also use molarity in osmotic pressure
problems. Examples of that are in the section on osmotic
pressure.
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