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W9 SolutionConcAndStoichiomV1

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					[SOLUTION MOLARITY AND STOICHIOMETRY / 6P.]                                           W9


MOLARITY PROBLEMS

The formula:         Molarity (M) = moles of solute / liters of solution

Recall that a solution is made up of two (or more) components: the solute and the
solvent. There is only one solvent. The solvent is that substance that is present in the
greatest quantity (by volume or mass).
There may be one, or many, solutes. The solute is any substance dissolved into the
solvent, and it is present in lesser quantity, relative to the solvent.
The solution is the volume of solvent + the volume of all of the solutes.

In aqueous solutions e.g. NaCl (aq), the (aq) indicates that water is the solvent and that
the compound or element that is in front of (aq) is dissolved in the water. NaCl(aq)
should be read as "sodium chloride, dissolved in water" or "aqueous sodium chloride".

How much salt can be dissolved in a glass of water? Well, a lot! True enough, but
if we have to get technical about it (and we do) the way we describe how much of one
substance is dissolved in another substance is through "concentration" which is a ratio
of solute to total solution. Just like a percent it is a ratio of "a part (solute) / whole".

All solutions that are described by "molarity" are homogeneous solutions. They are
mixed at the molecular level, and so the concentration (ratio of solute to solution) is the
same throughout.

                     A MOLE (mol)         MOLARITY (M)

Example 1a)
If 25.3 g Na2CO3 is dissolved in enough water to make 250. mL of solution, what is the
                          
concentration, in molarity, of Na2CO3?




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[SOLUTION MOLARITY AND STOICHIOMETRY / 6P.]                                         W9


When working with Molarity in math problems, with the exception of Dilution
problems, it is strongly advised to always break down the units into moles / L,
because the dimensional analysis is clearer.

Example 1b)
Using the information from the previous problem, what is the concentration of sodium
ions?




Example 1c)
In part 1a of this problem, 250 mL of this solution was made. If 80.0 mL were poured
out of the volumetric flask, and into a graduated cylinder, what would be the molarity of
Na2CO3 in the volumetric flask?




Example 1d)
If 80.0 mL were poured out of the volumetric flask, and into a graduated cylinder, how
many grams of Na2CO3 are in the volumetric flask?




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[SOLUTION MOLARITY AND STOICHIOMETRY / 6P.]                                       W9


Example 2
Another common use of molarity: making a solution to a specified concentration.

A scientist needs 2.00 liters of a 0.065 M NaOH solution to run an experiment. How
much NaOH, in grams, must he weigh out?




DILUTION PROBLEMS

             CcVc = CdVd

Cc stands for Concentration of the more concentrated solution.

Vc stands for the Volume of the more concentrated solution.

Cd stands for Concentration of the more dilute solution.

Vd stands for the Volume of the more dilute solution.

This formula assumes that the moles are equal...which they always are when
making a dilution...it does NOT replace the mole ratio in stoichiometry!! It works
sometimes, when the mole ratio is 1:1 (or 2:2 etc...) do NOT use except for
dilution problems.




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[SOLUTION MOLARITY AND STOICHIOMETRY / 6P.]                                       W9


Example 1

If you dilute 4.00 mL of 0.0250 M CuSO4 to 10.0 mL using distilled water, what is the
molar concentration of CuSO4?




Example 2

You have 250 mL of a 4.00 M NaOH solution, but what you need is 5.00 liters of a
0.0320 M NaOH solution. How much of the more concentrated solution, in mL, do you
need to pipet out to make the dilution? (think about this...)




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[SOLUTION MOLARITY AND STOICHIOMETRY / 6P.]                                         W9


TITRATION PROBLEMS

These are used to determine the concentration of some solute in an "unknown" solution.

Equivalence Point: the volume at which the moles of the reactant in the buret that have
been added to the Erlenmeyer flask equal the moles of the second reactant in the
Erlenmeyer flask.

 In the case of acid-base reactions: when moles of acid (H+) = moles of base (OH-).

End Point: the volume at which the indicator in the Erlenmeyer flask changes color and
tells you to stop the titration reaction.

The Equivalence Point and the End Point are not always perfectly matched, but they
are always pretty close. In calculations based on experimental data, it is assumed that
the End Point is perfectly matched to the Equivalence Point. The equivalence point can
only be arrived at theoretically, (on paper), experimentally, we must work with an
indicator and human error with regard to titration technique.

Example 1

If 38.55 mL of HCl is required to titrate 2.150 g of Na2CO3 according to the following
equation, what is the concentration, in molarity, of the HCl solution?

2 HCl (aq) + Na2CO3 (aq)  2 NaCl(aq) + CO2(g) + H2O (l)




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[SOLUTION MOLARITY AND STOICHIOMETRY / 6P.]                                       W9


Example 2

In the photographic developing process, sliver bromide is dissolved by adding sodium
thiosulfate, according to the reaction below. If you want to dissolve 0.225 g of AgBr,
what volume of 0.0138 M Na2S2O3, in mL, should be used?

AgBr (s) + 2 Na2S2O3 (aq)  Na3Ag(S2O3)2 (aq) + NaBr (aq)




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