# titration_redoxreview by stariya

VIEWS: 10 PAGES: 2

• pg 1
```									     Extra Practice: Titration Problems…plus free bonus: Redox Recap at no extra charge!!!

Many students seem to feel that the titration problems are the most challenging of
Chapter 5. These problems usually have several numbers given, and I think it likely that
those that struggle are unsure when and how to use these numbers. This review will
attempt to illuminate this type of problem.

Titration problems are simply stoichiometry problems, and as such require a balanced
equation. If one is not written in the problem, you should write the equation down, as you
will need it for the mole ratios. Titrations involve a very specific laboratory technique: A
volume of titrant (one of the reactants, contained in a buret) is added to a flask that
contains the other reactant. Depending on the problem, we have complete information
about one of the two reactants. In other words, we have info that will allow us to
determine moles of that reactant. We are given incomplete information about the other
reactant; we’ll need to do a stoich problem to finish up.

Common types of titration problems:
1) You are given molarity and volume of titrant delivered; you need to find
concentration/molar mass/identity of the chemical in the flask.
2) You are given (or can find) the moles of a chemical in the flask, you are asked
to determine the molarity of the titrant given the volume of it used to perform
the titration.

Example: What is the molarity of a hydrochloric acid solution if 36.7 mL of the HCl
solution is needed to react with 43.2 mL of 0.236M sodium hydroxide solution?

Answer: It’s not clear which chemical is in the buret, which is in the flask—it doesn’t
what we know and work toward what we don’t know. Our “roadmap” of the problem:
Vol NaOH mol NaOH mol HCl Molarity HCl
Since we know we’re looking for molarity of HCl, we know the last calculation will be
Molarity HCl = (moles HCl)/(L HCl), and we are given liters of HCl, so our real goal is
HCl + NaOH  NaCl + H2O

0.236 mol NaOH    1 mol HCl
0.0432 L NaOH                                  = 0.0102 mol HCl
L NaOH       1 mol NaOH

Then MHCl = mol HCl/L HCl = 0.0102 mol/0.0367L = 0.278M HCl


Example: Calculate the molarity of a solution of H2SO4 if 40.0 mL of the solution reacts
exactly with 0.364 g of sodium carbonate
H2SO4 + Na2CO3  Na2SO4 + H2O + CO2

Try a couple example problems:

Ex 1: Benzoic acid, C6H5COOH, is sometimes used as a primary standard for the
standardization fo solutions of bases. A 1.862-g sample of the acid is neutralized by
33.00 mL of NaOH solution. What is the molarity of the base solution?

C6H5COOH + NaOH  C6H5COONa + H2O

Ex 2: Calcium carbonate tablets can be used as an antacid and a source of dietary
calcium. A bottle of generic antacid tablets states that each tablet contains 500. mg
calcium carbonate. What volume of 6.0M HNO3 can be neutralized by the calcium
carbonate in one tablet?

Ex 3: What is the molarity of a solution of sodium hydroxide, NaOH, if 36.2 mL of this
solution is required to react with 25.0 mL of 0.0513 M nitric acid solution according to
this equation?

HNO3 + NaOH  NaNO3 + H2O

If it’s still not happening for you, see Mr. Langellier ASAP

Redox Recap:
• You must be able to assign oxidation numbers to elements; see page 207.
• When an element loses electrons as it becomes a product, it is oxidized.
Its reactant substance is the reducing agent.
• When an element gains electrons as it becomes a product, it is reduced.
Its reactant substance is the oxidizing agent.
• Any element that is in its pure elemental form on one side of the equation but is
in a compound on the other side has been oxidized or reduced!
• It helps to know elements that are commonly reduced/oxidized. These include:
Mn, Cr, S, P, N, Cl,
• When faced with a redox equation, first figure the oxidation numbers for the
elements involved, then determine who’s been reduced, who’s been oxidized.
Then you can determine oxidizing agent/reducing agent.

Example: determine what element has been reduced, which has been oxidized, which
substance is the reducing agent, and which is the oxidizing agent.

Zn (s) + HClO4 (aq)  Zn(ClO4)2 (aq) + H2 (g)

Ox’d                                    Red’d

OA                                      RA

```
To top