Chemistry 121A Hanson
Ch5(handout)
Due Friday, Oct. 17, 2014
Dilution Calculations Using Conversion Factors
In these problems do not use M
V
= M
V
unless you are absolutely sure you know what you
1
1
2
2
are doing. Instead, use conversion factors that use “mol/L” instead of “M” and are very clear in
what that is of. For example, in Problem 1, you could start with
2−
0.02
1
3
2
3
( 0.100 ) ×
×
× …
2−
1
3
using, perhaps, “dilute soln” and “stock soln” to refer to the two solutions being used.
2-
1. How many L of 0.10 M K
CO
are required to make 0.100 L of 0.02 M CO
solution by
2
3
3
dilution?
+
2. How many mL of 0.10 M K
CO
are required to make 100 mL of 0.02 M K
solution by
2
3
dilution? [Careful! M
V
= M
V
will not work here!]
1
1
2
2
Titration and Titration Ratios
Let’s start with a simple acid-base problem:
Q: What volume of 0.150 M NaOH solution is required to react completely with 50.0 mL of
0.200 M HCl solution?
A: The equation for the reaction is NaOH + HCl NaCl + H
O. We write:
2
1
0.200
1
1
50.0 mL HCl soln ×
×
×
×
1000
1
0.150
= 0.0667
s
This is just a classic mL mol mol L problem, right? All
we need to know is the balanced equation and then use its
stoichiometry and (in this case) a couple of molarities.
A key component of a titration that makes it different from a
dilution is that in a titration you are doing a chemical reaction for
which you must know the balanced chemical equation, exactly
like this.
The goal of a titration is to use a known amount or concentration
of one substance in one solution to determine the unknown
amount or concentration of another substance in another solution.
While it may be tempting to use M
V
= M
V
for titrations, that
1
1
2
2
turns out rarely to be useful. A much more powerful method is
just to express the problem in the form you need and then do
conversion factors to get there.