Chemistry 121A Hanson
Ch5(handout)
Due Friday, Oct. 17, 2014
Dilution Calculations Using Conversion Factors
In these problems do not use M1V1 = M2V2 unless you are absolutely sure you know what you
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
(0.100 𝐿 𝑑𝑖𝑙𝑢𝑡𝑒 𝑠𝑜𝑙𝑛) ×
0.02 𝑚𝑜𝑙 𝐶𝑂3 2−
𝐿 𝑑𝑖𝑙𝑢𝑡𝑒 𝑠𝑜𝑙𝑛
×
1 𝑚𝑜𝑙 𝐾2 𝐶𝑂3
1 𝑚𝑜𝑙 𝐶𝑂3 2−
× …
using, perhaps, “dilute soln” and “stock soln” to refer to the two solutions being used.
1. How many L of 0.10 M K2CO3 are required to make 0.100 L of 0.02 M CO32- solution by
dilution?
2. How many mL of 0.10 M K2CO3 are required to make 100 mL of 0.02 M K+ solution by
dilution? [Careful! M1V1 = M2V2 will not work here!]
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 + H2O. We write:
50.0 mL HCl soln ×
1𝐿
1000 𝑚𝐿
×
0.200 𝑚𝑜𝑙 𝐻𝐶𝑙
𝐿 𝐻𝐶𝑙 𝑠𝑜𝑙𝑛
×
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
1 𝑚𝑜𝑙 𝐻𝐶𝑙
×
1 𝐿 𝑁𝑎𝑂𝐻 𝑠𝑜𝑙𝑛
0.150 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
= 0.0667 𝐿 𝑁𝑎𝑂𝐻 𝑠𝑜𝑙𝑛
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 M1V1 = M2V2 for titrations, that
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.
In a titration, the experiment always gives us a titration ratio. In the questions below, first ask
yourself, “Do I want a ratio in the end, or do I want a quantity?” If you want a ratio, start with
the titration ratio. There’s really no single “right” way to do these problems. The suggestion to
start with a ratio is just a suggestion. Here’s a classic titration problem. I will solve it first, then
explain what I did:
A 0.5302 g sample of KHSO4 is titrated with 35.0 mL of an NaOH solution. The acid-base
reaction involves the following chemical equation:
HSO4– + OH–  SO42– + H2O
Q: What is the concentration of the NaOH solution?
A:
0.5302 𝑔 𝐾𝐻𝑆𝑂4
35.05 𝑚𝐿 𝑁𝑎𝑂𝐻 𝑠𝑜𝑙𝑛
= 0.1111
1 𝑚𝑜𝑙 𝐾𝐻𝑆𝑂4
136.18 𝑔 𝐾𝐻𝑆𝑂4
×
×
1 𝑚𝑜𝑙 𝐻𝑆𝑂4 −
1 𝑚𝑜𝑙 𝐾𝐻𝑆𝑂4
×
1 𝑚𝑜𝑙 𝑂𝐻 −
1 𝑚𝑜𝑙 𝐻𝑆𝑂4 −
×
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
1 𝑚𝑜𝑙 𝑂𝐻 −
×
1000 𝑚𝐿
𝐿
𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
𝐿 𝑁𝑎𝑂𝐻 𝑠𝑜𝑙𝑛
The titration ratio here is “0.5302 g KHSO4 per 35.05 mL of NaOH solution.” Everything after that is a
conversion factor you have seen before:
0.5302 𝑔 𝐾𝐻𝑆𝑂4
35.05 𝑚𝐿 𝑁𝑎𝑂𝐻 𝑠𝑜𝑙𝑛
(titration ratio)
×
1 𝑚𝑜𝑙 𝐾𝐻𝑆𝑂4
136.18 𝑔 𝐾𝐻𝑆𝑂4
(molar mass)
×
1 𝑚𝑜𝑙 𝐻𝑆𝑂4 −
1 𝑚𝑜𝑙 𝐾𝐻𝑆𝑂4
(dissociation)
×
1 𝑚𝑜𝑙 𝑂𝐻 −
1 𝑚𝑜𝑙 𝐻𝑆𝑂4 −
×
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
1 𝑚𝑜𝑙 𝑂𝐻 −
×
1000 𝑚𝐿
𝐿
(titration eqn) (dissociation) (units)
A titration just puts the puzzle pieces together in a new way. Notice that in this case, we start with a
statement of the titration ratio. This always exists. We start with a ratio because we know
that in the end we want a ratio – “mol NaOH / L NaOH solution.” And in this case, since we
know we want volume of NaOH solution on the bottom, it makes good sense to start with that
on the bottom.
The titration data give us “g KHSO4 per mL of NaOH solution,” and we want “mol NaOH per
L of NaOH solution.” Do you see that basically we are doing a conversion from “g KHSO4” to
“mol NaOH”? That’s just a classic g  mol  mol problem. You know how to do that; all
you need is a molar mass and one or more balanced chemical equations. (This example
includes a few more ratios to cover the dissociation of the reactants into ions.)
3. A 25.00 mL sample of a 0.208 M K2C2O4 solution is titrated with dark purple KMnO4 solution
of unknown concentration. As the KMnO4 is added, a reaction with the following equation
occurs almost instantly, causing the purple color to persist for just a split second:
5 C2O42– + 2 MnO4– + 16 H+  10 CO2 + 2 Mn2+ + 8 H2O
(purple)
(colorless)
After 18.76 mL of KMnO4 solution is added, the purple color does not go away, indicating that
all the C2O42– is gone, and the “equivalence point” has been reached. What is the concentration
of the KMnO4 solution? [Hint: What is the titration ratio in this case? How close does it look
like “mol KMnO4 per liter of KMnO4 solution”?]
4. Exactly 25.34 mL of 0.1023 M NaOH solution is required to titrate 0.5020 grams of an
unknown acid “HA”. What is the molar mass of HA? [Hint: What units do you want in end?]