Dilution Calculations Using Conversion Factors Worksheet (Page 2 of 2) in pdf

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:

–

 SO

2–

HSO

+ OH

+ H

Q: What is the concentration of the NaOH solution?

−

0.5302 ��

1 ��

1000 ��

−

35.05 ��

136.18 ��

1 ��

��

= 0.1111

��

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 ��

1 ��

1000 ��

−

35.05 ��

136.18 ��

1 ��

��

(titration ratio)

(molar mass)

(dissociation) (titration eqn) (dissociation) (units)

Notice that in this case, we start with a

A titration just puts the puzzle pieces together in a new way.

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 KHSO

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 KHSO

” 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 K

solution is titrated with dark purple KMnO

solution

of unknown concentration. As the KMnO

is added, a reaction with the following equation

occurs almost instantly, causing the purple color to persist for just a split second:

–

 10 CO

2–

5 C

+ 2 MnO

+ 16 H

+ 2 Mn

+ 8 H

(purple)

(colorless)

After 18.76 mL of KMnO

solution is added, the purple color does not go away, indicating that

is gone, and the “equivalence point” has been reached. What is the concentration

2–

all the C

of the KMnO

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?]

Dilution Calculations Using Conversion Factors Worksheet Page 2

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