Chapter 16 – Solutions
Mr.Yeung
Lesson 5 - Objectives
• Take up questions
• Dilutions (Super important)!
Dilutions
• When you buy a can of
juice concentrate from the
supermarket you are
expected to dilute the
juice concentrate.
• What would a spoonful
of concentrate taste like
compared to a spoonful
of the diluted juice?
• Yuck?
– The concentrate would be
a lot stronger
Think of the orange
concentrate are molecules
Concentrate
Diluted
What if …
• We think
• 1 can of concentrate = 1 mol
• And we need to make 1L of orange juice?
• 1 can / 1 L = 1M
• What about 0.5L?
• 1 can / 0.5L = 2M
• What about 0.75L?
• 1 can / 0.75L = 1.33M
• More water = less concentrated
Now..
• Say we took the
– 2M concentrated orange juice we
made (1 can / 0.5L) (Stock solution)
– We used 2L of this concentration (2M)
and we want to dilute this solution to
make 6L of orange juice
– What would you do?
Dilution
Before dilution
After dilution
Total volume of 6L
2M of 2L
2L is 3 times less than 6L so the 2M will be 3 times as less as well
= 0.67M
Formula
• Basically the formula for dilution
– C1 V1 = C2 V2
•
•
•
•
•
C1 = initial concentration (mol/L)
V1 = initial volume (L)
C2 = final concentration (mol/L)
V2 = final volume (L)
From last example
– 2mol/L * 2L = C2 * 6L
– C2 = 2/3 mol/L
Examples
• Example:
• What volume of concentrated sulfuric
acid, 18.0 M, is required to prepare 5.00 L
of 0.150 M solution by dilution with water?
– C1 = 18M
– V1 = ? (*required)
C2 = 0.150M
V2 = 5.0L
• 18M * (?) = 0.150M * 5.0L
• (0.150M * 5.0L) / 18M = 0.042L
– Does that make sense? 18.0M is A LOT more
concentrated than 0.150M so using a little
amount is expected!
Example
• How much in grams would you need
to prepare 500.0 mL of a 0.100 M
standard solution of KNO3?
– Find mols
• 0.5L * 0.1mol/L = 0.05mol
– Find mol mass
• 101.1g/mol
– Find grams
• 101.1g/mol * 0.05mol = 5.05g
Example
• If I have 340 mL of a 0.5 M NaBr solution,
what will the concentration be if I add 560
mL more water to it?
– C1V1 = C2V2
•
•
•
•
V1 = 340ml or .34L
C1 = 0.5M
V2 = 560ml or .56L
C2 = ?
0.34L * 0.5M = 0.56L * (?)
• C2 = .30M
Real life example
• How would you prepare 500 ml of 3 M HCl using 6 M HCl from the
stockroom. In other words how much water and how much 6 M HCl
would you mix to accomplish this dilution?
– Determine the volume of 6M HCl to use applying the dilution
equation 6 M ( Volume of 6M) = 3 M HCl ( 500 ml)
– Volume of 6 M HCl = 3 M HCl ( 500 ml) / 6 M HCl = 250 ml 6M
HCl
• So what is the 250ml 6M HCL?
– That is the amount you need but you still need to dilute this…
• To what volume?
– Total = 500ml – 250ml
Example in chem labs
• Example 1
• Let us consider 5.00 mL of a solution
labeled “6.00 M HCl” and we now
add enough water to give a total
volume of 14.00 mL. What is the
concentration of the new solution?
Summary
• Dilution Problems