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?

DilutionBefore dilution After dilution

2M of 2LTotal volume of 6L

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 C2 = 0.150M– V1 = ?(*required) 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?

2

(6.00M)(5.00mL)M = = 2.14M

14.00mL

Summary

• Dilution Problems