Free Powerpoint Presentations

Volumetric
Page
3

DOWNLOAD

PREVIEW

WATCH ALL SLIDES

A solution is prepared by dissolving a solute in a solvent.

Slide 17

When a solution is diluted, more solvent is added to it, the number of moles of solute stays the same.

When a solution is diluted, more solvent is added to it, the number of moles of solute stays the same.

i.e. n1 = n2

Recall, C = n ÷ V,

Make n the subject and substitute, it follows that

C1V1 = C2V2

where C1=original concentration of solution

V1=original volume of solution

C2=new concentration of solution after dilution

V2=new volume of solution after dilution

n1 = no of mol of solute before dilution

n2 = no of mole of solute after dilution

Slide 18

To calculate the new concentration (C2) of a solution given its new volume (V2) and its original concentration (C1) and original volume (V1).

To calculate the new concentration (C2) of a solution given its new volume (V2) and its original concentration (C1) and original volume (V1).

Note: V2 = V1 + vol. of water added.

Slide 19

Examples

Examples

Calculate the new concentration (molarity) if enough water is added to 100cm3 of 0.25M sodium chloride to make up 1.5dm3.

C2=(C1V1) ÷ V2

C1 = 0.25M

V1 = 100cm3 = 100 ÷ 1000 = 0.100dm3 (volume must be in dm3)

V2 = 1.5dm3

[NaCl(aq)]new = C2 = (0.25 x 0.100) ÷ 1.5 = 0.017M

(or 0.0.017 mol/dm3)

Slide 20

More

More

If 280cm3 of a 3moldm-3 sodium hydroxide solution is diluted to give 0.7moldm-3 soln.

What is the vol. of the resulting diluted solution?

What is the vol. of distilled water added to the original soln.?

Slide 21

Let’s do it

Let’s do it

V1 = 280cm3 ,C1 = 3moldm-3 ,C2 = 0.7moldm-3

V2 = ?

C1V1 = C2V2

V2 = 3 X 280 = 1200cm3 0.7

To know the vol. of distill water added

V2 = V1 + vol. of distill water added.

vol. of distill water added.= 1200 – 280

= 920cm3

Slide 22

One more!

One more!

Calculate the vol. of a 12.0moldm-3 HCl that should be diluted with distilled water to obtain 1.0dm3 of a 0.05moldm-3 HCl.

Soln.

C1 = 12moldm-3, V1 = ?

C2 = 0.05moldm-3 , V2 = 1.0dm3

I’ve done my own part, do yours!

Slide 23

PRActice problems

PRActice problems

Slide 24

Acid-Base Titrations

Acid-Base Titrations

Acid-base titrations are lab procedures used to determine the concentration of a solution. We will examine it's use in determining the concentration of acid and base solutions.

Titrations are important analytical tools in chemistry.

Slide 25

During the titration

During the titration

Go to page:
 1  2  3  4  5  6  7 

Contents

Last added presentations

© 2010-2024 powerpoint presentations