Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

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Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

Transcript of Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

Page 1: Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

Concentration of SolutionsCh 14.2-14.3 (RB) / Chapter 16 (BB)

Page 2: Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

Heterogeneous Mixtures

A heterogeneous mixture is a mixture that does not have a uniform composition and in which the individual substances remain distinct.

Suspensions are mixtures containing particles that settle out if left undisturbed.

Page 3: Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

Solutions are homogeneous mixtures that contain two or more substances called the solute and solvent.

Most solutions are liquids, but gaseous and solid solutions exist.

Homogeneous Mixtures

Page 4: Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

Homogenous Mixtures

Page 5: Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

Solutions

solution – a substance dissolved in a liquid.

solvent – the liquid, solute – the substance

being disolved.

Page 6: Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

Water—The Super Solvent!

Water is an universal solvent. Water dissolves more solutes than anything

else! A dissolved solid falls apart into its individual

constituent particles, either ions or molecules. This explains why they are usually transparent.

http://www.youtube.com/watch?v=EBfGcTAJF4o&playnext_from=TL&videos=gRLBCOaBXS0

Page 7: Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

Factors that affect solubility

Like tends to dissolve like. Polar solvent dissolves polar solutes Non-polar solute dissolves in non-polar solvents

Molecular Size: Big, heavy molecules tend to be less soluble than small

light molecules. It is harder for the solvent to “cage” the solute.

Temperature: As temperature rises, agitated molecules or ions break

their bonds more easily. So as temperature goes up, so does solubility.

Page 8: Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

Solution Equilibrium

The maximum concentration of solute that a solvent can hold at a certain temperature is called saturated solution.

If large amounts of solute are added to a solution at high temperature then allowed to slowly cool down, then a super saturated solution may occur.

Page 9: Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

Concentration

The concentration of a solution is a measure of how much solute is dissolved in a specific amount of solvent or solution.

Concentration can be described as concentrated or dilute.

Page 10: Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

Concentration

Concentrated vs. Dilute Concentrated: lots of solute per unit of

volume. Dilute: little solute per unit of volume.

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Page 12: Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

Concentration Units

Dilute and Concentrated can be ambiguous. What is dilute? What is concentrated? Chemists like their numbers!

Page 13: Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

Example Problem p. 149

If 18.0 grams of sodium sulfate are dissolved in 207.0 grams of water, what is the percent concentration of this solution?

X 100

= 8.00% = x 100

% by mass = g solute g solution

18.0g solute

225g solution

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Molarity

Molarity is the number of moles of solute dissolved per liter of solution.

Page 15: Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

Example Problem #1 p. 142

Calculate the molarity of 1500 ml of solution that contains 200.0g MgCl2.

200.0g MgCl2 1molMgCl2 1000 ml 1500ml 95.2g MgCl2 1 L

=1.40mol/L or M

Page 16: Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

Example Problem #2 p. 142

Calculate the molarity of a 500 cm3 solution that contains 10.0 g of sodium hydroxide.

To solve: Convert 10.0 g of NaOH per 500 cm3 to moles of NaOH per dm3 of soultion.

10.0g NaOH 1 mol NaOH 1000 cm3

500 cm3 40.0 g NaOH 1dm3

Answer: 0.500 mol/dm3 = 0.500 M

Page 17: Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

Dilutions

If you want a solution of lesser concentration you will need to dilute it! Add more solvent! The total number of moles of solute does not

change

M1 x V1 = M2 x V2

M = Molarity & V = VolumeM1 and V1 = “old” or stock

M2 and V2 = “new” or dilute

Page 18: Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

Example #1

It is necessary to make a .500M solution of HCl from 250.0 ml of a 2.00M solution of HCl. What is the volume of the new solution?

M1 x V1 = M2 x V2

2.00M x 0.250 L = 0.500M x V2

V2 = 1.00 L

So… you add enough solvent to bring the new volume up to 1L. How much do you add?

1.00 L – 0.250 L = 0.750 L

solve for V2 = 2.00M x 0.250ml 0.500M

Page 19: Concentration of Solutions Ch 14.2-14.3 (RB) / Chapter 16 (BB)

THE END

Questions??? HW

BB Chpt 16: 1 (a-c), 2 (a-c), 10 (a-c), 13 Chpt 18: 1-3 a-c, 10 a-c