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CHAPTER 4B Solutions

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Chapter 4B. Solutions. Solutions. Homogeneous mixture Solvent – substance present in the largest amount Solute – substance present in the smallest amount Aqueous solution – solutions with water as the solvent. Types of Solutions. Solubility. - PowerPoint PPT Presentation

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Page 1: Chapter 4B

CHAPTER 4BSolutions

Page 2: Chapter 4B

SolutionsHomogeneous mixture

Solvent – substance present in the largest amount

Solute – substance present in the smallest amount

Aqueous solution – solutions with water as the solvent.

Page 3: Chapter 4B

Types of SolutionsSolute Solvent ExampleGas Gas Air ( O2, CO2 in N2)

Liquid Gas Water in airGas Liquid Carbonated beverages (CO2 in

water)

Liquid Liquid Vinegar (acetic acid in water)

Solid Liquid Sugar watersolid Solid Stainless steel (Cr and Ni in Fe

Page 4: Chapter 4B

SolubilityThe maximum amount of a solute that

can be dissolved in a given amount of solvent at a constant temperature.

Factors that affect rate of Solubility (increase the rate of dissolving).

Pressure Temperature Polarity

Page 5: Chapter 4B

SolubilitySolutions can be classified into groups by how much

solute is in solution.

Saturated – contains as much solute as the solvent can hold at a given temperature.

Unsaturated – less the maximum amount of solvent at a given temperature.

Supersaturated – contains more solute than the solvent can hold at a given temperature.

Page 6: Chapter 4B

Solution by Mass PercentSolutions are described by mass percent

mass of solutemass percent= 100mass of solution

x

grams of solutemass percent= 100grams of solute +grams of solvent

x

Page 7: Chapter 4B

Solutions by Mass PercentA solution is prepared by mixing 2.50

grams of calcium chloride with 50.0 grams of water. Calculate the mass percent of calcium chloride in this solution.

Page 8: Chapter 4B

Solutions by Mass PercentA 75.0 gram sample of a solution is known

to contain 23.8 grams of glucose. Calculate the mass percent of glucose in this solution.

Page 9: Chapter 4B

Solutions by Mass PercentVinegar is a solution of acetic acid in water

and typically contains 5.00% acetic acid by mass. Calculate the mass of acetic acid in a 250.0 gram sample of vinegar.

Page 10: Chapter 4B

Solutions by Mass PercentConcentrated hydrochloric acid solution

contains 37.2 % by mass HCl. What mass of HCl is contained in 35.5 grams of concentrated HCl.

Page 11: Chapter 4B

MolarityMolarity – concentration of a solution

moles of solute mM=molarity=liter of solution Lt

Page 12: Chapter 4B

MolarityCalculate the molarity of a solution

prepared by dissolving 15.6 gram of solid KBr in enough water to make 1.25 L of solution.

Page 13: Chapter 4B

MolarityCalculate the molarity of a solution

prepared by dissolving 2.80 grams of solid NaCl in enough water to make a 135 ml of solution

Page 14: Chapter 4B

MolarityWhat mass of NaNO3 is required to make

2.50 Liter of a 0.150 M solution of NaNO3?

Page 15: Chapter 4B

MolarityWhat mass of (NH4)2SO4 is required to

make 1.25 liter of 0.250 M solution of (NH4)2SO4?

Page 16: Chapter 4B

Molarity-Ion ConcentrationGive the concentrations of all ions in each

of the following solutions

1.20 M Na2SO4 0.750 M K2CrO4

Page 17: Chapter 4B

Molarity-Ion ConcentrationHow many moles of Na+ ions are present

in 42.0 ml of a 0.350 M NaCl solution?

Page 18: Chapter 4B

Molarity-Ion ConcentrationHow many moles of Na+ ions are present

in 42.0 ml of a 0.350 M of Na3PO4 solution?

Page 19: Chapter 4B

DilutionsProcess of adding more solvent to a solution.

M1=molarity before dilutionV1=volume before dilutionM2=molarity after dilutionV1=volume after dilution

1 1 2 2M V M V

Page 20: Chapter 4B

DilutionWhat volume of 19 M NaOH must be used

to prepare a 1.0 Liter of a 0.15 M NaOH solution?

Page 21: Chapter 4B

DilutionWhat volume of water is needed to

prepare 500.0 ml of a 0.250 M Ca(NO3)2 solution from a 5.00 M solution of Ca(NO3)2?

Page 22: Chapter 4B

TitrationsCombining a solution of know

concentration with a solution of unknown concentration to determine the unknown.

Equivalence point – when the reactants are equal molar.

Indicator – dyes used in titrations to indicate equivalence point.

Page 23: Chapter 4B

Titrations What volume of 0.115 M HCO4 solution is needed

to neutralize 50.00 ml of 0.0875 M NaOH?

15.00 ml of acetic acid is titrated with 0.500 M NaOH. It takes 22.59 ml of NaOH. What is the molarity of the acetic acid?

How many grams of AgBr are formed when 35.5 ml of 0.184 M AgNO3 are mixed with an excess of aqueous hydrobromic acid?