Solvent

Solute

Solution

- What you are dissolving the stuff in (most commonly water)

- The stuff you are dissolving

- The combination of the solvent and the solute

We normally think of a solution as dissolving a solid into water but a solvent can be a mixture of liquid-liquid or gas-liquid as well.

Solution

Solvent

Solute

1. Alloys – a mixture of two solids (normally metals)

Examples Brass (copper and zinc), Steel ( iron and mainly carbon) Wood’s metal (50%bismuth, 10% cadmium,

13% tin, and 27% lead. melts at 70°C (160°F)) (Very Low)

Where would you use Woods metal?

2. Suspension – a solution with solute particles that are large in size and will not dissolve. If left alone, particles will fall out of solution (seperate).

Examples: sand and wateroil and water

3. Colloids - a solution where the particles maynot be fully mixed or dissolved but they will not settle out

Examples: smoke (solid in a gas)fog (liquid in a gas)mayo (oil in water)

1. You can look at the solute size. The smallest size of the molecules are in solution, then alittle bigger size are in colloids, and then the largest are in suspensions.

2. Suspensions will settle out so the real problem is telling colloids from solutions.

You can exam the mixture for the tyndall effect. Only colloids exhibit the Tyndall effect

Tyndall Effect is the scattering of light due to particles in a mixture not being completely dissolved.

 How Things DissolveIf a liquid dissolves in another liquid, we say they are

miscible or soluble (opposite is immiscible or insoluble)

Ionic compounds & polar molecules dissolve in water because polar soultes will always dissolve in polar solvents. We commonly refer to this as “likes dissolves likes”

Animation of Dissolving

Any substance that when put into a polar solvent will dissociate into ions.

ex. NaCl Na+ (aq) + Cl-(aq)

If a solution is an electrolyte, it will conduct

electricity in a solution.The more ions it dissociates into, the stronger

the electrolyte, the more it will conduct.Which would conduct easier, magnesium

chloride or aluminum nitrate?

Equations that show what ions the solute will dissolve into.

Rules1: Ionic bonds (metal and nonmetal) will dissociate

Rule 2: Covalent bonds (nonmetals – nonmetals or polyatomic ions) will NOT dissociate

MgO

BaF2

FeSO4

Ni2(CO3)3

WHICH COMPOUND WILL ACT AS THE BEST ELECTRYOLYTE IN TERMS OF CONDUCTIVITY

What is solvation?

Rates of solvation (dissolving) depend on:

1. Surface area-

small particles have a greater amount of surface exposed to the solvent, so dissolution is quicker.

2. Stirring- clears out already dissolved solute, increasing the solvent’s ability to access the solute. Stirring also increases collisions between solute and solvent.

3. Temperature- an increase in temperature increases rate of collision between solute and solvent.

Also sometimes called dissolution, is the attraction of a solvent with ions of a solute

Measuring Concentrations of Solutions

General terms to classify solutions

Dilute or concentrated

More detailed classifications

Unsaturated - solvent could hold more solute

Saturated – Solvent can hold no more solute so if any additional solute is added it will not dissociate and will just collect at the bottom.

Supersaturated- unstable solution that has more solute than can theoretically be dissolved.

This graph is the

solubility curves for a handful of ionic solutes.

Lets focus on NaNO3

If we are below the solubility curve, the solution would be described as unsaturated

If we are above the solubility curve, the solution would be described asSuper saturated

Ex:What is the solubility of a NaNO3 solution at 50˚C?

Ans: ~115gNaNO3

What happens to that solution if the temperature is decreased to 10˚C?

Ans: the solubility of NaNO3 is 80g per 100g of water at 10˚C. Initially, 115g were dissolved, so we have 35g more than we should (super saturated solution)

What word describes a NaNO3 solution with 80 g solute/100 g water at 20°C?How about 100 g same temp?

Ans: Unsaturated and Supersaturated

How many grams of KCl would make a saturated solution at 80 degrees?

Ans:50 grams of KCl

How many grams of KCl would make a saturated solution at 80 degrees if the solvent volume was increased to 200 grams?Ans: 100 g of KCl

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Solubility• Depends on temp.• Most solid’s

solubility• as temp

increases.• Gases’ solubility• as

temp increases.

increase

decrease

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Why do fish die in water that is too warm?

Because O2 gas is

less soluble in warm water, fish cannot obtain the amount of O2 required for

their survival.

O2

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A. Why could a bottle of carbonated drink possibly burst (explode) when it is left out in the hot sun ?

The pressure in a bottle increases as the gas leaves solution as it becomes less soluble at high temperatures. As pressure increases, the bottle could burst.

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Henry’s Law states • the solubility of a

gas in a liquid is directly related to the pressure of that gas above the liquid.

• higher pressures = more gas molecules dissolved in the liquid.

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Molarity (M)

A concentration that expresses the

moles of solute in 1 L of solution

Molarity (M) = moles of solute

1 liter solution

Two types of problems:

1. You are asked to calculate the concentration (Molarity)

- To solve you find mols of solute and divide it by the liters of

solution.

2. You are given the concentration and asked to find either mass or volume

- You need to use stoichiometry.

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Practice problem

A KOH solution with a volume of 400 mL contains 2 mole KOH. What is the molarity of the solution?

1) 8 M

2) 5 M

3) 2 M

Dra

no

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Practice problem

A KOH solution with a volume of 400 mL contains 2 moles of KOH. What is the molarity of the solution?

2) 5 M

M = 2 mole KOH = 5 M

0.4 L

Dra

no

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Molarity Calculation

NaOH is used to open stopped sinks, to treat cellulose in the making of nylon, and to

remove potato peels commercially.

If 4.0 g NaOH are used to make 500. mL of NaOH solution, what is the molarity (M) of the solution?

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Calculating Molarity

1)Find mols of NaOH

4.0 g NaOH x 1 mole NaOH = 0.10 mole NaOH

40.0 g NaOH

2) Find Liters of solution

500. mL x 1 L _ = 0.500 L

1000 mL

3.Divide the two

0.10 mole NaOH = 0.20 M NaOH

0.500 L

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Practice problem

A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose

has a molar mass of 180. g/mole, what is the molarity of the glucose solution?

1) 0.20 M

2) 5.0 M

3) 36 M

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Practice problem

A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose

has a molar mass of 180. g/mole, what is the molarity of the glucose solution?

1) 72 g x 1 mole = .400 x 1 = 0.20 M

180. g 2.0 L

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Problem type number 2

A solution is a 3.0 M NaOH.. Write the molarity in the form of conversion factors.

3.0 moles NaOH or 1 L NaOH soln

1 L NaOH soln 3.0 moles NaOH

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Practice problem

Stomach acid is a 0.10 M HCl solution. How many moles of HCl are in 1500 mL of stomach acid solution?

1) 15 moles HCl

2) 1.5 moles HCl

3) 0.15 moles HCl

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Practice

3) 1500 mL x 1 L x 0.10 mole HCl =

1000 mL 1 L

Molarity used as a conversion factor in the grid

.15 mol HCl

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Practice

How many grams of KCl are present in 2.5 L

of 0.50 M KCl?

1) 1.3 g

2) 5.0 g

3) 93 g

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Practice

3)

2.5 L x 0.50 mole x 74.6 g KCl = 93 g KCl

1 L 1 mole KCl

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Practice

How many milliliters of stomach acid, which is 0.10 M HCl, contain 0.15 mole HCl?

1) 150 mL

2) 1500 mL

3) 5000 mL

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Practice

2) 0.15 mole HCl x 1 L soln x 1000 mL

0.10 mole HCl 1 L

(Molarity inverted)

= 1500 mL HCl

DILUTION FORMULAThe act of diluting a solution is to simply add more solvent ( most of the time water) thus leaving the amount of solute unchanged.

Since the amount or moles of solute before dilution (n1) and the moles of solute after the dilution (n2) are the same: n1 = n2

A relationship can be established such that

M1V2 = n1 = n2 = M2V2

Or simply : M1V1 = M2V2

Calculate the molarity of a solution prepared by diluting 25.0 mL of 0.05 M potassium iodide with 50.0 mL of water

M1 = 0.05 mol/L M2 = ?

V1 = 25.0 mL V2 = 50.0 + 25.0 = 75.0 mL

M1V1 = M2V2

M1 V1 = M2 = (0.05 mol/L) (25.0 mL) = 0.0167 M of KI

V2 75.0 mL

MOLARITY& Dilution

Calculate the molarity of a solution prepared by diluting 40.0 mL of 0.325 M LiF to 50.0 mL of water

MOLARITY& Dilution

M2 = (40 mL) x (0.325 M)

(50.0 mL) = .260 M LiF

Calculate the molarity of a solution prepared by diluting 40.0 mL of 0.325 M LiF with 50.0 mL of water

M2 = (40 mL) x (0.325 M)

(50.0 mL + 40.0 mL) = .14 M LiF

Given a 6.00 M HCl solution, what volume would you need to prepare 250.0 mL of 0.150 M HCl?

M1 = 6.00 mol/L M2 = 0.150

V1 = ? mL V2 = 250.0 mL

M1V1 = M2V2

M2 V2 = V1 = (0.150 mol/L) (250.0 mL) = 6.25 mL of 6 M HCl

M1 6.00 mol/L

MOLARITY& dilution

1. If asked to find molarity. Find mols and then find liter and divide to the two. ( NO LONG GRIDS)

Recap

2. If given the molarity and asked to find volume or amount (mols or grams) then USE A GRID.

3. If given a problem with two molarities or two volumes then you should think DILUTION FORMULA

PRACTICE PROBLEMS_________1. What is the concentration of 250.0 mL of 0.60 moles of HCl?_________ 2. What is the concentration of 35.0 mL of 0. 0556 moles of KCl?_________ 3. How many grams of KCl is needed to prepare 50.0 mL of a 0.10 M solution?_________ 4. How many milliliters of water must be added to 30.0 mL of 9.0 M KCl to make a solution that is 0.50 M KCl?_________ 5. What volume of 0.7690 M LiOH will contain 55.3 g of LiOH? _________ 6. How many liters of water must be added to 100.0 mL of 4.50 M HBr to make a solution that is 0.250 M HCl?

2.4 M2.4 M

3.00 L3.00 L

1.80 L1.80 L

0.37 g

1.59 M

540 mL

In double displacement reactions, we make two new products. Sometimes one of those products is insoluble so with turns to a

. Instead of remaining aqueous.solid

In order to determine what product will do this, we need a chart called solubility rules.

Solubility Rules:

1. All chlorates, perchlorates, nitrates, and acetates are soluble

2. All alkali metal ions and ammonium are soluble

3. All halides (group 7) except Ag, Pb(II), and Hg(I) are soluble

4. Most sulfates are soluble except Ca, Sr, Ba,Hg(I), Ag, and Pb(II)

5. All hydroxides are insoluble except group I, Ba, and Sr (Ca is slightly soluble)

6. All sulfides are insoluble except group I and II, and ammonium

7. All phosphates, chromates, carbonates are insoluble except group I and ammonium

a) Ca(NO3)2

Soluble (salts containing NO3- are soluble)

b) FeCl2

Soluble (all chlorides are soluble)c) Ni(OH)2

Insoluble (all hydroxides are insoluble)d) AgNO3

Soluble (salts containing NO3- are soluble)

e) BaSO4

Insoluble (Sulfates are soluble, except … Ba2+)f) CuCO3

Insoluble (containing CO32- are insoluble)

g) PbCl2 Insoluble

Predict the insoluble product of the following reactions

1. CuSO4(aq) + BaCl2(aq) CuCl2( ) + BaSO4( )

2. Fe(NO3)3(aq) + LiOH (aq) ______ (s) + LiNO3 (aq)

3. Na3PO4 (aq) + CaCl2 (aq) _________ (s) + (aq)

4. Na2S (aq) + AgC2H3O2 (aq) __ ____ (s) + (aq)

saq

Fe(OH)3

Ca3(PO4)2 NaCl

NaC2H3O2Ag2S

Mix barium nitrate and sodium carbonate a solid is made. Write a balanced equation that shows the formation of the solid

BaCO3(s)

1. Molecular equation – normal balanced equation with states of matter.

Write the molecular equation of lead (II) nitrate being added to sodium chloride.

Pb(NO3)2 (aq) + 2NaCl (aq) PbCl2(s) + 2NaNO3 (aq)

2. Complete ionic equation – all aqueous compounds are shown in their ion form

Pb+2(aq) + 2 NO3

-1(aq) + 2Na+1

(aq) + 2 Cl-1(aq)

PbCl2(s) + 2Na+1(aq) + 2 NO3

-1(aq)

V

3. Net Ionic Equations- We just cancel out everything that appears on both sides of the complete ionic.- We should be left with only the ions needed to make the precipitate (solid).

Pb+2(aq) + 2 NO3

-1(aq) + 2Na+1

(aq) + 2 Cl-1(aq)

PbCl2(s) + 2Na+1(aq) + 2 NO3

-1(aq)

Pb+2(aq) + 2 Cl-1(aq) PbCl2(s)

Rxn between silver one nitrate and nickel (II) chloride.

Ag+1(aq) + Cl-1(aq) AgCl(s)

Colligative properties–

properties of a solution which depend only on relative numbers of particles, not on their molecular weights or identities

These are directly proportional to the molality of the solution

t(f) = i kf m t(b) = i kb m

i = Van’t Hoff factor = how many ions form from the solute

Kf and kb

are freezing point constants.

t(f) and t(b) are the change in the normal freezing and boiling

point. (not necessarily the new freezing and boiling point)

What is the I value for the following:

LiF(s)

FeBr3

Ca(NO3)2

C11H22O11

A solution is made by dissolving 24.0 g of sucrose (mw=342) in 175 g water.Calculate the molality of the solution.

Kb for water is 0.51°C kg/mol. Calculate the boiling point of the solution

Kf for water is -1.86 °C kg/mol. Calculate the freezing point of the solution

Sample

Estimate the freezing point of an aqueous solution of 0.106 m MgCl2, assuming complete dissociation.

A solution was made by dissolving 18.00 g of glucose in 150.0 g of water. The resulting solution boiled at 100.34°C. (Kb for water is 0.51°C kg/mol)

Calculate the molality of the solution based upon boiling point elevation.

Calculate the molecular mass of glucose using the molality.

What is the freezing point of adding 3.76 grams of NaOH into 320 g of chloroform?

What is the boiling point of adding 96.2 grams of AlCl3 into 677g of ethanol?

How many grams of Calcium nitrate would you have added if 500. grams of benzene did not freeze until -1.32 °C?

 Solution StoichiometryWhat volume of 0.200 M CuSO4 solution is required to react with

50.0 ml of 0.100 M NaOH? 1. Write a balanced equation.CuSO4 (aq) + 2NaOH(aq) ---->Cu(OH)2(s) + Na2SO4(aq)

2. Use stoichiometric starting with the compound that we have a volume for and work a grid to the compound in question.

50 mL ( 1 L ) ( 0.100 mol NaOH) ( 1 mol CuSO4 ) ( 1 L )

( 1000ml) ( 1 L ) ( 2 mol NaOH ) ( 0.2mol)

Molarity of NaOH

Mol to mol ratio

Molarity of CuSO4

 How many moles of barium sulfate can form when

20.0 ml of 0.600 M barium chloride is mixed with excess magnesium sulfate?

How many moles of barium sulfate can form when 30.0 ml of 0.250 M magnesium sulfate is mixed with excess barium chloride?

How many moles of barium sulfate can form when 20.0 ml of 0.600 M barium chloride is mixed with 30.0 ml of 0.250 M magnesium sulfate ?