let’s study solutions solutions homogeneous mixtures of two or more substances solvent & one...

Let’s study solutionsSolutions• homogeneous mixtures of two or more

substances• solvent & one or more solutes

Solutes• spread evenly throughout• cannot separate

• filtration• Separated

• evaporation. • not visible

• unless colored

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Most common solvent “water”

• polar molecule.• forms hydrogen bonds

• hydrogen atom in one molecule - oxygen atom of different water molecule

Na+ and Cl- ions• surface of a NaCl crystal attracted

• polar water molecules.

• hydrated solution • many H2O molecules surrounds

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Like dissolves like

Two substances form solution • an attraction between

• particles of solute and solvent.• polar solvent dissolves polar solutes

• water and sugar• water ionic solutes (NaCl)

• non-polar solvent hexane (C6H14)• dissolves nonpolar solutes

• (oil or grease)Substances in water are:• strong electrolytes produce ions, conduct an electric current

• NaCl(s) + H2O Na+(aq) + Cl− (aq) • weak electrolytes produce few ions

• HF(g) + H2O(l) H3O+(aq) + F- (aq) • non-electrolytes do not produce ions

Water and polar solute

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Solubility• maximum amount of solute

• dissolves in specific amount of solvent• expressed grams of solute in 100 grams of

solvent water g of solute100 g water

Unsaturated solutions • less than maximum amount of solute• dissolve more soluteSaturated solutions • contain maximum amount of solute than

dissolves • undissolved solute at bottom of container

Dissolved solute

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Effect of Temperature on SolubilitySolubility• Depends on temperature.• Of most solids increases as temperature

increases.• Of gases decreases as temperature

increases.

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How to calculate percent concentration?Solubility

amount of soluteamount of solution

• mass or volume of solute in a solution expressed in “100”

percent = amt.solute x 100 amt. solute + amt. solvent

• a. mass percent (m/m) = g of solute x 100 100 g of solution

b. volume % (v/v) = mL of solute x 100

100 mL of solutionc. mass/volume % (m/v) = g of solute x 100

100 mL of solution

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Example calculating percentages!

calculation of mass percent• grams of solute (g KCl) and• grams of solution (g KCl solution).

g of KCl = 8.00 gg of solvent (water) = 42.00 gg of KCl solution = 50.00 g

8.00 g KCl (solute) x 100 = 16.0% 50.00 g KCl solution

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Using percent concentration as conversion factors!

How many grams of NaCl are needed to prepare

225 g of a 10.0% (m/m) NaCl solution?STEP 1 Given: 225 g solution; 10.0% (m/m)

NaCl Need: g of NaCl

STEP 2 g solution g NaClSTEP 3 Write the 10.0% (m/m) as conversion

factors.10.0 g NaCl and 100 g solution

100 g solution 10.0 g NaCl

STEP 4 Set up using the factor that cancels g solution.225 g solution x 10.0 g NaCl = 22.5 g NaCl 100 g solution

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Learning Check

A solution is prepared by mixing 15.0 g Na2CO3 and 235 g of H2O. Calculate the mass percent (%m/m) of the solution.

A) 15.0% (m/m) Na2CO3

B) 6.38% (m/m) Na2CO3

C) 6.00% (m/m) Na2CO3

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Solution

C) 6.00% (m/m) Na2CO3

STEP 1 mass solute = 15.0 g Na2CO3

mass solution = 15.0 g + 235 g = 250. g

STEP 2 Use g solute/ g solution ratio

STEP 3 mass %(m/m) = g solute x 100 g solution

STEP 4 Set up problem mass %(m/m) = 15.0 g Na2CO3 x 100 = 6.00%

Na2CO3 250. g solution

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What is Molarity (M)?Molarity (M)

• concentration term for solutions• gives moles of solute in 1 L solution• moles of solute

liter of solution M = m v

• or m = M x v or v = m M

1.00 M NaCl solution prepared• weigh 58.5 g NaCl (1.00 mole) and• add water to make 1.00 liter of solution

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Steps for calculation of molarityWhat is the molarity of 0.500 L NaOH solution if

itcontains 6.00 g NaOH?STEP 1 Given 6.00 g NaOH in 0.500 L solution

Need molarity (mole/L)STEP 2 Plan g NaOH mole NaOH

molaritySTEP 3 Conversion factors 1 mole NaOH =

40.0 g1 mole NaOH and 40.0 g NaOH40.0 g NaOH 1 mole NaOH

STEP 4 Calculate molarity.6.00 g NaOH x 1 mole NaOH = 0.150

mole 40.0 g NaOH

0.150 mole = 0.300 mole = 0.300 M NaOH

0.500 L 1 L

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Learning Check

What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO3?

A) 0.557 M B) 1.44 MC) 1.71 M

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Solution

C) 1.71 M 46.8 g NaHCO3 x 1 mole NaHCO3 = 0.557

mole NaHCO3

84.0 g NaHCO3

0.557 mole NaHCO3 = 1.71 M NaHCO3

0.325 L

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Learning Check

What is the molarity of 225 mL of a KNO3 solution containing 34.8 g KNO3?

A) 0.344 MB) 1.53 MC) 15.5 M

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Solution

B) 1.53 M34.8 g KNO3 x 1 mole KNO3 = 0.344 mole KNO3

101.1 g KNO3

M = mole = 0.344 mole KNO3 = 1.53 M

L 0.225 LIn one setup:

34.8 g KNO3 x 1 mole KNO3 x 1 = 1.53 M 101.1 g KNO3 0.225 L

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Molarity Conversion Factors

The units of molarity are used as conversion factors in calculations with solutions.

TABLE 7.8

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Calculations Using Molarity

How many grams of KCl are needed to prepare 125mL

of a 0.720 M KCl solution?

STEP 1 Given 125 mL (0.125 L) of 0.720 M KCl

Need Grams of KCl

STEP 2 Plan L KCl moles KCl g KCl

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Learning Check

How many grams of AlCl3 are needed to prepare

125 mL of a 0.150 M solution?

A) 20.0 g AlCl3

B) 16.7g AlCl3

C) 2.50 g AlCl3

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Solution

C) 2.50 g AlCl3

0.125 L x 0.150 mole x 133.5 g = 2.50 g AlCl3

1 L 1 mole

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Learning Check

How many milliliters of 2.00 M HNO3 contain 24.0 g

HNO3?

A) 12.0 mLB) 83.3 mLC) 190. mL

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Solution

24.0 g HNO3 x 1 mole HNO3 x 1000 mL =

63.0 g HNO3 2.00 mole HNO3

Molarity factor inverted

= 190. mL HNO3

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Dilution

In a dilution• water is added.• volume increases.• concentration decreases.

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Comparing Initial and Diluted Solutions

In the initial and diluted solution,• the moles of solute are the same.• the concentrations and volumes are related

by the following equations:For percent concentration:C1V1 = C2V2

initial diluted

For molarity:M1V1 = M2V2

initial diluted

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Dilution Calculations with Percent

What volume of a 2.00% (m/v) HCl solution can be

prepared by diluting 25.0 mL of 14.0% (m/v) HCl

solution?Prepare a table:

C1= 14.0% (m/v) V1 = 25.0 mL

C2= 2.00% (m/v) V2 = ?

Solve dilution equation for unknown and enter values:C1V1 = C2V2

V2 = V1C1 = (25.0 mL)(14.0%) = 175 mL

C2 2.00%

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Learning Check

What is the percent (% m/v) of a solution prepared

by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?

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Solution

What is the percent (%m/v) of a solution prepared

by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?

Prepare a table:C1= 9.00 %(m/v) V1 = 10.0 mL

C2= ? V2 = 60.0 mL

Solve dilution equation for unknown and enter values:C1V1 = C2V2

C2 = C1 V1 = (10.0 mL)(9.00%) = 1.50% (m/v)

V2 60.0 mL

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Dilution Calculations with Molarity

What is the molarity (M) of a solution prepared by diluting 0.180L of 0.600 M HNO3 to 0.540 L?

Prepare a table:M1= 0.600 M V1 = 0.180 L

M2= ? V2 = 0.540 L

Solve dilution equation for unknown and enter values:M1V1 = M2V2

M2 = M1V1 = (0.600 M)(0.180 L) = 0.200 M

V2 0.540 L

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Learning Check

What is the final volume (mL) of 15.0 mL of a 1.80 M

KOH diluted to give a 0.300 M solution?

A) 27.0 mLB) 60.0 mL C) 90.0 mL

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Solution

What is the final volume (mL) of 15.0 mL of a 1.80 M

KOH diluted to give a 0.300 M solution?Prepare a table:

M1= 1.80 M V1 = 15.0 mL

M2= 0.300 M V2 = ? Solve dilution equation for V2 and enter

values:M1V1 = M2V2

V2 = M1V1 = (1.80 M)(15.0 mL) = 90.0 mL

M2 0.300 M

How colloids and suspension different than solutions?Solutions • contain small particles (ions or molecules).• are transparent.• do not separate.• cannot be filtered.• do not scatter light.

Colloids

• have medium size particles.

• cannot be filtered.

• can be separated by semipermeable membranes.

• scatter light (Tyndall effect).

Suspensions• have very large particles.• settle out. • can be filtered. • must be stirred to stay suspended, blood platelets, muddy water

Examples

• Fog

• Whipped cream

• Milk

• Cheese

• Blood plasma

• Pearls

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What is osmosis?• water (solvent) flows from lower solute concentration into

higher solute concentration. • level of solution with higher concentration rises• concentrations of two solutions become equal with timeOsmotic pressure • produced by solute particles

• dissolved in solution.• equal to pressure that

• prevent flow of additional water • into more concentrated solution.

• greater as number of dissolved particles in solution increases.isotonic solution• same osmotic pressure hypotonic solution • has a lower osmotic pressurehypertonic solution• has a higher osmotic pressure