© copyright 1995-2007 r.j. rusay aqueous solutions concentration / calculations dr. ron rusay fall...

© Copyright 1995-2007 R.J. Rusay© Copyright 1995-2007 R.J. Rusay

Aqueous SolutionsConcentration / Calculations

Dr. Ron RusayDr. Ron Rusay

Fall 2007Fall 2007


Solutions Homogeneous solutions are comprised of Homogeneous solutions are comprised of solutesolute(s), (s),

the substance(s) dissolved, [The lesser amount of the the substance(s) dissolved, [The lesser amount of the component(s) in the mixture], andcomponent(s) in the mixture], and

solventsolvent, the substance present in the largest amount., the substance present in the largest amount. Solutions with less solute dissolved than is physically Solutions with less solute dissolved than is physically

possible are referred to as “possible are referred to as “unsaturatedunsaturated”. Those with ”. Those with a maximum amount of solute are “a maximum amount of solute are “saturatedsaturated”.”.

Occasionally there are extraordinary solutions that Occasionally there are extraordinary solutions that are “are “supersaturatedsupersaturated” with more solute than normal.” with more solute than normal.

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Concentration and Temperature

Relative SolutionConcentrations:SaturatedUnsaturatedSupersaturated


DHMO, dihydromonoxide : “The Universal” Solvent

04_40

H

O2δ−

δ+

δ+

105 °

H

http://www.dhmo.org


Water : “The Universal” Solvent

Generally, likes dissolve likes, i.e. polar-polar and nonpolar-nonpolar. If polar and nonpolar mix , eg. oil and water:

The oil (nonpolar) and water (polar) mixture don’t mix and are immiscible. If liquids form a homogeneous mixture, they are miscible.


Aqueous Reactions & Solutions Many reactions are done in a homogeneous liquid or Many reactions are done in a homogeneous liquid or

gas phase which generally improves reaction rates.gas phase which generally improves reaction rates. The prime medium for many inorganic reactions is The prime medium for many inorganic reactions is

water which serves as a solvent (the substance water which serves as a solvent (the substance present in the larger amount), but does not react present in the larger amount), but does not react itself.itself.

The substance(s) dissolved in the solvent is (are) the The substance(s) dissolved in the solvent is (are) the solute(s). Together they comprise a solution. The solute(s). Together they comprise a solution. The reactants would be the solutes.reactants would be the solutes.

Reaction solutions typically have less solute Reaction solutions typically have less solute dissolved than is possible and are “unsaturated”.dissolved than is possible and are “unsaturated”.

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Salt dissolving in a glass of water


Water dissolving an ionic solid


Solution Concentration

A solution’s concentration is the measure of the amount of A solution’s concentration is the measure of the amount of solute dissolved. solute dissolved.

Concentration is expressed in several ways. One way is Concentration is expressed in several ways. One way is mass percent.mass percent.

Mass % = Mass solute / [Mass solute + Mass solvent ] Mass % = Mass solute / [Mass solute + Mass solvent ] x100x100

What is the mass % of 65.0 g of glucose dissolved in 135 g What is the mass % of 65.0 g of glucose dissolved in 135 g of water?of water?

Mass % = Mass % = 65.0 g / [65.0 + 135]g x 10065.0 g / [65.0 + 135]g x 100

= 32.5 % = 32.5 %

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Solution Concentration Concentration is expressed more importantly as Concentration is expressed more importantly as

molarity molarity (M)(M)..

Molarity (M) = Moles solute / LiterMolarity (M) = Moles solute / LiterSolutionSolution

An important relationship is An important relationship is M x VM x Vsolutionsolution= mol= mol This relationship can be used directly in mass This relationship can be used directly in mass

calculations of chemical reactions.calculations of chemical reactions. What is the molarity of a solution of 1.00 g KCl in What is the molarity of a solution of 1.00 g KCl in

75.0 mL of solution?75.0 mL of solution?M M KCl KCl == [1.00g [1.00g KClKCl / 75.0mL ][1mol / 75.0mL ][1molKClKCl / 74.55 g / 74.55 g KCl KCl ][ 1000mL / L]][ 1000mL / L]

= 0.18 mol= 0.18 molKClKCl / L / L

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Preparation of Solutions


Preparing a Standard Solution


QUESTIONHow many grams of NaCl are contained in 350. mL of a 0.250 M solution of sodium chloride? 1) 41.7 g 2) 5.11 g 3) 14.6 g 4) 87.5 g 5) None of these


Solution Concentration

The following formula can be used in dilution The following formula can be used in dilution calculations:calculations:

MM11VV1 1 = M= M22VV22

A concentrated stock solution is much easier to A concentrated stock solution is much easier to prepare and then dilute rather than preparing a prepare and then dilute rather than preparing a dilute solution directly. Concentrated sulfuric acid dilute solution directly. Concentrated sulfuric acid is 18.0M. What volume would be needed to is 18.0M. What volume would be needed to prepare 250.mL of a 1.50M solution? prepare 250.mL of a 1.50M solution?

VV1 1 = M= M22VV2 2 / M/ M1 1

VV1 1 = 1.50 M x 250. mL / 18.0 M = 1.50 M x 250. mL / 18.0 M

VV1 1 = 20.8 mL= 20.8 mL

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Solution Dilution


Solution ApplicationsSolution Applications

A solution of barium chloride was prepared by A solution of barium chloride was prepared by dissolving 26.0287 g in water to make 500.00 mL of dissolving 26.0287 g in water to make 500.00 mL of solution. What is the concentration of the barium solution. What is the concentration of the barium chloride solution? chloride solution? MMBaCl2BaCl2 = ? = ?

MMBaCl2BaCl2 =

= [26.0287g BaCl2 / 500.00mL ][1mol BaCl2 / 208.23g BaCl2 ] [1000mL / L]

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= 0.25000 mol / L


Solution ApplicationsSolution Applications

10.00 mL of this solution was diluted to make 10.00 mL of this solution was diluted to make exactly 250.00 mL of solution which was then exactly 250.00 mL of solution which was then used to react with a solution of potassium used to react with a solution of potassium sulfate. What is the concentration of the sulfate. What is the concentration of the diluted solution. diluted solution. MM22 = ? = ?

MMBaCl2BaCl2 = MM11

M2 = M1V1 / V2

M2 = 0.25000 M x 10.00 mL / 250.00 mL

M2 = 0.010000 M

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QUESTIONWhat volume of 18.0 M sulfuric acid must be used to prepare 15.5 L of 0.195 M H2SO4? 1) 168 mL 2) 0.336 L 3) 92.3 mL 4) 226 mL 5) None of these


Solution ApplicationsSolution Applications20.0020.00 mL of the barium chloride solution required mL of the barium chloride solution required

15.5015.50 mL of the potassium sulfate solution to mL of the potassium sulfate solution to react completely. react completely. MMK2SO4K2SO4 = ? = ?

BaClBaCl22(aq)(aq) + K + K22SOSO44(aq)(aq) ? + ? ? + ?

BaClBaCl22(aq)(aq) + K + K22SOSO44 (aq)(aq) 22 KCl KCl(aq)(aq) + BaSO + BaSO44 (s)(s)

??MMK2SO4K2SO4 = [MBaCl2x VBaCl2 / VK2SO4K2SO4 ] [? ] [? molmolK2SO4K2SO4 / ? / ? molmolBaCl2 BaCl2 ]]

0.010000 molBaCl2 x 0.02000 LBaCl2 x 1 1 molmolK2SO4K2SO4

LLBaCl2BaCl2 x 0.01550 0.01550 LLK2SO4K2SO4 x 1 1 molmolBaCl2 BaCl2

1

??MMK2SO4K2SO4 =

??MMK2SO4K2SO4 = 0.01290 0.01290 molK2SO4K2SO4 / L/ LK2SO4 K2SO4 = 0.01290 0.01290 MK2SO4K2SO4


Solution ApplicationsSolution ApplicationsHow many grams of potassium chloride are How many grams of potassium chloride are

produced?produced?

BaClBaCl22(aq)(aq) + K + K22SOSO44(aq)(aq) ? ? + ?+ ?

BaClBaCl22(aq)(aq) + K + K22SOSO44 (aq)(aq) 22 KCl KCl(aq)(aq) + BaSO + BaSO44 (s)(s)1

?g?gKClKCl = 0.010000 = 0.010000 molmolBaCl2 BaCl2 / L/ LBaCl2 BaCl2 x x 0.02000 0.02000 LLBaCl2BaCl2 X X 22 molmolKClKCl / / 11 molmolBaCl2 BaCl2 X X

74.5574.55 g gKCl KCl /mol/molKClKCl

11

= 0.029820.02982 ggKClKCl


Solution ApplicationsSolution ApplicationsIf If 20.0020.00 mL of a mL of a 0.100.10 M solution of barium chloride was M solution of barium chloride was

reacted with reacted with 15.0015.00 mL of a mL of a 0.200.20 M solution of M solution of potassium sulfate, what would be the theoretical potassium sulfate, what would be the theoretical yield of barium sulfate?yield of barium sulfate?

BaClBaCl22(aq)(aq) + K + K22SOSO44 (aq)(aq) 2 KCl 2 KCl(aq)(aq) + BaSO + BaSO44 (s) (s) 1

MolMolBaCl2BaCl2 = M = MBaCl2BaCl2x Vx VBaCl2BaCl2

== 0.100.10 molmolBaCl2 BaCl2 / L/ LBaCl2BaCl2 x x 0.020000.02000 LLBaCl2BaCl2

1 mol1 molBaCl2BaCl2

= = 2.0 x 102.0 x 10-3-3

MolMolK2SO4K2SO4= M= MK2SO4K2SO4x Vx VK2SO4K2SO4

== 0.200.20 molmolK2SO4 K2SO4 / L/ LK2SO4K2SO4 x x 0.015000.01500 LLK2SO4K2SO4

1 mol1 molK2SO4K2SO4

= = 3.0 x 103.0 x 10-3-3

Which is the Limiting Reagent?

2.0 x 10-3 < 3.0 x 10-3

2.0 x 10-3 mol is limiting


Solution ApplicationsSolution ApplicationsIf If 20.0020.00 mL of a mL of a 0.100.10 M solution of barium chloride was M solution of barium chloride was

reacted with reacted with 15.0015.00 mL of a mL of a 0.200.20 M solution of M solution of potassium sulfate, what would be the theoretical potassium sulfate, what would be the theoretical yield of barium sulfate?yield of barium sulfate?

BaClBaCl22(aq)(aq) + K + K22SOSO44 (aq)(aq) 2 KCl 2 KCl(aq)(aq) + BaSO + BaSO44 (s) (s) 1

Must use the limiting reagent:Must use the limiting reagent:

0.100.10 molmolBaCl2 BaCl2 x x 0.020000.02000 LLBaCl2 BaCl2 xx 1 mol1 molBaSO4 BaSO4 x 233.39 g x 233.39 g BaSO4BaSO4

LLBaCl2BaCl2 1 mol 1 molBaCl2 BaCl2 molmolBaSO4BaSO4

=

= 0.47 g


QUESTIONWhat mass of NaOH is required to react exactly with 25.0 mL of 1.2 M H2SO4? 1) 1.2 g 2) 1.8 g 3) 2.4 g 4) 3.5 g 5) None of these

© copyright 1995-2007 r.j. rusay aqueous solutions concentration / calculations dr. ron rusay fall...

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