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Solution Chemistry - PowerPoint PPT Presentation


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Solution Chemistry. solution. homogeneous mix. of two or more substances. solvent =. substance present in major amount. substance present in minor amount. solute =. gas . air . solvent = N 2. solute = O 2 , Ar, CO 2 , etc. solid . steel . solvent = Fe. solute = C.

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slide1

Solution Chemistry

solution

homogeneous mix

of two or more substances

solvent =

substance present in major amount

substance present in minor amount

solute =

gas

air

solvent = N2

solute = O2, Ar, CO2, etc.

solid

steel

solvent = Fe

solute = C

liquid

solvent = H2O

solute = salts, covalent compounds

slide2

+

+

H2O

-

but not equally

O and H share electrons

separation of charge

dipole

dipole moment

polar solvent

hydrogen bonds

H-bond

need donor

H-O

acceptor

O

-

+

H-N

N

+

-

H-F

F

+

-

slide3

solvent

Aqueous solutions

solute

H2O

NaCl

H-bond

Ion-dipole

Ion-ion

Cl-

Na+

Cl-

H+

O-

H+

Na+

O-

solvation

Na+ (aq)

+ Cl- (aq)

+ H2O (l)

NaCl (s)

slide4

Non-ionic solutions

solvent

solute

H2O

glucose

C6H12O6

H-bond

H-bond

H+

O-

H+

O-

C6H12O6 (aq)

+ H2O (l)

C6H12O6 (s)

“Likes dissolve likes”

slide5

Non-ionic solutions

solvent

solute

H2O

octane

C8H18

non-polar

H-bond

H+

O-

no reaction

+ H2O (l)

C8H18 (l)

slide6

Properties of aqueous solutions

ionic

covalent

conduct electricity

do not conduct electricity

NaCl

C6H12O6

electrolytes

produce ions

non-electrolytes

mobile, charged

salts

produce other anions and cations

bases

produce OH- in aqueous solutions

produce H+ in aqueous solutions

acids

slide7

Electrolytes

Rule

Exceptions

HBr

HI

1. Most acids are weak electrolytes

HCl

HNO3

H2SO4

HClO4

2. Most bases are weak electrolytes

LiOH – CsOH

Ca(OH)2 – Ba(OH)2

3. Most salts are strong electrolytes

HgCl2

Hg(CN)2

slide8

Strong Electrolytes

dissociate completely

form hydrated ions

strong acids

 H+ (aq)

+ Cl- (aq)

HCl (g)

+ H2O (l)

strong bases

NaOH (s) + H2O (l)

Na+ (aq) + OH- (aq)

salts

MgSO4 (s) + H2O (l)

Mg2+(aq) + SO42-(aq)

slide9

Weak Electrolytes

do not dissociate completely

weak acids

equilibrium

all species present

H+ (aq)

+ F- (aq)

HF (g)

+ H2O (l)

weak bases

NH4+ (aq)

+ OH- (aq)

NH3 (g)

+ H2O (l)

weak electrolytic salts

Hg2+ (aq)

+ 2Cl- (aq)

HgCl2 (s)

+ H2O (l)

slide10

Non- Electrolytes

do not dissociate to form ions

 CH3CH2OH (aq)

+ H2O

CH3CH2OH (l)

slide11

Solution Composition

molarity

= mol

= M

[ ]

concentration

= amount of solute

volume of solution

L

What is the molarity of a solution prepared

by dissolving 23.4 g sodium sulfate in

enough water to give 125 mL of solution?

23.4 g Na SO4

1 mol Na2SO4

= 0.165 mol Na2SO4

2

142.0 g Na2SO4

125 mL

1 L

= .125 L

M =

0.165 mol Na2SO4

= 1.32 M

1000 mL

0.125 L

[Na2SO4]

= 1.32 M

slide12

Solution Composition

= mol

= M

[ ]

concentration

= amount of solute

volume of solution

L

How many moles of HNO3 are present

in 2.0 L of 0.200 M HNO3 solution?

2.0 L

= 0.40 mol HNO3

0.200 mol HNO3

L

slide13

Solution Composition

= mol

= M

[ ]

concentration

= amount of solute

volume of solution

L

How many grams of Na2SO4 are required

to make 350 mL of 0.500 M Na2SO4?

0.500 mol Na2SO4

= 24.9 g Na2SO4

0.350 L

142.0 g

1 mol Na2SO4

L

slide14

Solution Composition

= mol

= M

[ ]

concentration

= amount of solute

volume of solution

L

stock solution

HCl = 12.0 M

moles solute before dilution

= moles solute after dilution

How would you prepare 1.5 L

of a 0.10 M HCl solution?

0.10 mol HCl

1.5 L

= 0.15 mol HCl

moles after dilution

L

0.15 mol HCl =

12.0 mol HCl

(x)

L

moles before dilution

L

(x)

= 0.0125 L

12.5 mL of 12.0 M HCl

= 1.50 L 0.10 M HCl

+ 1.4875 L H2O

slide15

How would you prepare 1.5 L

of a 0.10 M HCl solution, using

a 12.0 M stock solution?

moles of solute before dilution =

moles of solute after dilution

Mi

x Vi

= Mf

x Vf

(mol/L)

(L)

x Vi

=

12.0 M HCl

0.10 M HCl

x 1.5 L

Vi =

0.0125 L

then add H2O to get to Vf

=1.37 L H2O