Solution Chemistry

solutionof two or more substanceshomogeneous mix

solvent =solute =

substance present in major amountsubstance present in minor amount

gassolute = O2, Ar, CO2, etc.

air solvent = N2

solidsolute = C

steel solvent = Fe

liquid solvent = H2Osolute = salts, covalent compounds

H2O

O and H share electrons

separation of charge dipole

dipole moment

polar solvent

hydrogen bonds H-bondneed donor H-O

H-NH-F

acceptor ONF

but not equally

-

+

+

+

+

+

-

-

-

Aqueous solutionsNaClH2O

solvent solute

H-bondO-H+

Ion-ionNa+ Cl-

Ion-dipoleCl- H+

Na+ O-

solvation

NaCl (s) + H2O (l) Na+ (aq) + Cl- (aq)

Non-ionic solutions

C6H12O6glucosesolvent soluteH2O

H-bondO-H+

H-bondO-H+

C6H12O6 (s) + H2O (l) C6H12O6 (aq)

“Likes dissolve likes”

Non-ionic solutions

C8H18octanesolvent soluteH2O

H-bondO-H+

non-polar

C8H18 (l) + H2O (l) no reaction

Properties of aqueous solutions

ionic covalent

conduct electricity do not conduct electricityNaCl C6H12O6

electrolytes non-electrolytes

acids produce H+ in aqueous solutions

bases produce OH- in aqueous solutions

salts produce other anions and cations

produce ions

mobile, charged

Electrolytes

Rule Exceptions

1. Most acids are weak electrolytes HCl HBr HI

HNO3

H2SO4

HClO4

2. Most bases are weak electrolytesCa(OH)2 – Ba(OH)2

LiOH – CsOH

3. Most salts are strong electrolytes HgCl2

Hg(CN)2

Strong Electrolytes

dissociate completely form hydrated ions

HCl (g)

strong acids

strong basesNaOH (s) + H2O (l) Na+ (aq) + OH- (aq)

saltsMgSO4 (s) + H2O (l) Mg2+(aq) + SO4

2-(aq)

+ H2O (l) H+ (aq) + Cl- (aq)

Weak Electrolytes

do not dissociate completely

HF (g)

weak acids

+ H2O (l) + F- (aq)H+ (aq)

equilibrium all species present

NH3 (g)

weak bases

+ H2O (l) + OH- (aq)NH4+ (aq)

HgCl2 (s)

weak electrolytic salts

+ H2O (l) + 2Cl- (aq)Hg2+ (aq)

Non- Electrolytes

do not dissociate to form ions

CH3CH2OH (l)

+ H2O CH3CH2OH (aq)

Solution Compositionconcentration = amount of solute

volume of solution= mol

L= M

molarity

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 SO421 mol Na2SO4

142.0 g Na2SO4

= 0.165 mol Na2SO4

125 mL 1 L1000 mL

= .125 L M = 0.165 mol Na2SO40.125 L

= 1.32 M

[ ]

[Na2SO4] = 1.32 M

How many moles of HNO3 are present in 2.0 L of 0.200 M HNO3 solution?

0.200 mol HNO3

L

2.0 L = 0.40 mol HNO3

Solution Compositionconcentration = amount of solute

volume of solution= mol

L= M[ ]

How many grams of Na2SO4 are requiredto make 350 mL of 0.500 M Na2SO4?

0.500 mol Na2SO4

L

0.350 L1 mol Na2SO4

142.0 g = 24.9 g Na2SO4

Solution Compositionconcentration = amount of solute

volume of solution= mol

L= M[ ]

stock solution HCl = 12.0 Mmoles solute before dilution = moles solute after dilution

How would you prepare 1.5 L of a 0.10 M HCl solution?

0.10 mol HClL

1.5 L = 0.15 mol HCl

0.15 mol HCl =

moles after dilution

moles before dilution12.0 mol HCl L

= 0.0125 L

12.5 mL of 12.0 M HCl + 1.4875 L H2O = 1.50 L 0.10 M HCl

L

(x)

Solution Compositionconcentration = amount of solute

volume of solution= mol

L= M[ ]

(x)

How would you prepare 1.5 L of a 0.10 M HCl solution, usinga 12.0 M stock solution?

Mi x Vi = Mf x Vf

12.0 M HCl x Vi 0.10 M HCl x 1.5 L

Vi = 0.0125 L

then add H2O to get to Vf

=

moles of solute after dilution moles of solute before dilution =

(mol/L) (L)

=1.50 L H2O