Solution Stoichiometry and Types of Reactions...4.4. Types of Chemical Reactions • Molecular...

Solution Stoichiometry and Types of Reactions

Chapter 4

Copyright © Houghton Mifflin Company. All rights reserved. 4 | 2

4.1 Water – A Most Versatile Solvent

• A soluteconsists of atoms, molecules or ions that dissolve in

the solvent.• A solvent

is the substance that dissolves the solute.• A solution

is a homogenous mixture made by dissolving a solute in a solvent.

• If a solute is dissolved in water, it forms an aqueous solution


Solution


Hydration Process

• Water molecules surround the ions.

• Note the positions of the positive and negative charges.


Electrolytes

• Strong Electrolytes– Good conductors of electricity in solution.– Strong acids, strong bases, soluble ionic compounds.– Completely ionized in solution.

• Weak Electrolytes– Poor conductors of electricity in solution.– Partially ionized in solution. CH3COOH, NH3

• Non-electrolytes– Do not conduct electricity in solution.– Do not form ions in solution.

: C6H12O6,


Electrolytes

• A strong electrolyte is completely dissociated in water.KCl(s) KCl(aq) K+(aq) + Cl−(aq)

• A weak electrolyte is partiallydissociated in water.NH3(aq) + H2O(l) NH4

+(aq) + OH−(aq)

• Non-electrolytes do not form ions.C12H22O11(s) C12H22O11(aq)

MSOffice4


Electrolytes


Sample Problem

Classify each of the following compounds as strong electrolyte, weak electrolyte or non-electrolyte.

• CaCl2• C6H12O6 (glucose)

• HF

• HNO3


4.2. The Concentration of Solutions

• The concentration of a solution is the amount of solute per volume of solution.

moles of soluteMolarity ( ) = liter of solution

M


Sample Problem

What is the molarity of 100 mL solution containing 15.6g of NaOH?

( ) 1 mol NaOH15.6g NaOH 0.390 mol NaOH40.0g NaOH

0.390 mol NaOH 3.90 NaOH1 L100 mL

1000 mL

=

=

M = M


Sample Problem

How many moles of HCl are contained in 250 mL of a 0.600 molar solution?

( ) =

1 L 0.600 mol HCl250. mL 0.150 mol HCl1000 mL 1 L


Other Concentration Units


Other Concentration Units

ppm ppb ppt


Sample Problem

What is the molarity of a solution that is 4.0 ppb Cr6+?

6+ 6+8 6+

6 6+

4.0 g4.0 ppbL soln.

4.0 g 1 g Cr 1mol Cr 7.7 10 CrL soln. 10 g 52.0 gCr

−

µ=

µ = × µ M


Diluting Solutions

moles in the concentrated solution = moles in the dilute solution

Cinitial × Vintial = Cfinal × Vfinal


Sample Problem

How many mL of 15.0 M nitric acid solution should be dilute to make 250. mL of a 0.50 M solution?

×=

×=

=

final finalinitial

initial

C VVC

0.50 250. mL15.0

8.3 mL

MM


4.3. Stoichiometric Analysis of Solutions

TitrationControlled addition of a solution of known

concentration to react with a solution of unknown concentration.

Equivalence pointThe point when all reactants are completely

consumed.End point

Close to the equivalence point marked by an indicator.


Titration


4.4. Types of Chemical Reactions

• Molecular equation (components written as compounds):

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

• Complete Ionic Equation (electrolytes written as ions):Ag+(aq) + NO3

−(aq) + Na+(aq) + Cl−(aq) →AgCl(s) + Na+(aq) + NO3

−(aq)

• Net ionic equation (ions participating in the reactionAg+(aq) + Cl−(aq) → AgCl(s)

(Na+ and NO3− are spectator ions.)


4.5. Precipitation Reactions


Precipitation Reactions


Solubility Rules

• How can a precipitation reaction be predicted ?


Sample Problem

How much AgCl will be formed from mixing 1.50 L of 0.500 M AgNO3 with 1.75 L of 0.300 M NaCl?

( )

( )

+

-

0.500 mol Ag1.50 L 0.750 mol Ag1L

0.300 mol Cl1.75 L 0.525 mol Cl

+

−

=

=


Sample Problem (cont)

The net ionic equation is:

Ag+(aq) + Cl−(aq) → AgCl(s)

Thus Cl- is the limiting reactant and 0.525 mol AgCl will be formed. In grams,

( ) 143.4g0.525 mol AgCl 75.3g AgCl1mol

=


4.6. Acid-Base Reactions


Acid-Base Reactions

Drain decloggers –contain s.base


Acid-Base Reactions

• Neutralization reaction for strong acids and strong bases:

H+(aq) + OH-(aq) → H2O(l)


Sample Problem

What volume of 0.200 M HCl solution is need to neutralize 25.0 mL of a 0.450 MKOH solution?

The balanced molecular equation is:

HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)


Sample Problem (cont)

( )

( )

-

-

- +

++

0.450 mol KOH 1mol OH 1 L25.0 mL1 L 1 mol KOH 1000 mL

0.01125 mol OHat neutralization, mol OH mol H

1mol HCl 1 L 1000 mL0.01125 mol H1 mol H 0.200 mol HCL 1 L

56.3 mL HCl

=

=

=


4.7. Oxidation-Reduction Reactions

Zn(s) + Cu2+(aq)→ Zn2+(aq) + Cu(s)

oxidation reaction (loss of electrons)

Zn → Zn2+ (aq) + 2e-

reduction reaction (gain of electrons)

Cu2+ (aq) + 2e- → Cu


Sample Problem

Assign oxidation numbers to all atoms in the following:

CO2, K2Cr2O7, PCl5

CO2 – oxygen is -2×2 = -4, making C +4.

K2Cr2O7 – oxygen -2×7 = -14.

K (group 1A) is +1×2 = +2

2Cr must equal +12 = -14+2

each Cr = +6

PCl5 – Cl (group VIIA) = -1×5 = -5

P = +5


Zinc and Iodine


Dry Ice and Magnesium


Reaction of Phosphorous

Solution Stoichiometry and Types of Reactions...4.4. Types of Chemical Reactions • Molecular...

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