It always seems impossible until it is done. Nelson Mandela
Idea Transcript
SCC-CH110/UCD-CH41C
Chapter: 16
Instructor: J.T.
P1
Solution: A solution is a homogeneous (uniform appearance) mixture. Solution States:
(I) Gas (i.e., Air) (II) Liquid (i.e., Ocean) (III) Solid (Ruby: Cr2O3/Al2O3)
Solvent (the large amount substance in solution) Solute (the small amount substance in solution)
+
→
Blue is solvent
+
→
Blue is solute
A solution is unsaturated when more solute will dissolve in it. A solution is saturated when no more solute will dissolve. Supersaturated Solutions: If the concentration of the solute is greater then the normal solubility limit of solvent. The solution is unstable.
SCC-CH110/UCD-CH41C
Chapter: 16
oil & water immiscible
Instructor: J.T.
P2
ethanol & water miscible
Solubility is a physical property that is associated with intermolecular forces. Like dissolves like: water/ methanol; hexane/ decane; …
Factors that determine Solubility: 1) Substances with similar intermolecular forces. 2) The greater the partial pressure * of a gas over a liquid solution. 3) Temperature: (solid/Liquid; gas/liquid) * The partial pressure of one gas in a mixture of gases is the pressure that one gas would exert if it alone occupied the same volume at the same temperature.
Ptotal = P1 + P2 + P3 + …
SCC-CH110/UCD-CH41C
Chapter: 16
Instructor: J.T.
A carbonated beverage bottle: Bubbles of CO2 escaping from the solution
PCO2 > Patm Carbo
Carbo
P3
SCC-CH110/UCD-CH41C
Chapter: 16
Instructor: J.T.
P4
Solution Concentration: How much solute is present per amount of solvent or solution. Mass ratio:
% by mass =
g solute × 100 g solution
Molarity:
M =
moles solute mol = liter solution L
Molality:
mol solute mol m= = Kg solvent Kg
===================== Example: How many grams of solute are in 65.0 g of a 13.0 % solution? Solution ≡ Solute 100 g ≡ 13.0 g 65.0 g ≡ ? 65.0 g solution ×
13.0 g solute = 8.45 g 100 solution
SCC-CH110/UCD-CH41C
Chapter: 16
Instructor: J.T.
P5
Example: Calculate the molarity of a solution made by dissoving 15 g of NaOH in water and diluting to 1.00 × 102 mL.
mol M = L mol =
mass 15 = = 0.38 mole molar mass 40.0
V = 1.00 ×10 2 mL
M =
1L = 0.100 L 1000 mL
0.38 mol = 3.8 M of NaOH 0.100 L
SCC-CH110/UCD-CH41C
Chapter: 16
Instructor: J.T.
P6
Example: Calculate the volume of concentrated ammonia solution, which is 15 molar, that contains 75.0 g of ammonia.
mol M = L mol L= = M
75.0
17.03 = 4.40 mol = 0.29 L NH OH 4 15 15 mol
Example: How many moles of solute are in 65.0 mL 2.20 M sodium hydroxide?
mol M = L
mol = M × L mol = 2.20 × 65.0 mL
1L mol = 2.20 × ( 65.0 mL ) = 0.143 mol NaOH 1000 mL
SCC-CH110/UCD-CH41C
Chapter: 16
Instructor: J.T.
Article I. Titration is an important lab operation in
Analytical chemistry. Titration can be used to standardize a solution. Buret shows the precise volume of reactant solution with known/Unknown concentration.
Flask contains a solution of a sample with known/unknown concentration.
A chemical indicator or solution color marks The end point of the titration.
Buret : ( M × V ) =
Flask : ( M × V )
mol M ×V = × L = mol L
M 1 . V1 = M 2 . V2
P7
SCC-CH110/UCD-CH41C
Chapter: 16
Instructor: J.T.
P8
Example: A chemist dissolves 1.18 g H2C2O4 . 2H2O (MM=126.07 g/mol) in water and titrates the solution with a solution of NaOH of unknown concentration. She determines that 28.3 mL NaOH (aq) is required to neutralize the acid. Calculate the molarity of the NaOH solution. H2C2O4(aq) + 2 NaOH(aq) → Na2C2O4(aq) + 2H2O
mole =
mass 1.18 = molar mass 126.07
acid 1 mol 9.36 × 10-3 mol ? = 9.36 ×10 −3 mol ×
Article II.
M =
1L = 1000 mL
2 mol 1 mol
mol 0.0187 = 28.3 L 1000
≡ ≡ ≡
= 9.36 × 10 −3
mol acid
base 2 mol ?
= 0.0187 mol NaOH
= 0.661 M
NaOH
SCC-CH110/UCD-CH41C
Chapter: 16
Instructor: J.T.
P9
Dilution of Concentrated Solution:
M C × VC = M d × Vd Example: A student adds 50.0 mL H2O to 25.0 mL 0.881 M NaOH. What is the concentration of the diluted solution?