1. Unit 13: Solutions

2. Solvation - Definitions
Solution - homogeneous mixture
Solute - substance being dissolved
Solvent – substance that dissolves the solute

3. A. Definitions
Solute - KMnO4
Solvent - H2O

4. B. Solvation First... Then... Solvation – the process of dissolving
solute particles are surrounded by solvent particles
First...
solute particles are separated and pulled into solution
Then...

5. B. Solvation Non- Electrolyte Weak Electrolyte Strong Electrolyte+
sugar - +
acetic acid - +
salt
Non-
Electrolyte
Weak
Electrolyte
Strong
Electrolyte
solute exists as
molecules only
solute exists as
ions and
molecules
solute exists as
ions only
DISSOCIATION
IONIZATION
View animation online.

6. B. Solvation
“Like Dissolves Like”
NONPOLAR
POLAR

7. B. Solvation Soap/Detergent polar “head” with long nonpolar “tail”
dissolves nonpolar grease in polar water

8. Solubility Unsaturated
When the solvent holds less solute than it normally can at a given temperature.

9. Solubility
Saturated
When the solvent holds as much of a solute as it normally can at a given temperature.

10. Solubility Supersaturated
When the solvent holds more dissolved solute than it normally can at that temperature.

11. C. Solubility UNSATURATED SOLUTION more solute dissolves
no more solute dissolves
SUPERSATURATED SOLUTION
becomes unstable, crystals form
concentration

12. C. Solubility Solubility
maximum grams of solute that will dissolve in 100 g of solvent at a given temperature
varies with temp
based on a saturated soln

13. C. Solubility Solubility Curve
shows the dependence of solubility on temperature

14. C. Solubility Solids are more soluble at...
high temperatures.
Gases are more soluble at...
low temperatures &
high pressures (Henry’s Law).
EX: nitrogen narcosis, the “bends,” soda

15. Solubility Curve

16. Things to remember The graph is set in 100 mL or 100 grams of water.
The line represents the saturation point at that temperature.
Anything above the line (at that temp) is supersaturated
Anything below the line (at that temp) is unsaturated.

17. Rules to solubility problems
1. Look up the amount of water given in the problem. (highlight it)
If it is 100 mL or 100 grams, look the answer up on the graph.

18. Example
What amount of NaCl would make a saturated solution in 100 mL of water at 800C?
Just look it up on the graph.
Answer : 40 grams
GRAPH

20. What if it is not 100 mL or 100 grams of water?
Set up a proportion:
solid solid =
liquid liquid
This will be x
From graph
From problem
Always
100 mL

21. Example
What amount of KNO3 would make a saturated solution in 177mL of water at 500C?
60 g x =
100mL mL
= g
GRAPH

22. What if there is no amount of water given?
Set up a proportion:
solid solid =
liquid liquid
From graph
From problem
Always
100 mL
This will be x

23. Example
What amount of water at 200C would make a saturated solution with 63g of KNO3?
20 g 63 g =
100mL X
= 315 mL
GRAPH

24. Stuff from yesterday a. Rate of Solubility depends upon
1.the size of the solute crystals
2.the vigor and duration of stirring
3.the temperature of the solvent
b. Substances may be soluble or insoluble 1. refers to substance than be dissolved into water
2. insoluble refers to substances which are only very, very slightly soluble in water (ex. Chalk is insoluble in water)

25. Precipitation Reactions: What are they?
Type of double replacement reaction.
Two solutions of ionic compounds are mixed
One of the products of the reaction is an insoluble salt called a precipitate (chunky milk)
Use solubility rules to determine the precipitate.

26. Solubility Rules Solubility (in water) Compounds Exceptions Soluble
Alkali metal (Group 1) salts
none
Ammonium salts (NH4+)
Acetates (C2H3O2 -)
Nitrates (NO3 -)
Chlorides (Cl -), Bromides (Br -), Iodides (I -)
Compounds of Ag, Hg, and Pb
Sulfates (SO4 2-)
Compounds of Sr, Ba, Hg, and Pb
**********
***************************
Insoluble
Carbonates (CO3 2-), Phosphates (PO4 3-), Sulfites (SO3 2-), Chromates (CrO42-)
Compounds of Alkali metal (Group 1) and NH4+
Sulfides (S 2-), Hydroxides (OH -)
Compounds of Alkali metal (Group 1) and NH4+, Ca, Sr, Ba

27. Example using chart Insoluble Pb(OH)2 NH4Cl K2SO4 Soluble Ba3(PO4)2
RULES

28. How to use it.
Write the products for the double replacement reactions and balance.
Then, using the solubility rules
identify the precipitate by a subscript (s) for solid (insoluble in water).
Identify the soluble compounds by marking them with a subscript (aq) for aqueous (soluble in water).

29. Example Ag(NO3)2 + LiCl  Ag(NO3)2 + 2 LiCl  AgCl2 + 2 LiNO3 AgCl2+1 -1 +2 -1
AgCl2 +
LiNO3 aq aq s aq
RULES

30. Practice Problems 1) Na2CO3 + Ca(NO3)2  2) Pb(C2H3O2)2 + Na2S 
3) K3PO4 + MgCl2 
4) (NH4)2SO4 + BaI2 

31. Practice #1 Na2CO3 + Ca(NO3)2  Na2CO3 + Ca(NO3)2 2NaNO3 +CaCO3 NaNO3+2 -2 -1 +1
NaNO3 +
CaCO3 aq aq s aq
RULES

32. Practice #2 Pb(C2H3O2)2 + Na2S  PbS NaC2H3O2-1 +1 +2 -2
PbS +
NaC2H3O2 s aq aq aq
RULES

33. Practice #3 K3PO4 + MgCl2  2K3PO4 + 3MgCl2 6 KCl + Mg3(PO4)2 KCl+2 -1 -3 +1
KCl +
Mg3(PO4)2 aq aq aq s
RULES

34. Practice #4 (NH4)2SO4 + BaI2  (NH4)2SO4 + BaI2  2 NH4I + BaSO4 NH4I+2 -1
NH4I +
BaSO4 +1 -2 aq aq s aq
RULES

35. Practice Problems
1) Na2CO3 (aq) + Ca(NO3)2(aq) 2 NaNO3 (S)+ CaCO3(aq)
2) Pb(C2H3O2)2 + Na2S PbS + 2NaC2H3O2
3) 2K3PO4 + 3MgCl2 6 KCl + Mg3(PO4)2
4) (NH4)2SO4 + BaI2  2 NH4I + BaSO4

36. A. Concentration The amount of solute in a solution.
Describing Concentration
% by mass - medicated creams
% by volume - rubbing alcohol
ppm, ppb - water contaminants
molarity - used by chemists
molality - used by chemists

37. Percent Concentration
Percent concentration is
solid
(solid + liquid)
(solid + liquid) x 1000 = ppt
(solid + liquid) x = ppm x
100 = pph

38. Example
What is the percent concentration (pph) of NaCl when 42 g are dissolved in 379g of water?
42g
(42g + 379g) x =
421
9.98% 0r 9.98 pph

39. B. Molarity
solvent only
1000mL = 1 L

40. B. Molarity
Find the Molarity of a solution containing .5 mol of glucose in 2 L of water.
GIVEN:
M = ?
Mol = .5 mol
V = 2 L
WORK:
M= mol solute/ liters solvent
M= .5 mol
2 L
= .25 M

41. C. Dilution
Preparation of a desired solution by adding water to a concentrate.
Moles of solute remain the same.

42. C. Dilution
What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution?
GIVEN:
M1 = 15.8M
V1 = ?
M2 = 6.0M
V2 = 250 mL
WORK:
M1 V1 = M2 V2
(15.8M) V1 = (6.0M)(250mL)
V1 = 95 mL of 15.8M HNO3

43. D. Preparing Solutions 1.54m NaCl in 0.500 kg of water
500 mL of 1.54M NaCl
mass 45.0 g of NaCl
add water until total volume is 500 mL
mass 45.0 g of NaCl
add kg of water
500 mL water
45.0 g NaCl
500 mL
mark
volumetric
flask