1. Solutions Chapter 14

2. solution
Homogeneous mixture of 2 or more substances in a single physical state
particles in a solution are very small
particles in a solution are evenly distributed
particles in a solution will not separate

3. solute
The substance that is dissolved
examples: sugar, salt

4. solvent Aqueous solutions-use water as solvent
Substance that does the dissolving
example: water, ethanol
Aqueous solutions-use water as solvent

5. Like dissolves like
A solute will dissolve best in a solvent with similar intermolecular forces.
If the intermolecular forces are too different the solute will not dissolve in that solvent.

6. Calculating the strength of a solution
Often the strength of a solution can be expressed in terms of percent.

7. Percent Solutions can be calculated 2 ways.
% by volume
This compares the volume of solute to the total volume of solution.
% by mass
This compares the mass of solute to the total mass of solution.
Percent Solutions
The percent composition of a solution can be expressed two possible ways.
% by volume or mass
Let’s first discuss percent by volume.
Percent by volume ( % (v/v)) =
Volume of solute / Volume of solution x 100%

8. Volume Percent Volume of solute present in a total volume of solution.
Volume Percent (v/v) = volume of solute
/ volume of solution x 100%

9. Calculating volume percent
A solution is prepared by dissolving 36 ml of ethanol in water to a final volume of
150 ml what is the solution’s volume percent?
% (v/v) ethanol = 36 ml ethanol / 150 ml total x 100%
Volume Percent ethanol = 24 %

10. Volume percent
If 15.0ml of acetone is diluted to 500ml with water what is the % (v/v) of the prepared solution?
% (v/v) = 15.0ml / 500ml x 100%
% v/v = 3.0% acetone

11. Mass percent Way to describe solutions composition
mass of solute present in given mass of solution
mass percent = mass of solute
mass of solution
grams of solute
grams of solute + grams of solvent
X 100
X 100

12. Mass percent
A solution is prepared by dissolving 1.0g of sodium chloride in 48 g of water. The solution has a mass of 49 g, and there is 1.0g of solute (NaCl) present. Find the mass percent of solute.

13. Mass Percent
A solution is prepared by mixing 1.00g of ethanol, C2H5OH, with g of water. Calculate the mass percent of ethanol in this solution.

14. Solubility The extent to which a solute will dissolve
expressed in grams of solute per 100g of solvent
‘likes dissolve likes’

15. Not every substance dissolves in every other substance
soluble- capable of being dissolved
salt
insoluble- does not dissolve in another
oil does not dissolve in water

16. Solubility & liquids
Miscible- two liquids that dissolve in each other completely
immiscible- liquids that are insoluble in one another
oil & vinegar

17. The compositions of the solvent and solute will determine if the substance will dissolve
stirring
temperature
surface area of the dissolving particles

18. A solution is prepared by mixing 2
A solution is prepared by mixing 2.8 g of sodium chloride with 100 g of water. What is the mass percent of NaCl?
What is the volume percent alcohol when you add sufficient water to 700mL of isopropyl alcohol to obtain 1000mL of solution?

19. saturated solution
contains the maximum amount of solute for a given quantity of solvent
no more solute will dissolve

20. unsaturated solution contains less solute than a saturated solution
could use more

21. Supersaturated solution
Solution contains more solute than it can ‘hold’
too much

22. Dilute solution- contains a small amount of solute
Concentrated solution- contains large amount of solute

23. Solubility
Table salt: at room temperature, 37.7 g can be dissolved in 100 ml of H2O
Sugar: at room temperature, 200 g can be dissolved in 100 ml of H2O

24. Solubility Curve

25. Solubility curve
Determines solubility of substances at specific temperatures
with raising temperature solids increase in solubility
with increase in temperature gases decrease in solubility
ex: fish die

26. on the line- saturated (can not hold anymore) above the line- supersaturated (holding more than it can) below the line- unsaturated (can hold more solute)
supersaturated
Solute (g) per g H2O
saturated
unsaturated
temperature

27. 92 g of NaNO3 are added to 100ml of water at 25°C and mixed
92 g of NaNO3 are added to 100ml of water at 25°C and mixed. What type of solution is it?
80 g of NaNO3 are added to 100ml of water at 25°C and mixed. What type of solution is it?

28. What is the solubility of NaNO3 in 100g of H2O at 20°C?

29. Concentration of solutions
Concentration of a solution is the amount of solute in a given amount of solvent
most common measurements of concentration are:
molarity
(mole fraction)

30. Concentration of solutions
Concentration of a solution is the amount of solute in a given amount of solvent
most common measurements of concentration are:
molarity
(mole fraction)

31. Molarity Number of moles of solute per volume of solution in liters
molarity (M) =
liters of solution
mol L

32. Molarity
Calculate the molarity of a solution prepared by dissolving 11.5g of solid NaOH in enough water to make 1.50 L of solution.
Given:
mass of solute = 11.5 g NaOH
vol of solution = 1.50 L
molarity is moles of solute per liters of solution

33. Molarity
Convert mass of solute to moles (using molar mass of NaOH). Then we can divide by volume
molar mass of solute = 40.0 g
11.5 g NaOH x 1 mol NaOH
40.0 g NaOH
0.288 mol NaOH
1.50 L solution
= mol NaOH
= M NaOH

34. Molarity
Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl into enough water to make 26.8 mL of solution.
Given: mass of solute (HCl) = 1.56 g
volume of solution = 26.8 mL

35. Molarity Molarity is moles per liters
we have to change 1.56 g HCl to moles HCl and then change 26.8 mL to liters
molar mass of HCl = 36.5 g
1.56 g HCl x 1 mol HCl
36.5 g HCl
= mol HCl
= 4.27 x 10-2 mol HCl

36. Molarity
Change the volume from mL to liters
1 L = 1000 mL
26.8 mL x 1 L
1000 mL
= L
= 2.68 x 10-2 L

37. molarity
Finally, divide the moles of solute by the liters of solution
molarity = x 10-2 mol HCl
2.68 x 10-2 L
= 1.59 M HCl

38. molarity
Calculate the molarity of a solution prepared by dissolving 1.00 g of ethanol, C2H5OH, in enough water to give a final volume of 101 mL.

39. molarity
One saline solution contains 0.90 g NaCl in exactly 1.0 L of solution. What is the molarity of the solution?

40. molarity
A solution has a volume of 250 mL and contains 0.70 mol NaCl. What is its molarity?

41. dilution Diluting a solution:
reduces the number of moles of solute per unit volume
the total number of moles of solute in solution does not change

42. Diluting solutions M1V1 = M2V2
M1 = molarity of stock solution (initial)
V1 = volume of stock solution (initial)
M2 = molarity of dilute solution
V2 = volume of dilute solution

43. M1V1 = M2V2
How many milliliters of aqueous 2.00M MgSO4 solution must be diluted with water to prepare mL of aqueous 0.400M MgSO4?
M1 = 2.00M MgSO4
M2 = 0.400M MgSO4
V2 = mL MgSO4
V1 = ?

44. M1V1 = M2V2 Solve for V1 V1 = M2 x V2 M1 0.400M x 100.00 mL 2.00M

45. M1V1 = M2V2
How many milliliters of a solution of 4.00M KI are needed to prepare L of 0.760M KI?