1. Chapter 7 Solutions 7.1 Solutions
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2. Solute and Solvent Solutions
are homogeneous mixtures of two or more substances.
consist of a solvent and one or more solutes.

3. Nature of Solutes in Solutions
spread evenly throughout the solution.
cannot be separated by filtration.
can be separated by evaporation.
are not visible, but can give a color to the solution.
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4. Examples of Solutions
The solute and solvent in a solution can be a solid, liquid, and/or a gas.
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5. Learning Check A. 2 g of sugar and 100 mL of water
Identify the solute in each of the following solutions.
A. 2 g of sugar and 100 mL of water
B mL of ethyl alcohol and 30.0 mL of
methyl alcohol
C mL of water and 1.50 g of NaCl
D. Air: 200 mL of O2 and 800 mL of N2

6. Water Water is the most common solvent. is a polar molecule.
forms hydrogen bonds between the hydrogen atom in one molecule and the oxygen atom in a different water molecule.
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7. Formation of a Solution
Na+ and Cl- ions
on the surface of a NaCl crystal are attracted to polar water molecules.
are hydrated in solution with many H2O molecules surrounding each ion.
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8. Equations for Solution Formation
When NaCl(s) dissolves in water, the reaction can be written as
H2O
NaCl(s) Na+(aq) + Cl- (aq)
solid separation of ions

9. Learning Check
Solid LiCl is added to water. It dissolves because A. the Li+ ions are attracted to the 1) oxygen atom ( -) of water. 2) hydrogen atom (+) of water. B. the Cl- ions are attracted to the

10. Like Dissolves Like Two substances form a solution
when there is an attraction between the particles of the solute and solvent.
when a polar solvent such as water dissolves polar solutes such as sugar, and ionic solutes such as NaCl.
when a nonpolar solvent such as hexane (C6H14) dissolves nonpolar solutes such as oil or grease.

11. Water and a Polar Solute

12. Like Dissolves Like Water (polar) Ni(NO3)2 CH2Cl2(nonpolar) (polar)
Solvents Solutes
Water (polar)
Ni(NO3)2
CH2Cl2(nonpolar) (polar)
I2 (nonpolar)

13. Learning Check 1) Na2SO4 2) gasoline (nonpolar) 3) I2 4) HCl
Which of the following solutes will dissolve in water? Why?
1) Na2SO4
2) gasoline (nonpolar)
3) I2
4) HCl

14. 7.2 Electrolytes and Nonelectrolytes

15. Solutes and Ionic Charge
In water,
strong electrolytes produce ions and conduct an electric current.
weak electrolytes produce a few ions.
nonelectrolytes do not produce ions.

16. Strong Electrolytes Strong electrolytes
dissociate in water, producing positive and negative ions.
conduct an electric current in water.
in equations show the formation of ions in aqueous(aq) solutions.
H2O % ions
NaCl(s) Na+(aq) + Cl−(aq)
H2O
CaBr2(s) Ca2+(aq) + 2Br−(aq)

17. Learning Check
Complete each for strong electrolytes in water. H2O A. CaCl2(s) 1) CaCl2(s) 2) Ca2+(aq) + Cl2−(aq) 3) Ca2+(aq) + 2Cl−(aq) B. K3PO4(s) 1) 3K+(aq) + PO43−(aq) 2) K3PO4(s) 3) K3+(aq) + P3−(aq) + O4−(aq)

18. Weak Electrolytes dissociates only slightly in water.
A weak electrolyte
dissociates only slightly in water.
in water forms a solution of a few ions and mostly undissociated molecules.
HF(g) H2O(l) H3O+(aq) + F-(aq)
NH3(g) + H2O(l) NH4+(aq) + OH-(aq)

19. Nonelectrolytes Nonelectrolytes dissolve as molecules in water.
do not produce ions in water.
do not conduct an electric current.

20. Equivalents
An equivalent (Eq) is the amount of an electrolyte or an ion that provides 1 mole of electrical charge (+ or -).
1 mole of Na+ = 1 equivalent
1 mole of Cl− = 1 equivalent
1 mole of Ca2+ = 2 equivalents
1 mole of Fe3+ = equivalents

21. Electrolytes in Body Fluids
In replacement solutions for body fluids, the electrolytes
are given in milliequivalents per liter (mEq/L).
Ringer’s Solution
Na mEq/L
Cl− mEq/L
K mEq/L
Ca mEq/L
The milliequivalents per liter of cations must equal the milliequivalents per liter of anions.

22. Electrolytes in Body Fluids

23. Learning Check A. In 1 mole of Fe3+, there are
1) 1 Eq. 2) 2 Eq ) 3 Eq.
B. In 2.5 moles of SO42−, there are
1) 2.5 Eq. 2) 5.0 Eq ) 1.0 Eq.
C. An IV bottle contains NaCl. If the Na+ is
34 mEq/L, the Cl− is
1) 34 mEq/L. 2) 0 mEq/L ) 68 mEq/L.

24. Chapter Solutions
7.3
Solubility

25. Solubility Solubility is
the maximum amount of solute that dissolves in a specific amount of solvent.
expressed as grams of solute in 100 grams of solvent, usually water.
g of solute
100 g water

26. Unsaturated Solutions
contain less than the maximum amount of solute.
can dissolve more solute.

27. Saturated Solutions Saturated solutions
contain the maximum amount of solute that can dissolve.
have undissolved solute at the bottom of the container.

28. Learning Check At 40 C, the solubility of KBr is 80 g/100 g of H2O.
Identify the following solutions as either
1) saturated or 2) unsaturated. Explain.
A. 60 g KBr added to 100 g of water at 40 C.
B g KBr added to 200 g of water at 40 C.
C. 25 g KBr added to 50 g of water at 40 C.

29. Effect of Temperature on Solubility
depends on temperature.
of most solids increases as temperature increases.
of gases decreases as temperature increases.

30. Solubility and Pressure
Henry’s law states
the solubility of a gas in a liquid is directly related to the pressure of that gas above the liquid.
at higher pressures, more gas molecules dissolve in the liquid.

31. Percent Concentration
Chapter 7 Solutions
7.4
Percent Concentration

32. Percent Concentration
The concentration of a solution
is the amount of solute dissolved in a specific amount of solution.
amount of solute
amount of solution

33. Mass Percent Mass percent (% m/m) is the
concentration by mass of solute in a solution.
mass percent = g of solute x 100
g of solute + g of solvent
amount in g of solute in 100 g of solution.
mass percent = g of solute a
100 g of solution

34. Mass of Solution Add water to give 50.00 g of solution 8.00 g KCl
50.00 g of KCl solution

35. Calculating Mass Percent
The calculation of mass percent (% m/m) requires the
grams of solute (g KCl) and
grams of solution (g KCl solution).
g of KCl = g
g of solvent (water) = g
g of KCl solution = g
8.00 g KCl (solute) x 100 = 16.0% (m/m)
50.00 g KCl solution

36. Learning Check 1) 15.0% (m/m) Na2CO3 2) 6.38% (m/m) Na2CO3
A solution is prepared by mixing 15.0 g of Na2CO3 and 235 g of H2O. Calculate the mass percent (% m/m) of the solution.
1) 15.0% (m/m) Na2CO3
2) 6.38% (m/m) Na2CO3
3) 6.00% (m/m) Na2CO3

37. Volume Percent The volume percent (% v/v) is
percent volume (mL) of solute (liquid) to volume (mL) of solution.
volume % (v/v) = mL of solute x mL of solution
solute (mL) in 100 mL of solution.
volume % (v/v) = mL of solute mL of solution

38. Mass/Volume Percent The mass/volume percent (% m/v) is
percent mass (g) of solute to volume (mL) of solution.
mass/volume % (m/v) = g of solute x mL of solution
solute (g) in 100 mL of solution.
mass/volume % (m/v) = g of solute mL of solution

39. Percent Conversion Factors
Two conversion factors can be written for each type of % value.
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40. Using Percent Concentration (m/m) as Conversion Factors
How many grams of NaCl are needed to prepare 225 g of a 10.0% (m/m) NaCl solution? STEP 1: Given: 225 g solution; 10.0% (m/m) NaCl Need: g of NaCl STEP 2: g solution g NaCl STEP 3: Write the 10.0% (m/m) as conversion factors g NaCl and 100 g solution 100 g solution 10.0 g NaCl STEP 4: Set up using the factor that cancels g solution.

41. Learning Check 1) 2.56 mL 2) 12.9 mL 3) 39.1 mL
How many milliliters of a 5.75% (v/v) ethanol solution can be prepared from mL of ethanol?
1) mL
2) mL
3) mL

42. Using Percent Concentration (m/v) as Conversion Factors
How many mL of a 4.20% (m/v) will contain 3.15 g of KCl? STEP 1: Given: 3.15 g of KCl(solute); 4.20% (m/v) KCl Need: mL of KCl solution STEP 2: Plan: g of KCl mL of KCl solution STEP 3: Write conversion factors g KCl and 100 mL solution 100 mL solution 4.20 g KCl STEP 4: Set up the problem

43. Learning Check
How many grams of NaOH are needed to prepare 125 mL of a 8.80% (m/v) NaOH solution?