Solutions
15.1 Solubility
 Solution: homogeneous mixture or mixture in which
components are uniformly intermingled
 Solute: substance that is being dissolved in solvent
 Solvent: substance that dissolves solvent and present in a
large amount
 Aqueous solutions: solutions with water as the solvent
Table 15.1 – Various Types of Solutions
15.2 Solution Composition: An Introduction
 Saturated: A solution in which the maximum amount
of solvent has been dissolved. Any more solute added
will sit as crystals on the bottom of the container
 Unsaturated: A solution in which more of solute can
be dissolved
 Concentrated: a relative large amount of solute is
being dissolved in solvent
 Diluted: a relative small amount of solute is being
dissolved in solvent
15.3 Solution Composition: Mass Percent
mass of solute
 Mass percent (m/m%) = ------------------------ x 100
mass of solution
Examples
 A solution is prepared by mixing 1.00 g of ethanol,
C2H5OH, with 100.0 g water. Calculate the mass percent of
ethanol in this solution
 A 135 g sample of seawater is evaporated to dryness, leaving
4.73 g of solid residue (the salts formerly dissolved in the
seawater). Calculate the mass percent of solute present in
the original seawater
Example: Determine Mass of Solute
 Although milk is not a true solution (it is really a suspension
of tiny globules of fat, protein, and other substrates in
water), it does contain a dissolved sugar called lactose.
Cow’s milk typically contains 4.5 % by mass of lactose,
C12H22O11. Calculate the mass of lactose present in 175 g
of milk
15.4 Solution Composition
 Molarity: the number moles of solute per volume of
solution in liters
moles of solute
 Molarity = -----------------------Liters of solution
 unit = moles/L or M (molar)
Standard solution: is a solution whose concentration is
accurately known.
Examples
 Calculate the molarity of a solution prepared by dissolving
11.5 g of solid NaOH in enough water to make 1.50 L of
solution
 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
 Determine how much volume (in ml) will be needed to
dissolved 2.50 g of solid NaCl to make 0.050M solution.
E.g Solution Composition: Calculating Ion
Concentration
 Give the concentration of all the ions in each of the
following solutions:
 0.50 M Co(NO3)2
 1.0 M FeCl3
E.g Solution Composition: Calculating
Number moles from Molarity
 How many moles of Ag+ ions are present in 25.0 mL of a
0.75 M AgNO3 solution?
 How many moles of Na+ ions are present in 42.0 mL of
0.350M NaCl?
Examples: Calculating mass from molarity
 To analyze the alcohol content of a certain wine, a
chemist needs 1.00L of an aqeuous 0.200 M K2Cr2O7
(potassium dichromate) solution. How much solid
K2Cr2O7 (molar mass = 294.2 g) must be weighed out
to make this solution?
 Formalin is an aqueous solutions of formaldehyde,
HCHO,, used as a preservative for biological
speciments. How many grams of formaldehyde must
be used to prepare 2.5 L of 12.3 M formalin?
15.5 Dilution
 Reducing the original concentration of a chemical solution
 A process of transferring solution to achieve a the desired
molarity by diluting with solvent
 Moles of solute after dilution = moles of solute before dilution
 Formula  M1 V1 = M2 V2
Examples
 What volume of 16 M sulfuric acid must be used to prepare
1.5L of 0.10 M of H2SO4 solution?
 Calculate the new molarity if a dilution is made for:
 25.0 ml of water is added to 10.0 mL of 0.251 M CaCl2
15.6 Stoichiometry of Solution
Reactions
 Steps for solving stoichiometric problems involving
solutions
 Step 1: Write a balanced equation for the reaction. For each
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reactions involving ions, it is best to write the net ionic
equation.
Step 2: Calculate the moles of reactant
Step 3: Determine which reactant is limiting
Step 4: Calculate the moles of other reactants or products, as
required
Step 5: Convert to grams or other units, if required
Examples
 When Ba(NO3)2 and K2CrO4 react in aqueous solution, the
yellow solid BaCrO4 is formed. Calculate the mass of
BaCrO4 that forms when 3.50 x 10-3 mole of solid
Ba(NO3)2 is dissolved in 265 mL of 0.0100 M K2CrO4
solution
Examples
 When aqueous solutions of Na2SO4 and Pb(NO3)2 are
mixed, PbSO4 precipitates. Calculate the mass of PbSO4
formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of
0.0250 M Na2SO4 are mixed
Neutralization Reaction
 Use stoichiometry to determine how much of acid or base
must be used to reach neutralization
 Strong acid: HCl(aq)  H+(aq) + Cl-(aq)
 Strong base: NaOH(s)  Na+(aq) + -OH(aq)
 Net equation: H+(aq) + -OH(aq)  H2O(l)
Example
 What volume of a 0.100 M HCl solution is needed to
neutralize 25.0 mL of a 0.350 M NaOH?
 Calculate the volume of 0.10 M HNO3 needed to neutralize
125 mL of 0.050 MKOH
15.8 Solution Composition: Normality
 Normality is another unit of concentration sometime used
when dealing with acid and base
 H+ and –OH
 Equivalent of an acid: the mount of acid that can be
furnish 1 mol of H+ ions
 Equivalent of a base: the amount of that base that can
furnish 1 mol of –OH ions
Equivalent
 E.g 1 mol HCl = 1 equiv HCl
molar mass (HCl) = equivalent weight (HCl)
1 mol KOH = 1 equiv KOH
molar mass KOH = 1 equiv KOH
½ mol H2SO4 = 1 equiv H2SO4
½ molar mass H2SO4 = 1 equiv H2SO4
Solution Stoichiometry: Calculating
Equivalent Weight
 Phosphoric acid, H3PO4 can furnish three H+ ions per
molecule. Calculate the equivalent weight of H3PO4.
 Calculate the equivalent weight of HBr
Normality (N)
 Normality (N) = number of equivalent of solute per liter
of solution
 Knowing Normality can help us calculate
 The number of equivalents
 The volume of solution
Calculating Normality
 A solution of sulfuric acid contains 86 g of H2SO4 per liter
of solution. Calculate the normality of this solution
 Calculate the normality of a solution containing 23.6 g of
KOH in 755 ml of solution
Neutralization
 The number of H+ ions furnished by the sample of acid is
the same as the number of –OH ions furnished by the
sample of base
reacts exactly with
n equiv acid --------------- n equiv base
Nacid x Vacid = Nbase x Vbase