Copyright Sautter 2003
SOLUTIONS & CONCENTRATIONS
• WHAT IS A SOLUTION ?
• WHAT IS CONCENTRATION & HOW IS
IT MEASURED ?
IS DISSOLVING A CHEMICAL OF A PHYSICAL
PROCESS ?
• IF SUGAR IS DISSOLVED IN WATER IS ITS CHEMICAL
NATURE CHANGED ?
• NO !!
• IF THE SOLUTION IS DRIED, A WHITE, SWEET
CRYSTALLINE SUBSTANCE (SUGAR) IS OBSERVED.
• THEREFORE DISSOLVING IS A PHYSICAL CHANGE. THE
CHEMICAL PROPERTIES OF A SUBSTANCE MUST BE
CHANGED IN ORDER FOR A CHEMICAL CHANGE TO
OCCUR !
A SOLUTION THEN IS THE RESULT OF THE
PHYSICAL CHANGE CALLED DISSOLVING, BUT
PRECISELY, WHAT IS A SOLUTION?
• THE OPPPOSITE OF A SOLUTION IS A SUSPENSION.
• MUDDY WATER IS A SUSPENSION.
• SUGAR DISSOLVED IN WATER IS A SOLUTION.
• CAN YOU NAME SOME DIFFERENTS BETWEEN THEM?
SOME OBSERVED DIFFERENCES
•
•
•
•
DIFFERENCES IN CLARITY !
DIFFERENCES IN UNIFORMITY !
DIFFERENCES IN SETTLING !
DIFFERENCES IN FILTRATION !
CLARITY
• SOLUTIONS ARE TRANSPARENT *
• SUSPENSIONS ARE OPAQUE *
•
* ALTHOUGH SOLUTIONS ARE TRANSPARENT (YOU CAN SEE THROUGH
THEM) THEY MAY BE COLORED
•
* OPAQUE MEANS CLOUDLY
UNIFORMITY
• UNFORMITY OR HOMOGENOUS MEANS THE SAME
THROUGHOUT
• SOLUTIONS ARE HOMOGENOUS ( THE SAME AMOUNT OF
DISSOLVED SUBSTANCE FOR THE SAME AMOUNT OF
SOLUTION EVERYWHERE IN THE SYSTEM)
• SUSPENSIONS ARE HETEROGENOUS (THE AMOUNT OF
SUSPENSED MATERIAL DIFFERS FROM PLACE TO PLACE)
• FOR EXAMPLE, THE AMOUNT OF MUD IS MORE
CONCENTRATED AT THE BOTTOM THAN AT THE TOP
SETTLING OUT
• SOLUTIONS NEVER SETTLE. THE
DISSOLVED MATERIAL WILL NOT FALL
OUT OF THE SOLUTION (UNLESS THE
TEMPERATURE IS CHANGED)
• IN SUSPENSIONS, EVENTUALLY THE
SUSPENDED MATERIAL WILL FALL TO THE
BOTTOM OF THE CONTAINER !
FILTRATION
• THE DISSOLVED MATERIAL IN A SOLUTION
CANNOT BE FILTERED OUT BY ORDINARY
MEANS (FOR EXAMPLE THE SALT CANNOT
BE FILTERED FROM SEA WATER WITH
FILTER PAPER)
• IN A SUSPENSION SUCH A MUDDY WATER,
THE SUSPENDED SUBSTANCE CAN BE
FILTERED LEAVING A CLEAR LIQIUD
SOLUTIONS VS. SUPSENSIONS
•
•
•
•
•
• SUSPENSIONS
SOLUTIONS
(1) TRANSPARENT
(2) HOMOGENOUS
(3) WILL NOT SETTLE OUT
(4) CANNOT BE FILTERED
•
•
•
•
(1) OPAQUE
(2) HETEROGENOUS
(3) WILL SETTLE OUT
(4) CAN BE FILTERED
FOG
COLLOIDAL
SUSPENSION
WHAT IS A SOLUTION ?
• DEFINITION: A SOLUTE DISSOLVED IN A
SOLVENT (FOR EXAMPLE A SUGAR / WATER SOLUTION)
• SOLUTE IS THE DISSOLVED SUBSTANCE (THE SUGAR)
• SOLVENT IS THE DISSOLVING MEDIUM IN WHICH
THE SOLUTE IS DISSOLVED (THE WATER)
WHAT COMBINATION OF PHASES CAN FORM
SOLUTIONS ?
• THINK OF SOME DIFFERENT KINDS OF SOLUTIONS. THEY MUST
HAVE THE PROPERTIES OF SOLUTIONS TO BE CONSIDERED.
• SOME COMMON COMBINATIONS: SOLID SOLUTE / LIQUID SOLVENT
(SUGAR DISSOLVED IN WATER)
• LIQUID SOLUTE / LIQUID SOLVENT (ANTIFREEZE DISSOLVED IN
WATER)
• GAS SOLUTE / LIQUID SOLVENT (SODA WATER – CARBON DIOXIDE
DISSOLVED IN WATER)
• GAS SOLUTE / GAS SOLVENT ( AIR – OXYGEN DISSOLVED IN
NITROGEN)
WHAT ARE THE GENERAL CLASSIFICATIONS OF SOLUTIONS
?
• THERE ARE GENERALLY TWO TYPES:
ELECTROLYTES AND NON ELECTROLYTES
• ELECTROLYTIC SOLUTIONS ARE ELECTRICALLY
CONDUCTIVE. THEY CONSIST OF IONIC SOLUTES
DISSOLVED IN POLAR SOLVENTS
• NON ELECTROLYTIC SOLUTIONS ARE NON
CONDUCTIVE AND THEY CONSIST OF
MOLECULAR SOLUTES DISSOLVED IN NON
POLAR SOLVENTS.
TYPES OF ELECTROLYTIC SOLUTIONS
• STRONG ELECTROLYTES
• SOLUTIONS IN WHICH ALL OF THE DISSOLVED
SOLUTE FORMS IONS
• WEAK ELECTROLYTES
• SOLUTIONS IN WHICH ONLY A PERCENTAGE OF
THE DISSOLVED SOLUTE FORMS IONS
OTHER WAYS TO CLASSIFY SOLUTIONS
• SATURATED SOLUTIONS
• NO MORE SOLUTE CAN BE DISSOLVED
(SOLUBILITY LIMIT HAS BEEN REACHED)
• UNSATURATED SOLUTIONS
• ADDITIONAL SOLUTE CAN STILL BE DISSOLVED
WHAT DETERMINES THE SATURATION POINT
OF A SOLUTION ?
• (1) THE TYPE OF SOLUTE AND SOLVENT
USED
• (2) THE TEMPERATURE OF THE SOLUTION
(Generally solids dissolve better at higher
temperatures while gases dissolve more poorly)
• (3) GAS PRESSURE WHEN A GAS IS THE
SOLUTE (HENRY’S LAW)*
• *THE SOLUBILITY OF A GAS IS DIRECTLY
RELATED TO THE PRESSURE OF THAT GAS
ABOVE THE SOLUTION
•
Solubility = a constant x Pressure of the gas
How Temperature Effects Solubility
Note that all substances (even solids), do not dissolve
better at higher temperatures although most do.
HOW CAN THE SOLUTIONS COMPOSED OF THE
SAME SUBSTANCES BE DIFFERENT
?
• FOR EXAMPLE, HOW CAN ONE AQUEOUS* SUGAR
SOLUTION BE DIFFERENTIATED FROM ANOTHER ?
• DIFFERENT SOLUTIONS CONSISTING OF THE SAME
SOLUTE / SOLVENT COMBINATIONS MAY BE
DIFFERENT IN CONCENTRATION !
• WHAT DOES CONCENTRATION MEAN ??
• * AQUEOUS MEANS THAT WATER IS THE SOLVENT
MEDIUM IN THE SOLUTION
CONCENTRATION
• CONCENTRATION REFERS TO A RATIO OF SOLUTE
AMOUNT TO SOLVENT OR SOLUTION AMOUNT.
• FOR EXAMPLE, CONCENTRATED ORANGE JUICE
MEANS THAT THE SOLUTE (THE ORANGE
COMPONENT) IS PRESENT IN LARGE QUANTITY
RELATIVE TO THE SOLVENT (THE WATER
COMPONENT).
METHODS OF MEASURING
SOLUTION CONCENTRATION
• DEPENDING ON THE UNITS OF MEASURE AND THE
WHETHER SOLVENT OR SOLUTION QUANTITIES ARE
MEASURED, CONCENTRATION CALCULATIONS VARY.
• WHEN THE SOLUTE QUANTITY IS MEASURED IN MOLES
AND THE SOLUTION VOLUME IS MEASURED IN LITERS,
THE CONCENTRATION IS EXPRESSED AS MOLARITY OR
MOLES PER LITER OF SOLUTION.
• MOLARITY = MOLES OF SOLUTE / LITER OF SOLUTION
MOLARITY CALCULATIONS
• WHAT IS THE MOLARITY OF A SOLUTION WITH A VOLUME
OF 2.0 LITERS AND CONTAINING 90.0 GRAMS OF GLUCOSE
(C6H12O6) ?
• THE DEFINITION OF MOLARITY IS MOLES OF SOLUTE PER
LITER OF SOLUTION !
• STEP I – FIND THE NUMBER OF MOLES OF SOLUTE PRESENT
• 90.0 GRAMS / 180 GRAMS PER MOLE OF C6H12O6 GIVES 0.50
MOLES OF GLUCOSE
• STEP II – MOLARITY = MOLES / LITERS
• 0.50 MOLES / 2.0 LITERS = 0.25 M (CAPITAL M = MOLARITY)
MOLARITY CALCULATIONS
(CONTINUED)
• HOW MANY MOLES OF ZINC CHLORIDE ARE
CONTAINED IN 500 ML OF A 0.20 M SOLUTION?
• SINCE M = MOLES / LITERS,
MOLES = M X LITERS
• 500 MLS = 0.500 LITERS
• MOLES = 0.20 M X 0.500 L = 0.100 MOLES
MOLARITY CALCULATIONS
(CONTINUED)
• HOW MANY GRAMS OF SODIUM CHLORIDE (NaCl) ARE
CONTAINED IN 250 MLS OF A 0.50 M SOLUTION ?
•
STEP I – CALCULATION MOLES AS IN THE PREVIOUS PROBLEM
MOLES = MOLARITY X LITERS
•
•
•
MOLES = 0.50 M X 0.250 L = 0.125 MOLES
STEP II – CONVERT MOLES TO GRAMS
NaCl IS 58.5 GRAMS PER MOLE
•
0.125 MOLES X 58.5 GRAMS PER MOLE = 7.31 GRAMS OF NaCl ARE
CONTAINED IN THE SOLUTION
MOLARITY CALCULATIONS
(CONTINUED)
• HOW MANY MILLILITERS OF A 0.40 M SOLUTION
ARE NEEDED TO OBTAIN 45.0 GRAMS OF GLUCOSE ?
• MOLARITY = MOLES / LITERS
• LITERS = MOLES / MOLARITY
• 45.0 GRAMS / 180 GRAMS PER MOLE = 0.25 MOLES
• LITERS = 0.25 MOLES / 0.40 M = 0.625 LITERS
• 0.625 LITERS = 625 MILLILITERS
SOLUTIONS AND DILUTION
• WHAT DOES DILUTION MEAN ?
• DILUTE MEANS LESS CONCENTRATED
• HOW CAN A SOLUTION BE DILUTED?
• BY THE ADDITION OF MORE SOLVENT (MOST
OFTEN WATER)
• WHAT HAPPENS TO THE ORIGINAL
CONCENTRATION OF THE SOLUTION ?
• IT IS REDUCED !
FINAL
VOLUME
V2
(ORIGINAL
VOLUME
PLUS ADDED
WATER)
M1V1 = M2V2
STARTING
VOLUME
V1
STARTING
MOLARITY
M1
FINAL
MOLARITY
M2
DILUTION CALCULATIONS
• THE DILUTION FORMULA:
• M1V1 = M2V2
•
•
•
•
M1
V1
M2
V2
= ORIGINAL CONCENTRATION OF SOLUTION
= ORIGINAL VOLUME OF SOLUTION (ML OR L)
= CONCENTRATION AFTER DILUTION
= VOLUME OF SOLUTION OF AFTER DILUTION (ML OR L)
(ORIGINAL VOLUME + VOLUME OF WATER ADDED)
DILUTION CALCULATIONS
• WHAT IS THE CONCENTRATION OF 200 ML OF A 0.50 M SUGAR
SOLUTION AFTER 100 ML OF WATER HAVE BEEN ADDED?
• M1 = 0.50 M , V1 = 200 ML
• V2 = 200 ML + 100 ML = 300 ML, M2 = ?
• M1V1 = M2V2
• (0.50) x (200) = M2 x (300)
• M2 = 0.33 M
DILUTION CALCULATIONS
(continued)
• HOW MUCH WATER MUCH BE ADDED TO 500 ML OF A 2.0 M SOLUTION
OF GLUCOSE TO DILUTE IT TO A CONCENTRATION OF 0.50 M ?
•
•
M1 = 2.0 M ,
V2 = ?
,
V1 = 500 ML
M2 = 0.50 M
• M1V1 = M2V2
•
•
•
•
(2.0) x (500) = (0.50) x V2
V2 = 2000 ML
NEW VOL OF SOLUTION – ORIGINAL VOL = VOL OF WATER ADDED
2000 ML - 500 ML = 1500 ML ADDED
OTHER CONCENTRATION MEASURES
MOLALITY
• The concentration of a solution can be measured in
molality(m) units as well as molarity (M).
• Molality is defined as moles of solute divided by
kilograms of solvent. Notice the differences between
molality and molarity. Molarity uses a volume measure
in liters, molality uses mass measures in kilograms.
Molarity is based on the quantity of solution, molality is
based on the quantity of solvent. Both however use
moles of solute as the numerator term.
• Molality = moles of solute / kilograms of solvent
• Remember grams must be divided by 1000 to obtain
kilograms!
Molality
• Problem: What is the molality of a solution
containing 20.0 grams of glucose (C6H12O6) in 100
grams of water?
• Solution: m = mole / Kg
• Moles of glucose = 20.0 / 180 = 0.111
• m = 0.111 moles / 0.100 Kg = 1.11 molal
• Molality is common used in calculations involving
the freezing and boiling points of solutions.
Mole Fractions
• Measure concentrations using mole fractions is used in
the determination of the vapor pressure of solutions. It
is the decimal percent of a component in mole terms.
• For example, a solution in which half the molecules
are water and half are alcohol would contain a mole
fraction of water equal to 0.50 and a mole fraction of
alcohol equal to 0.50. The sum of the mole fractions
present in a solution must equal 1.00.
• Mole fraction of X = moles of X / total moles present
• Capital X is used to represent mole fraction.
• Mole fraction contains no unit values but is simply a
number value.
Mole Fractions
• Problem: A solution contains 25.0 grams of ethyl
alcohol (C2H5OH) and 150.0 grams of water. What
is the mole fraction of each component?
• Solution: X = moles / total moles
• For C2H5OH, moles = 25.0 / 46.0 =0.543
• For H2O , moles = 150.0 / 18.0 =8.33
• XC2H5OH = 0.543 / (0.543 + 8.33) = 0.0612
• X H2O = 8.33 / (0.543 + 8.33) = 0.939
• Check: 0.0612 + 0.939 = 1.000
Percent Solution by Mass
• Percent solution by mass is percent of the solution which
is the solute. It follows the usual percent calculation of
parts divided by the whole times 100 %
• % solution = (gram of solute/gram of solution) x 100%
• Problem: What is the percent solution of the solution in
the previous problem? (25.0 grams of alcohol in 150.0
grams of water)
• Solution: % = (g solute / total grams) x 100 %
• % alcohol = (25.0 g / (25.0 + 150.0 g)) x 100 %
• % alcohol = 14.3 %
Normality
• Often, when dealing with acids and bases especially,
concentrations are expressed in normality terms. Normality, for
acids and bases is the molarity of the hydrogen or hydroxide ion
in solution.
• One mole of hydrogen or hydroxide ions is called an equivalent.
The weight of acid or base which contained one mole of
hydrogen or hydroxide ions is called the equivalent weight of
that substance.
• Examples:
• A 1.0 molar solution of HCl is a 1.0 normal solution
• A 1.0 molar solution of H2SO4 is a 2.0 normal solution because
each sulfuric acid releases two hydrogen ions.
• A 1.0 molar solution of H3PO4 is a 3.0 normal solution because
each phosphoric acid releases three hydrogen ions.
Normality
• A 1.0 molar solution of NaOH is a 1.0 normal solution
• A 1.0 molar solution of Ca(OH)2 is a 2.0 normal solution because each
calcium hydroxide releases two hydroxide ions.
• A 1.0 molar solution of Al(OH)3 is a 3.0 normal solution because each
aluminum hydroxide releases three hydroxide ions.
• The equivalent weight of HCL is 36.5 grams (its molecular weight)
• The equivalent weight of H2SO4 is 49 grams (98 / 2 or half of its
molecular weight)
• The equivalent weight of H3PO4 is 32.7 grams (98 / 3 or one third of
its molecular weight)
• Normality = equivalents / liter
• Normality of acids = moles of H+ ions / liter
• Normality of bases = moles OH- / liter
Concentration Unit Conversions
• Percent Solution to Molality
• Problem: What is the molality of 10.0 % NaCl
solution?
• Solution: A 10.0 % solution contains 10.0 grams
of solute in 100.0 grams of solution. The solvent
must have a mass of 90.0 grams (100.0 g – 10.0 g)
or 0.090 kg
• Molality = moles / kg
• Moles NaCl = 10.0 / (23.0 + 35.5) = 0.171
• m = 0.171 moles / 0.090 kg = 1.89 molal
Concentration Unit Conversions
• Percent solution to molarity
• Problem: Concentration HCl is 36.0 %. The density of the
solution is 1.19 grams per milliliter. Find its molarity.
• Solution: Find the mass of 1.0 liter of solution.
• Density x volume = mass
• (1.19 g/ml) x 1000 ml = 1190 g
• Find the mass of solute
• 36.0 % of 1190 = 428.4 grams is HCl, the rest is water.
• Convert mass of solute to moles
• Moles of HCl = 428.4 g / (1 + 35.5) =11.7
• There are 11.7 moles of HCl in one liter of solution
therefore the solution is 11.7M
Concentration Unit Conversions
• Molality to mole fraction
• Problem: Find the mole fractions of each
component in a 4.57 molal solution of CaCl2.
• Solution: A 4.57 molal solution of CaCl2 contains
4.57 moles of calcium chloride in 1.0 kg (1000
grams) of water
• Moles of water = 1000 g / 18.0 g per mole = 55.5
• X = moles / total moles
• XCaCl2 = 4.57 / (4.57 + 55.5) = 0.0761
• Xwater = 55.5 / (4.57 + 55.5) = 0.924
• Check: 0.0761 + 0.924 = 1.000
Concentration Unit Conversions
• Molarity to Molality
• Problem: A solution of HCl has a molarity of 11.8 M and a density of
1.19 g per ml. Find its molality.
• Solution: Find the mass of 1.0 liter
• Mass = density x volume
• Mass = 1.19 g per ml x 1000 ml = 1190 grams
• Find the mass of solute in one liter
• Mass = mole x molar mass
• Mass = 11.8 x (1 + 35.5) = 431 grams HCl, the rest is water
• Mass of water = mass of solution – mass of solute
• Mass of water = 1190 g – 431 g = 569 grams
• Molality = moles / kg = 11.8 moles / 0.569 kg = 20.7 molal
Concentration Unit Conversions
• Molality to molarity
• Problem: A solute with molar mass of 72.0 grams per mole is
dissolved in water to make a 2.00 m solution. The density of the
solution is 1.25 g / ml. Find its molarity.
• Solution: Find the mass of 1.0 liters of solution
• Mass = density x volume
• Mass = (1.25 g /ml) x 1000 ml = 1250 g
• Find the mass of solution containing 2.00 moles of solute
• A 2.00 molal solution consists of 142.0 grams of solute (2.00 x 72)
plus 1000 g of solvent = 1420 g
• Set up a ratio to find the moles in 1.0 liter or 1250 grams of solution
• (2.00 moles / 1420 grams solution) = x moles / 1250 grams solution
• Solving the proportion gives 1.76 moles in 1.0 liter or 1.76 M
SOLUTION
VOLUME
SOLVENT
MASS
THE NUMERATORS IN ALL OF THE
EQUATIONS REFER TO THE SOLUTE
MOLES OF H+ OR OH-
Summary
•
•
•
•
•
•
•
•
(1) Molarity (M) = moles solute / liters of solution
(2) M1V1 = M2V2 (Dilution formula)
(3) Molality (m) = moles solute / kg of solvent
(4) Mole Fraction (XA) = (moles A / total moles)
(5) Percent Sol’n = (grams solute / total grams) x 100 %
(6) Normality = equivalents / liter
(7) Normality of acids = moles of H+ ions / liter
(8) Normality of bases = moles OH- / liter