Chapter 12. Physical Properties of Solutions
~
The Solution Process (Sections 12.1 - 12.2)
Concentration Units (Section 12.3)
Temperature and Pressure Effects on Solubility (Sections 12.4- 12,5)
tive Properties (Sections 12.6- 12.7)
SUMMARY
The Solution Process (Sections 12.1 - 12.2)
Some Definitions. The main subject of this c....~ter concerns the formati n ag~.a_&~
properties of liquid solutions: Recall from Chapter 4 that a solution is a #omofileneous
~~mor stances. T_.hhe,.~qmgoonent i#.qr~at-~r quantity is called the"~’~so vent
and the.£c.om_...~com onent ~n lesser..,am°u" - --- -nt. is called the__~solute. In ~ater ~s thea ueous s61utions water is the
solvent. Solutes can be liquids, sohds, or gases, Several te
rms are used to describe thE-~
~"~--~
degree to which a solute will dissolve in a solve~.___,t:
(~. When two liquids are complete.ly soluble in each other in all propor[ions, they are said to
be misc~r example, ethanol and water are miscible. If the Ii_quids do not mix,, they
are said to be ir~. Oil and water, for examp e~, are immiscible.,
~contains the maximum concentration of a solute..j~t a’given temperature is
called a saturated solution. For example, the solubility of NaCl is 6.1 moles per liter of
water. An unsatur~’~ution has a concentration of solute that is less than the
.j~ conce~aturated solutions have a concentration of solute that is
.greater than thatof a s__aturated solution;
"
The Solution Process. Dissolving is a process that takes place at the molecular level
and can be discussed~ecular terms. W~,D,~:L~.~bstance dissolves in another, the
particles of !he so__lute disperse uniformly thr~t. The s
.positions that are normally taken by solwnt molecules, The ease with which a solute particle
may replace a solvent molecule depends on thee strengths of three typ~s of
interactions:
¯ Solvent-solvent interaction ~¯ Solute-solute interaction" ~
o Solvent-solute interaction
Imagine the solution process as taking place in three steps as shown in Figure 12.1,
Step 1 is the separation of solveo~ Step 2 is the separation of...~_solute molecules.
~t~o over~om~-~ractive int~rmolecular forces, S~
t~ of solvent and so{ute motec.u!es; it ma~ermic or endothermic. Accer,1~qq,
.to Hess’s law (see Section 6.6 of the texl.) the heat of solution is given by ~d~um of~e
enthalpies of the three steps:
233
12: Physical Properties of Solutions
234
"~ AHsoln = AH~ + AH2 + AH3
The solute will be soluble in the solvent if the solut~r c ~on is stronger than
s°luti rt_p[ocess is ,
~n is weaker th~~ solution
~~ic, SoI~~~N~ a solute " ~, (an ioa er
molecule) i& ~urround~d b~~o str6~e-solvent a~ractive forces.
When water is the solvent the process is called hydration.
.~
0 000
~0000~0,~ Stepl:~..OO
Oo0:O
~-, O
000(~
0000
OoO oO
Ooo 8~ ~H,
Solvent
~@ @ ~ .,~.Step 2
@~@ ~
AH~
Solute
Step 3
Solution
Figure 12.1. A molecular view of the solution process. Think of the solute
molecules and solvent molecules first being spread apart, and then being
mixed together. The relative strength of forces holding solvent molecules
together AH~, holding solute particles together. ±H2, and the forces
between solvent and solute molecules ,&Ha in the solution are important in
determining the solubility of a solute in a solvent.
Here we see that the solution process is assisted by.an exothermic heat of solution, Yet,
there are a number of soluble compounds with endothermic heats of solution. In addition,
every substance is somewhat soluble no matter what the value of AHso~n. It turns out that all
chemical processes are governed by two factors. The first of these is the energy factor. In
other words, does the solution process absorb energy or release energy?
Disorder or randomness is the other factor that must be considered, Processes that
increase randomness or disorder are also favored. In the pure state, the solvent and solute
possess a fair degree of order. Here we mean the ordered arrangement of atoms,
molecules, or ions in a three-dimensional crystal. Where order is high, randomness is low.
The order in a crystal is lost when the solute dissolves and its molecules are dispersed
in the solvent, The solution process is accompanied by an increase in disorder or
randomness, It is the increase in disorder of the system that favors the solubility of any
substance.
Chemistry, Ch. 12: Physical Properties of Solutions 235
¯
A_ Ge~lSl_u_b!![~, Rule. The mos!~g~.t.."li_ke dissolyes like."
In this rule, the term like ~s to m~ec(Jl~~lves like"~mea~t
substances of like, or similar, polarity ~nd ~t
--~-olari{y will be immiscible, or will tend only slightly to form solutions. The rule predicts that
two p~~form a solution, and two nonpolar subst~ a
~ su s ance and~ ~
Examples
~-~or~x-am~le,
respectively, 12.1,12.2
~ ......
Fo ~,-. ..... water and oil, beino oolar and ~bstances,
are immiscible. Water and ethanol, both being~oolar, are m~sclD~e. ~J~t w~
diss~’~’arbon tetrachloride because both substances are nonpolar.
Exercises
Many ionic compounds tend to be quite soluble in water (a polar solvent).
Concentration Units (Section 12.3)
Concentration. The term con_~centration refers to how much of one component of a
sol~n a given amount of solution. In Chapter 4 the concentration unit molarity
..(c. _ mo, sofso,ute
Thr e new concentration units are introduced in this ~hapter.
.
\~’~-~
The percent by mass of solute
_,.
.
mass of solute
Percent by mass of so~ute = ~" . x~uu~o
~
mass SOlUte + mass so~vent
~ mass of s@~ion
¯ ~ The mole fraction of a component of the solution
.. --
J
~~~
mo~es o~
The molality of a solute in a solution
.
Exercises
/~ ~perature and Pressure Effects on Solubility (Section, 12.4 - 12.5}
~Tem era~ Effect on the Solubility of Solids. T~[ur~a~r~ affectr
~~ost solid solutes. The effects of ~erature on the ~e
~i~ of a solid in r
common salts is shown in Figure, 12.3 (text). In most cases the solubility
increases~increasing temperature. HoWever, this is not alwa~ true. Table 12.1 lists
~whose solubility d~ses with inc~. In general, it is not
~ ~ predict ju~ ’~empe~a~re will affect the solubility of a g~ven substance.
Temperature effe~ determined experimentally.
236 Chemistry, Ch. 12: Physical Properties of Solutions
TABLE 12.1 Compounds With Solubilities That
Decrease with Increasing Temperature
CaSO4
Na2SO4.10H20
0e2(SO4)3
Ca(OH)2
Ca(C2H302)2
Li2SO4
~and the Solubility of Gases. The solubility of cases in liquids is directly
proport~ona to [ne g~as~ t~enry’s ~aw relates gas concentration c, in moles per liter,
to the gas pressure p in atmospheres.
~ c=kP
~~
where k is a_consta0t for each gas and has.~of mol/L.atm. The r..g.E~.~er th_____e value o! k,.the
greater the solubility of the ,~a§: A few of~for ~solved in waterer at
25 °C are listed in Table 12.2
’
,~TABLE 12.2 Some Henry’s Law Constants at 25 °C
Gas
k(mol/L,atm)
02
N2
CO2
1.3 x 10-3
6,8 x 10-4
3.4 x 10-2
olubility and Temperature. A co.£~mon observation i~s that a ~ass of cold water,
when warm~~rn-pe~at-u-~,, shows the f_orm~tion of ma~y small air bubbles. This "
phenomenon results in part from~’he decreased solubility of gase~
temperature.
The ~ of an unreact ve gaseous solute s due to ntermo ecu ar
temPeratur# of such a solution !s increased more solute molecules attain sufficient
kinetic energy to break awa~y from these attract ve forces and enter t~he gas phase, | ¯
and so the solubility d~reases. [~,-~,-,-,-,-,-,-,~[oColligative Properties (Sections 12.6- 12.~..~
_q~Properties. The ro~es of solutions that depend on the number of solu_~
articles in solution re called colligative properties. The four colligative properties of interest
5ere are:
::~ ~ Vapor-pressure lowering, z~P._
is given by: ~ ’~:~.,.~,~.3£~
where~X ,~
is the
lracti0n
of the
arid "° "’ ~ is ~ne v~por pressure of~ J
,,,~mole
,,,u,~
,~uvn
v~solute,
u~ ~u~u~
solvent. Recall that the vapor pressure of a liquid is the pressure exe~ed by ~ vapor in
Chemistry, Ch. 12: Physical Properties of Solutions 237
e uilibrium with its li~For a ~tile solute the vapor
~ due to thegn ~e amount of Iowen~g
of th~essure can be seen to de~Eend on X2, the mole fraction of the solu.t_&e.
~~: the boiling point of a solution is hi~her than that of pure solven~-because th
or
~~s than the vapor pressure of pure
~olvent. Thus, a solution must be hotter than a pure so yen ,~ 0 v
ssures are
~ arm. Figure 12.11 in the text shows the effect that lowering the vaPor pressure
has on the boiling p_o~boiling-point elevation z&Tb of a solution of molality m is
where ~ a constant called the molal boiling-point elevation constant and has
°C/m.~There is a Kb for each solvent and T~able 12.1 (text) lists Kb values for six
,~ts. The magnitude of A, Tb, ~the increase in boiling point of the solution over the
~~, is P_r0por~ional ent
to the
concentration
m._The
of thesolute
solute, but
requires that the solute
be
nonvolatile.
~Freezing-point depression, ATf, is givenby:~-~-~"~=" ~:= ~. t~. ~ ~ ~
where Kr i__~s the molal freezing-point de .p.{essron constant and has the units °C/m, As is
the case of E’b, .~.L[s ~lifferer{t for each-solvent, ._~nd ~ must be foun~ble
such as Ta I
xtboo~. The equation a~, ev.~en volatile ones,
because freezing points are rather insensitive to vapor pressure.
he osmotic pressure, rr, is given by:
where M is the molar concentration,~E~,Js ~he ideal gas, constant (0.0821 L.atm/K.mol),
and T is the’abs~peratur~. Osmosi~Ts the net flow ~-cules~hrodgh
~e mem r~ne rom a more d~lute solution to a more concentrated
solution. Nole that in this process the s~ -s region of greater solvent
concentration to a region of lesser solvent concentration. The osmotic pressure of a
solution is the.pressure required to stop~osmosis. __ __
Molar Mass Determination. Colli a~we ro ert~de a means~termine the
mo~r mass of the solute. For instance, the freezing-point depression equation is
m01 solute
AT~f = Kr__.~m = Kf..£x ~
Since ATf can be measured and .Kf is known then the rnol~Iity, m, of the solution can be
calculated. Recall that molality is .... .
238 Chemistry, Ch. 12: Physical Properties of Solutions
tool solute mass solute/molar mass of solute
"~ molality =
kg solvent
~
kg solvent ’If you already know the molality..of a solution and a known of solute is dissolved in a
known mass of solvent, this leav_~es the molar mass of the solute as the only unknow~n
quan ,
Freezin -point de ression is more sensitive to the number of moles of solute than the
boiling-point elevation because for the same solvent Kf is larger then Kb. The most sensitive
method for the d t rmination of the molar mass of a solute is to measure the osmotic~
._pressure. The equation for osmotic pressure then becomes usefbl in the following form:
ff = MRT - (g solute/molar ma._ss Of _sol~ute)RT
L solution
nonelectrolyte solutions. For a 0.1 m methanol solu~,.ATf is the same as the calcul~
~ATf is 1,9 times greater than calculated. And for NazSO4, ATr is 2,7
~ter than the c~table
for several solutez~, This ratio is always greater than one for solutions of electrol~es. Since
~ative’ prope~ies depend on the number of solute~a~icles, the most likely explan~
of this behavior is that~ssociate into ions and this produces a greater ’
concentration of solute pa~icle~, Therefore ~~n ~ I~ ~icles in
0.10 m NaCI is essentially 2 x 0.10 m, and in Na~O4 it is almost 3 x 0,10 ~The f~g~ depression depends ~n~oncen~tion of a~ solute £adicles. Each ion acts as
an independent pa~i£ie..
TABLE 12.3 van’t Hoff Facto_r i fo.r Several Solutes
Solution
0.10 m CHaOH
0.10 m NaCI
0,10 m Na2SO4
ATf ~_-~ ~
0,186 °C
0,353 °C
0,502 °C
ATf (calc)
0.186 °C
0.186 °C
0.186 °C
1.0
1.9
2.7
Examples
12.6 - 12.9
~~ ..~~) is called the van’t Hoff factor i. For 1:1 electrolyte.s,
such as~NaC~l~ KCI, and MgSO4, i ~ 2; for 2~ e~ectrolytes s~ch as M CI2 and
Exercises
K2SO4, i is a li~~e less than ~. - ~e fi~’mber resu~ the 12-16-12-2~
"-’~6mporary forrn~tiOn of sorn~ ion pairs such as "{q~-Cq~r~he solution.
~
WORKED EXAMPLES
EXAMPLE 12.1 Solubility
Which would be a better solvent for molecular 12(s): CCI,= or H20?
¯ Solution
Using the like-dissolves-like rule, first we identify 12 as a nonpolar molecule. Therefore it will
be more soluble in the nonpolar solvent CCI4 than in the polar solvent
EXAMPLE 12.2 Solubility
In which solvent will NaBr be more soluble, benzene or water?
¯ Solution
Benzene is a nonpolar solvent and water Is a polar solvent. Since NaBr is an ionic
compound, it cannot dissolve in nonpolar benzene, but is very soluble in water,
EXAMPLE 12.3 Concentration Units
The dehydrated form of Epsom salts is magnesium sulfate.
a. What is the percent MgSO4 by mass in a solution made from 16.0 g MgSQ4 and 100 mL
of H20 at 25 °C? The density of water at 25 °C is 0.997 g/mL,
b. What is the mole fraction of each component?
c. Calculate the molality of the solution.
Chemistry, Ch. 12: Physical Properties of Solutions 243
¯ Solution for (a)
Write the equation for percent by mass.
percent MgSO4 =
mass MgSO4
mass MgSO4 + mass water
mass H20 = 100 mL x--0997 g
lmL
x 100%
= 99.7 g H2O
percent MgSO4- 16.0 g16.0g
+ 99.7 g x100%- 16.0g x100%
115.7 g
percent MgSO4 = 13.8%
¯ Solution for (b)
Write the formula for the mole fraction of MgSO4.
XMgSO4
tool MgSO4
mol MgSO4 +mol
H20
Convert the masses in part (a) into moles to substitute into the equation.
16.0 g MgSO4 x 1 mol MgSO4 - 0.133 mol MgSO4
120.4 g MgSO4
1 mol H~O
99.7gH~Ox =5.54molH20
I8.0 g H2O
The mole fractions of MgSO4 End H20 are
0.133 mol
5.54 mol
XMgS°4= 0.133mol+5.54mo! =0.0235 XH2O- 5,67mol 0.977
Notice that the sum of the two mole fractions is 1.000. The sum of the mole fractions of a~l
solution components is always ! .00.
¯ Solution for (c,)
Write the formula for molality:
molality = moles of MgSO4
kilograms of H20
Substitute into this equation the quantities previously calculated in parts (a) and (b).
molality = 0.133 mol MgSO
g
x -=31.33
m
4 10
99.7 g
1 kg
244 Chemistry, Ch. 12: Physical Properties of Solutions
EXAMPLE 12.4 Molality and Molarity of a Solution
Concentrated hydrochloric acid is 36.5 percent HCI by mass. Its density is 1.18 g/mL,
Calculate:
a. The molality of HCI.
b. The molarity of HCI.
¯ Solution for (a)
Write the formula for molality,
mo! HCI
kg H20
find the number of moles of HCl per kilogram of H20. We take 100 g of solution, and
determine how many moles of HCl and how many kilograms of the solvent it contains. A
solution that is 36.5 percent HCI by mass corresponds to 36.5 g HCI/IO0 g solution. Since
100 g of solution contains 36.5 g HCl, the difference 100 g - 36.5 g must equal the mass of
water which is 63.5 g. We have two ratios:
molality =
36.5 g HCI and
100 g soln
36.5 g HCI
63.5 g H2O
The moles of HCI is given by:
moles HCI = 36.5 g HCI x
1 mol
= 1.00 mol HCl
36.5 g HCI
The kilograms of water is given by:
kg
kgg - 0.0635 kg
H20 = 63.5 g H20 x 1 x 1
103
Then we calculate molality:
molality -
1.00 mol HCI
- 15.7 m
0,0635 kg H2O
¯ Solution for (b)
Write the formula for molarity:
molarity =
moles HCI
liters soln
Find the number of moles of HCI and liters of solution are present in 100 g of solution
36.5 g HCl
100 g soln
Convert 100 g of solution to volume of solution using the density. (Note: 36.5 g HCI = 1.00
mol.)
Chemistry, Ch. 12: Physical Properties of Solutions 245
Volume of soln = (100 g soln) x
molarity -
1 mL
x 10-3L = 0.0847 L
1.18 g soln 1 mL
1.00 mol HCI
- 11.8 M
0.0847 L soln
EXAMPLE 12.5 Henry’s Law
What is the concentration of 02 in air-saturated water at 25 °C and atmospheric pressure of
645 mmHg? Assume the mole fraction of oxygen in air is 0.209.
¯ Solution
Henry’s law states that the concentration of dissolved 02 (Co2) is proportional to its partial
pressure (Po;)in atm.
Co2 = k P 02
The partial pressure of 02 in air is found by using Dalton’s law of partial pressures.
1 atm
= 0.177 atm
Po~ = XO2PT = 0.209 (645 mmHg) x
760 mmHg
Taking k from Table :i2.2, the concentration of dissolved oxygen is:
Co2 = kPo2 = (1.28 x 10-3 mol/L.atm) (0.177 atm)
= 2.27 x 10-4 mol/L
-EXAMPLE 12.6 Boiling Point Elevation
What is the boiling point of an "antifreeze/coolant" solution made from a 50-50 mixture (by
volume) of ethylene glycol, 02H602 (density ! .11 g/mL), and water?
¯ Solution
The boiling point depends on the molality of the 50-50 mixture. AT = Kbm. F.or simplicity
assume 100 mL of the solution. Then using the density of water, the mass of 50 mL
(50 vol %) of H20 is 50 g. The mass of 50 mL (50 vol %) of ethylene glycol using the
density given above is 55.5 g.
m=
mol C2H602
kg H20
mol C2H602 = 55.5
1 moi
g x -- = 0.895 mol
62.0 g
m= 0.895 mol=C2H602
17.9 mol/kg = 17.9 m
0.050 kg
ATb = Kbm = (0.52 °CIm) x 17.9 m = 9.3 °C
Therefore the boiling point of the solution is 9.3 °C above the normal boiling point of water.
Boiling point = 109.3 °C.
246 Chemistry, Ch. 12: Physical Properties of Solutions
EXAMPLE 12.7 Vapor Pressure Lowering
Calculate the vapor pressure of an aqueous solution at 30 °C made from 100 g of sucrose
(C12H22Oll), and 100 g of water. The vapor pressure of water at 30 °C is 31.8 mmHg.
¯ Solution
Sucrose is a nonvolatile solute, so the vapor pressure over the solution will be due to H20
molecules. The problem can be worked in two ways.
a. The vapor-pressure lowering is proportional to the mole fraction of sucrose )(’2, and P~,
the vapor pressure of pure water at 30 °C,
P=X2P
First, we calculate the mole fraction of sucrose, X2.
)(’2 = n2
n1 + n2
1 tool
n2 = !00 g x
- 0.292 mol
342 g
100 g x
X2=
1 mol
- 5.55 mol
18.0 g
0.292
0.292
0.292 + 5.55 5.84
X2 = 0,0500
The vapor pressure lowering is:
&P = (0.0500) (3!.8 mmHg) = 1.59 mmHg
The vapor pressure, P~, is
P~ = 31.8- 1.59
= 30.2 mmHg
b. Alternatively Raoult’s law can be used to calculate the vapor pressure of the solvent:
P1= X! P~
where P~ and P~ are the vapor pressures of pure solvent and of the solvent in solution
respectively. From part a, X2 = 0.0500, therefore XI = 1.00 - X2 = 0.95, and
P~ = (0.95)(31.8 mmHg)
= 30 mmHg
Chemistry, Ch. 12: Physical Properties of Solutions 247
EXAMPLE 12.8 Finding the Molar Mass of a Solute
Benzene has a normal freezing point of 5.51 °C. The addition of 1.25 g of an unknown
compound to 85.0 g of benzene produces a solution with a freezing point of 4.52 °C. What is
the molecular mass of the unknown compound?
Solution
We need to find the number of moles in 1.25 g of unknown compound X. The freezing-point
depression is proportional to the number of moles of X per kilogram of solvent.
The freezing-point depression is:
z3Tf = 5.51 °C - 4.52 °C
= 0.99 °C
We can calculate the molality of the solution because &Tf is given above.
where Kf is the freezing-point depression constant for benzene (Table 12.2 textbook).
m=
0,99 °C
= 0.19 m
5.12 °C/m
This means that there are 0.19 mol of X per kg of benzene. The number of moles of X in
0.085 kg of the solvent benzene (the given amount) is:
0,085kgx 0.19molX =1,6x10-2molX
1 kg benzene
Therefore 1.25 g X = 1.6 x 10-2 tool X, and
molar mass of X =
1.25 g
= 78 g/mol
1.6 x 10k2 mot
EXAMPLE 12.9 Colligative Propereties of Electrolytes and Nonelectrolytes
List the following aqueous solutions in the order of increasing boiling points: 0.10 rn glucose
(C6H1206); 0.10 m Ca(NO3)2; 0.10 rn NaCl.
¯ Solution
Ethanol is a nonelectrolyte, whereas NaCI and Ca(NO3)2 are electrolytes. NaCI dissociates
into 2 ions per formula unit, and Ca(NO3)2 dissociates into three ions per formula unit. The
effective molalities are approximately:
glucose 0.10 m
NaCI ~ 0.20 m
Ca(NO3)2 ~ 0,30 m