Solutions are usually thought of as a solid solute dissolved in a liquid solvent, as in sugar water. However solutions in a wider sense can be formed by combinations of all states of matter.
| Solid | Solid | alloys; stained glass |
| Solid | Liquid | salt water; sugar water; tincture of iodine |
| Solid | Gas | moth flake odor in air |
| Liquid | Liquid | vodka; gasoline; perfume |
| Liquid | Solid | mercury amalgam; crystal hydrates |
| Liquid | Gas | moist air |
| Gas | Gas | air; natural gas |
| Gas | Liquid | carbonated water; household ammonia |
| Gas | Solid | hydrogen in platinum or palladium |
When we describe a solution, we need to specify its composition in terms of how much solute there is in a unit of solvent or solution. This is called the concentration of the solution.
There are a number of different ways of defining the concentration of a solution. We will discuss:
Solution concentration can be expressed as percent by mass. This means that there are so many mass units of solute in each 100 mass units of solution.
Percent by mass, %m/m can be calculated as:
(mass solute)
%m/m = --------------- X 100
(mass solution)
Expressed as a conversion factor, percent by mass is:
x grams solute
x %m/m = --------------
100 g solution
or the equivalent.
If the solution is measured in volume instead of mass we get percent mass/volume. Percent mass/volume, %m/v, is the number of mass units of solute for each 100 volume units of solution. Thus:
(mass solute)
%m/v = ----------------- X 100
(volume solution)
or:
x g solute
x %m/v = ----------------
100 ml solution
If both solute and solvent are liquids, it is common to use percent by volume, %v/v. Percent by volume is the volume of solute dissolved to give 100 units of volume of solution. Thus:
(volume solute)
%v/v = ----------------- X 100
(volume solution)
or:
x ml solute
x %v/v = ---------------
100 mL solution
Note that since volume depends on the temperature, %m/v and %v/v also depend on the temperature of the solution. However %m/m does not depend on temperature. When we refer to concentrations as just percentages we will understand that to mean by mass.
Sugar (35.73 g) was dissolved in 256.00 of water. What is the % by mass of the resulting solution?
Note that masses are additive and thus the solution weighs 35.73 + 256.00 = 291.73 g.
35.73 g sugar
----------------- X 100 = 12.25 % sugar
291.73 g solution
Used as a conversion factor, percent allows us to convert between an amount of solution and the amount of solute dissolved in it.How much salt is there in 345 g of a 10.0% aqueous salt solution?
This is percent by mass so:
10.0 g salt
345 g solution X -------------- = 34.5 g salt
100 g solution
How much of a 55.4 %v/v alcohol solution will contain 24.7 mL of alcohol?
100 mL solution
24.7 mL alcohol X --------------- = 44.6 mL solution
55.4 mL alcohol
One approach to this problem is to use solutions with their concentrations in molarity. Molarity, M, is the number of moles of solute in every liter of solution. Thus it can be used to convert between moles of solute and volume of solution. The advantage of this is that one can use reagents in the form of solutions and measure out a given number of moles by measuring out the equivalent volume of reagent.
The symbol M is read "molar". Thus 2.3 M NaOH is read "2.3 molar sodium hydroxide". The units of molarity are moles/L, i.e. "moles per liter".
x moles solute
x M = --------------
1 L solution
How many moles of hydrogen chloride are there in 326 mL of a 0.100 M solution of hydrochloric acid?
0.100 moles HCl
326 mL solution X --------------- = 0.0326 moles HCl
103 mL solution
How many mL of a 0.361 M NaOH solution will contain 0.125 moles of NaOH?
103 mL solution
0.125 moles NaOH X ---------------- = 346 mL solution
0.361 moles NaOH
x moles solute
x m = --------------
1 kg solvent
Note that m is defined in terms of the amount of solvent rather than the amount of solution! It is useful for problems involving the mixing of a solution from given amounts of solute and solvent. Note that the total mass of the solution is the sum of the masses of solute and solvent. The total volume of solution can then be calculated from the density of the final solution.
What is the molality of a solution of 34.7 g of glucose, C6H12O6, in 476 g of water? What is the final volume if the solution has a density of 1.02 g/mL?
1 mole glucose
34.7 g glucose X ---------------- = 0.193 moles glucose
180.18 g glucose
0.193 moles glucose
------------------- = 0.405 m
0.476 kg water
To calculate the volume we need the total mass of solution:
476 g + 34.7 g = 511 g solution
Then use the density to calculate the volume:
1 mL
511 g solution X ------ = 501 mL solution
1.02 g
(moles total) = (moles A) + (moles B) + (moles C) + (moles S)
(moles A)
XA = -------------
(moles total)
(moles B)
XB = -------------
(moles total)
(moles C)
XC = -------------
(moles total)
(moles S)
XS = -------------
(moles total)
Note that the solvent is considered just one more component. It should be obvious that any mole fraction will be less than 1 and the sum of the mole fractions for all components is equal to 1. Since mole fraction is moles divided by moles, it has no units. Mole fraction can be used for any mixture of components.
A solution consists of 51.3 g of glucose (C6H12O6), 48.7 g of sucrose (C12H22O10), and 1.00 kg of water. What is the mole fraction of each component?
1 mole glucose
51.3 g glucose X ---------------- = 0.285 mole glucose
180.18 g glucose
1 mole sucrose
48.7 g sucrose X ---------------- = 0.142 mole sucrose
342.34 g sucrose
1 mole water
1.00 X 103 g water X ------------- = 55.5 mole water
18.02 g water
55.5 + 0.142 + 0.285 = 55.9 moles total
0.285 moles glucose
Xglucose = ------------------- = 0.00510
55.9 moles total
0.142 moles sucrose
Xsucrose = ------------------- = 0.00254
55.9 moles total
55.5 moles water
Xwater = ------------------ = 0.993
55.9 moles total
** !IMPORTANT! **
Symbols such as M or % represent not a single quantity but a ratio of quantities. Therefore one should never use them as single quantities in conversion problems but only as conversion factors.
For instance, you should never see something like this in a calculation:
1.0 M 10% NaCl 40.00 g
---------- or -------- or 5.6 m X ------- X .... etc.
3.55 moles g NaCl 1 mole
In a calculation, M should be written always as mole/L, m as moles/kg, etc., so they will cancel correctly.
A good rule of thumb is that if a solute can form better than a 0.1 M solution it is soluble in that solvent. If it saturates at significantly less than a 0.1 M solution it is considered insoluble. There is a grey area in the mid range where compounds are considered slightly soluble.
When a nonvolatile solute is added to a solvent the vapor pressure of the solution is less than that of the pure solvent. This is called vapor pressure depression. The amount of depression for a given solvent is the same irregardless of what solute is used. Vapor pressure depression depends only on the solvent and the total concentration of solutes. That is it is a colligative property.
The amount of vapor pressure depression is summarized by Raoult's Law, named after François-Marie Raoult. Raoult found that the vapor pressure of a solution was proportional to the mole fraction of the solvent:
Psoln = XsolvPsolv
or:
DP = XsolPsolv
where Psoln is the vapor pressure of the solution, Xsolv is the mole fraction for the solvent, Xsol is the total mole fraction for solutes, DP is the amount of depression, and Psolv is the normal vapor pressure of the pure solvent. Again note that the amount of depression depends on the solvent used but not on the identity of the solutes.
What is the vapor pressure of a solution of 10.5 g of sucrose, C12H22O11, in 100 mL of water at 25oC?
We need to know the moles of water and sucrose to calculate the mole fraction of water:
1 mole sucrose
10.5 g sucrose X --------------- = 0.0.306 mole sucrose
342.3 g sucrose
1 g water 1 mole water
100 mL water X ---------- X ------------- = 5.55 mole water
1 mL water 18.02 g water
5.55 mole water
Xsolv = ------------------ = 0.995
5.55 + 0.0306 mole
Then by Raoult's Law:
Psoln = XsolvPsolv = (0.995)(23.8 torr) = 23.7 torr
What is the vapor pressure depression at 25oC when 6.44 g of NaCl is dissolved in 100 mL of water?
Note here that NaCl is an ionic salt and dissolves to form sodium ions and chloride ions. The vapor pressure depression depends on the total amount of dissolved nonvolatile materials so:
100 mL water = 5.55 moles water (see above)
1 mole NaCl
6.44 g NaCl X ------------ = 0.110 mole NaCl
58.44 g NaCl
0.110 mole NaCl = 0.110 mole Na+ ion = 0.110 mole Cl- ion
Thus the total moles of ions dissolved is 0.220 moles.
0.220 moles ions
Xsol = ------------------------ = 0.0381
5.55 moles + 0.220 moles
DP = XsolPsolv = (0.0381)(23.8 torr) = 0.906 torr
That is, the vapor pressure of water goes down by 0.906 torr when this amount of salt is added.
** !IMPORTANT! **
Colligative properties depend on the total concentration of solute species dissolved. Thus one has to consider what species are formed and how many when a given solute dissolves.Boiling point elevation is governed by the formula:
| Acetone | 1.71 | 56.2 |
| Benzene | 2.53 | 80.1 |
| Naphthalene | 5.80 | 217.7 |
| Water | 0.51 | 100 |
Note that calcium chloride will dissolve to give one calcium ion and two chloride ions! Therefore the total molality of solutes is three times the molality of CaCl2. The molality of CaCl2 is:
1 mole CaCl2
118 g CaCl2 X -------------- = 1.07 mole CaCl2
110.98 g CaCl2
And so:
1.07 mole CaCl2
msol = 3 X ---------------- = 3.78 m
0.846 kg water
DTb = kbmsol = (0.51 oC kg/mol)(3.78 mol/kg) = 1.93 oC
BP = 100 + DTb = 101.93 oC
It should be noted that not only is the freezing point depressed when a solute is added but also the range of temperatures over which freezing occurs is broadened. A pure solvent will freeze, or melt, over a small range. It will begin to freeze as it is cooled when the temperature reaches the freezing point. By the time all the liquid is frozen the temperature will have dropped only one or two degrees. For a solution the temperature will have dropped by a greater amount by the time all the solution has frozen.
This is a consequence of the fact that as a solution freezes it tends to form pure crystals of solvent leaving the solute behind in the remaining liquid. This means that as it freezes the remaining liquid portion becomes more concentrated. That means that the freezing point of the remaining liquid is even lower than before. Thus the temperature will have to drop to freeze the entire solution. This behavior is shown in the cooling curve formed when the temperature is plotted against time as the material is cooled.
The freezing point depression is governed by a formula similar to the one used for boiling point elevation:
| Acetone | 2.40 | -95.35 |
| Benzene | 5.12 | 5.5 |
| Naphthalene | 6.94 | 80.5 |
| Water | 1.86 | 0 |
1 mole benzoic acid
45.7 g benzoic acid X ---------------------
122.13 g benzoic acid
= 0.374 mole benzoic acid
0.374 mole benzoic acid
msol = ----------------------- = 1.50 m
0.250 kg benzene
DTf = kfmsol = (5.12 oC kg/mol)(1.50 mol/kg) = 7.68 oC
FP = 5.5 - DTf = -2.18 oC
What is happening here. First the cellulose acetate is a semipermeable membrane. That is it contains small holes that allow small molecules, such as water, to pass but prevents the passage of larger molecules, such as sugar. The rise in liquid level in the tube is due to the net flow of water into the bag. When the level reaches a certain height the weight of the liquid in the tube counteracts the force that is causing the flow inward. We can say that there must be a pressure across the membrane that can support the column of liquid in the tube. This pressure is called osmotic pressure and this phenomenon is called osmosis.
What causes this pressure? Something is causing more molecules of water to move through the small holes in the membrane going into the bag rather than out of the bag. One explanation is that there is a higher concentration of water outside than there is inside and therefore it is more likely that a water molecule will enter a membrane hole from the outside rather than from the inside. Another explanation is that higher concentrated solutions have higher potential energy. Therefore any process that will dilute a solution is spontaneous. Therefore there is more of a tendency for water to enter the bag and dilute the sugar solution rather than the opposite. This tendency can be counteracted by the higher potential energy of the column of liquid as it rises in the tube. This balancing of potential energies is similar to a balancing of pressures, thus the name osmotic pressure.
Osmosis occurs whenever a semipermeable membrane separates two solutions of different total concentration. This total concentration is called the osmolarity of the solution when expressed as a molarity, mol/L. When expressed as other forms of concentration it is refered to as the tonicity of the solution. The net flow is always from the region of lower osmolarity to the higher.
An important example of osmotic pressure is the behavior of living cells placed in various solutions. The cell membrane is semipermeable. Therefore when the cell is placed in a solution of different tonicity than the cytoplasm inside the cell there will be a pressure difference across the membrane. If the cell is placed in a solution of lower tonicity, a hypotonic solution, there is a net flow of water into the cell causing it to swell until it bursts. If the solution is higher in tonicity, hypertonic, water flows out of the cell and it shrivels up.
Both situations can of course lead to the death of the cell. Therefore keeping a balance in tonicity between the cells in living tissues and their surroundings is necessary for the health of the organism. A good example of this is the behavior of plants in a drought. The cells on the surface of the roots normally have a tonicity higher than that of the soil moisture. Therefore there is a net flow of water into the roots and from there the other tissues of the plant. If the soil drys out, the tonicity around the roots increases until water flows out of the plant into the soil. The cells in the plant tissues shrivel up and cause the plant to wilt. Without water the plant will eventually die.
Osmotic pressure is a colligative property in that the pressure depends only on the total concentration of solutes and not their identity. It is governed by the formula:
where P is the osmotic pressure across the membrane, R is the gas constant (0.08206 L atm/mol K), T is the Kelvin temperature, and M is the molarity of the solution or more precisely the difference in molarity between the two sides of the membrane. Since molarity is moles per unit volume, M = n/V, we can rearrange the formula as:
which is suspiciously like the ideal gas law. That suggests that the behavior of solute molecules dissolved in a solvent has similarities with the behavior of gas molecules moving through an open space.
Since the formulas for the colligative properties include a concentration term, determination of a property like freezing point depression or osmotic pressure for a solution can allow one to determine its concentration or related values.
P = MRT M = P/RT = (1.55 atm)/(0.08206 L atm/mol K)(296 K) M = 0.0638 mol/LIf we assume that the unknown does not dissociate into multiple ions or other species on solution then the total molarity is the moles of unknown per liter of solution:
0.0638 mole
0.254 L X ----------- = 0.0162 mole
1 L
1.55 g
----------- = 95.7 g/mol
0.0162 mole