Types Concentration And Solubility of Solutions

Solutions are homogenous mixtures of two or more substances in a single phase. Most of solutions can be considered as having a majority ingredient called a solvent and one more minority ingredients called solutes. For the sake of simplicity in this unit, we shall consider only binary solutions. Here each component may be in solid, liquid or in gaseous state and, therefore, several types of possible solutions are summarized in the following table:

DIFFERENT TYPES OF SOLUTIONS

CONCENTRATION UNITS

The concentration of a solute is the amount of solute dissolved in a given quantity of solvent or solution. The quantity of solvent or solution can be expressed in terms of volume or in terms of mass or molar mass. Thus there are several ways of expressing the concentration of a solution.

(a) Molarity (M): Moles of solute present in one litre solution.

$M=\frac{no.ofmolesofsolute}{volumeofsolutioninlitre}$

(b) Molality (m): Moles of solute present in one kilogram of solvent.

$M=\frac{no.ofmolesofsolute}{massofsolventinkg}$

(c) Normality (N): No. of equivalents present in one litre solution.

$N=\frac{no.ofequivalentsofsolute}{volumeofsolutioninlitre}$

(d) Mole fraction: The mole fraction of a component substance A(X_A) in a solution is defined as the moles of component substance divided by the total moles of solution.

$X_A=\frac{molesofsubs\tan ceA}{totalmolesofsolution}$

Mass percent: The mass percent of a component A in solution is defined as

Mass % of A = $\frac{massofA}{Totalmass}\times 100$

Part per million (PPM): It is defined as the parts of given component in one million parts of solution. Mathematically

$ppm=\frac{massofsolute}{massofsolution}\times {10}^6$

Illustration 1.

5 g of NaCl is dissolved in 1000 g of water. If the density of the resulting solution is 0.997 g per cc, calculate molality, molarity, normality and mole fraction of the solute.

Solution:

Mole of NaCl = $\frac{5}{58.5}$ = 0.0854 (Mol. wt. of NaCl = 58.5)

Molality = $\frac{Moles}{Wt.ofsolventingram}\times 1000$

= $\frac{0.0854}{1000}\times 1000$ = 0.0854 m

Volume of the solution= $\frac{Wt.ofsolutioningram}{Densityingram/cc}$= $\frac{1005}{0.997}cc$

Again by definition

Molarity = .

= $\frac{0.0854}{1.008}$ = 0.085 M

$\therefore$ Normality = 0.085 N (for NaCl, eq. wt. = mol. wt)

Further, Mole of H₂O = $\frac{1000}{18}$ = 55.55

(1000 gram of water = 1000 ml of water, because density = 1 g/cc)

Total mole = Mole of NaCl + Mole of H₂O

= .0854 + 55.55 = 55.6354

Mole fraction of NaCl = $\frac{7.85}{98.66}$ = $\frac{0.0854}{55.6409}$ = 1.535 ´ 10^–3

SOLUBILITY OF GASES

We are familiar that gases are completely miscible with each other. Gases also dissolve in liquids and solids. For example, soda-water contains carbon dioxide dissolved in water under high pressure. Oxygen is sufficiently soluble in water to allow survival of aquatic life in lakes, rivers and oceans. An example of dissolution of gas in a solid is the solubility of hydrogen gas in palladium.

The solubility of a gas in a liquid is determined by several factors. In addition to the nature of the gas and the liquid, solubility of the gas depends on the temperature and pressure of the system. The solubility of a gas in a liquid is governed by Henry's law which states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas. Dalton, a contemporary of Henry, also concluded independently that the solubility of a gas in a liquid solution is a function of the partial pressure of the gas. If we use the mole fraction of the gas in the solution as a measure of its solubility, then:

Mole fraction of the gas in a solution is proportional to the partial pressure of the gas.

Or, partial pressure of the gas in solution = K_H x mole fraction of the gas in solution

Here K_H is Henry's law constant

or, p = K_Hx

If we draw a graph between partial pressure of the gas versus mole fraction of the gas in solution, then we should get a plot of the type as shown in figure.

Different gases have different K_H values at the same temperature. This suggests that K_H is a function of the nature of the gas. Table gives K_H values of some common gases at specified temperature

Values of Henry's law constant (K_H) for some selected gases in water

It is obvious from figure that the higher the value of K_H at a given pressure, the lower is the solubility of the gas in the liquid. It can be seen from table that K_H value for both N₂ and O₂ increases with increase in temperature indicating that solubility of gases decreases with increase of temperature. It is due to this reason that aquatic species are more comfortable in cold waters rather than warm waters.

Illustration 2.

If N₂ gas is bubbled through water at 293 K, how many millimoles of N₂ gas would dissolve in 1 litre of water. Assume that N­₂ exerts a partial pressure of 0.987 bar. Given that Henry's law constant for N₂ at 293 K is 76.84 kbar.

Solution:

The solubility of gas is related to its mole fraction in the aqueous solution. The mole fraction of the gas in the solution is calculated by applying Henry's law. Thus,

$x_{N_2}=\frac{p_{N_2}}{K_H}=\frac{0.987bar}{76480bar}=1.29\times {10}^{-5}$

As 1 litre water contains 55.5 mol of it, therefore, if n represents number of moles of N₂ in solution,

$x_{N_2}=\frac{nmol}{nmol+55.5mol}=\frac{n}{55.5}=1.29\times {10}^{-5}$

Thus, n = 1.29 ´ 10^-5 x 55.5 mol = 7.16 x 10^-4 mol

$=7.16\times {10}^{-4}mol\times \frac{1000mmol}{1mol}$

= 0.716 m mol

Henry's law finds several applications in industry and explains some biological phenomenon. Notable among these are:

(i) To increase the solubility of CO₂ in soft drinks and soda water, the bottle is sealed under high pressure.

(ii) To minimize the painful effects accompaynig the decompression of deep sea divers, oxygen diluted with less soluble helium gas is used as breathing gas.

(iii) In lungs, where oxygen is present in air with high partial pressure, haemoglobin combines with oxygen to form oxyhaemoglobin. In tissues where partial pressure of oxygen is low, oxyhaemoglobin releases oxygen for utilization in cellular activities.

SOLID SOLUTIONS

Solid solutions are formed by mixing two solid components in the molten state in appropriate proportion and then cooling the molten mass. Solid solutions are of two types: substitutional solid solutions and interstitial solid solutions. In a substitutional solid solution, atoms, molecules or ions of one substance take the place of similar species of other substance in its crystal lattice as shown in figure (a). Brass, bronze, monel metal and steel are familiar examples of this type of solid solution.

Interstitial solid solutions constitute the other type and are formed by placing atoms of one kind into voids or interstices, that exist between atoms in the host lattice. This is illustrated in figure (b).

Tungsten carbide, WC, an extremely hard substance, is an example of interstitial solid solution. Here tungsten atoms are arranged in a face-centred cubic pattern with carbon atoms in the octahedral holes, where these are surrounded by six tungsten atoms placed at the vertices of the octahedron. Tungsten carbide has many industrial uses in making of cutting and grinding tools.