SOLUTION

S

SUBMITTED BY:- TANMAYEE BASUCLASS:- XII ‘B’

SUBMITTED TO:-MRS. JATINDER KAUR MAM{PGT CHEMISTRY}

In chemistry, a solution is a homogeneous mixture composed of only one phase. In such a mixture, a solute is a substance dissolved in another substance, known as a solvent. The solvent does the dissolving. The soln. more or less takes on the characteristics of the solvent including its phase, and the solvent is commonly the major fraction of the mixture.

CHARACTERISTICS:-

A soln. is a homogeneous mixture.

A soln. is a single phase system. The particles of solute in soln. can’t be seen by naked eye.

The soln. does not allow beam of  light to scatter.

A soln. is stable. The solute from the soln. can’t be separated by filtration (or mechanically).

TYPES:-

1.Gas If the solvent is a gas, only gases are dissolved under a given set of conditions. An example of a gaseous soln. is air (oxygen and other gases dissolved in nitrogen). Since interactions b/w molecules play almost no role, dil. gases form rather trivial solns. In part of the literature, they are not even classified as solns., but addressed as mixtures.

2.LiquidIf the solvent is a liquid, then gases,

liquids, and solids can be dissolved. Here are some examples:

Gas in liquid: Oxygen in water Carbon dioxide in water – the soln. is accompanied by a chemical rxn (formn. of ions). The visible bubbles in carbonated water is not the dissolved gas, but only an effervescence of carbon dioxide that has come out of soln; the dissolved gas itself is not visible since it is dissolved on a molecular level.

Liquid in liquid: The mixing of two or more substances of the same chemistry but diff conc.s to form a const. (Homogenization of soln.s)

Alcoholic beverages are basically soln.s of ethanol in water.

Solid in liquid: Sucrose (table sugar) in water Sodium chloride (table salt) or any other salt in water, which forms an electrolyte: When dissolving, salt dissociates into ions.

3. Solid If the solvent is a solid, then gases,

liquids and solids can be dissolved. Gas in solids:

Hydrogen dissolves rather well in metals, especially in palladium.

Liquid in solid: Mercury in gold, forming an amalgam Hexane in paraffin wax

Solid in solid: Steel, basically a soln. of carbon atoms

in a crystalline matrix of iron atoms. Alloys like bronze and many others. Polymers containing plasticizers.

1. Mass percentage2. Volume percentage3. Mass by volume

percentage4. Parts per million5. Mole fraction6. Molarity7. Molality

EXPRESSING

CONCENTRATION

OF SOUTIONS

1. MASS PERCENTAGE

Mass percentage is calculated as the mass of a component div. by the total mass of the mixture, multiplied by 100%.

Also Known as: mass percent, (w/w)%

Example:Ordinary bleach is 5.25% NaOCl by mass, which means each 100 g of bleach contains 5.25 g NaOCl.

Mass Percent Formula is expressed as :-

 It is mathematically expressed as :-

2. VOLUME-VOLUME PERCENTAGE

 Volume-Volume percentage (v/v %) is a measure of conc. of a substance in soln. expressed as the ratio of the volume of the soln. to the total volume of the soln. multiplied by 100%.

Examples: Wine has a typical value of alcohol content v/v % of 12%. This means there is 12 mL of ethanol for every 100 mL of wine.

Volume Percent Formula is expressed as :-

3. WEIGHT/VOLUME PERCENT Another variation on percentage conc. is weight/volume percent or mass/volume percent. This variation measures the amount of solute in grams but measures the amount of soln. in millilitres.

An example would be a 5%(w/v) NaCl solution. It contains 5 g of NaCl for every 100. mL of solution.

Per cent of solute mass by volume

=Mass of solute/Volume of solution × 100

4. PARTS PER MILLION When the solute is present in trace quantities, it is convenient to express the conc. in parts per million (ppm).

It is defined as the quantity of the solute in grams present in 106 grams of the soln.

Example:- 10 ppm of SO2 in air means 10 mL of SO2 is present in 106 mL of air.

 ppm = Mass of solute/Mass of solute × 106

5. MOLE FRACTION

This method is used when the soln. is constituted by mixing two or more components.

It is defined as the ratio of no. of moles of one component to the total no of moles of the soln. (i.e., all the components).

In a binary soln.,Mole fraction of solute + Mole fraction of solvent = 1

Let n moles of solute (A) and N moles of solvent (B) be present in a solution.

Mole fraction of solute = n/N+n = XA

Mole fraction of solvent = N/N+n = XB

Thus,             XA + XB = 1

6. MOLARITY(Molar Concentration)

It is defined as the no. of moles of the solute per litre or per dm3 of the soln.

Let wA g of the solute of molecular mass mA be dissolved in V litre of soln. 

Molarity of the soln. = wA/mA×V

The unit of molarity is mol litre−1 or mol dm−3

                   

Molarity (M) = Number of moles of solutes/Number of litres of solution

 Molarity of the solution = wA/mA×V × 1000

7. MOLALITY

It is defined as the no. of the moles of the solute present in 1 kg of the solvent, It is denoted by m.

Let wA grams of the solute of molecular mass mA be present in wB grams of the solvent, then,

                Molality (m) = wA/mA×wB × 1000

Molality (m) = Number of moles of solute/Number of kilograms of the solvent

HENRY'S

LAW

According to this law at a const. temp., the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.

» The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.

Henry's law can be put into mathematical terms (at constant temperature) as:-

where p is the partial pressure of the solute in the gas above the soln., c is the conc. of the solute and kH is a constant with the dimensions of pressure divided by conc.

The const., known as the Henry's law const., depends on the solute, the solvent and the temp.

SOLUBILITY

Solubility is the property of a solid, liquid, or gaseous chemical substance  called solute to dissolve in a solid, liquid, or gaseous solvent to form a homogeneous soln. of the solute in the solvent.

The solubility of a substance fundamentally depends on the physical and chemical properties of the used solute and solvent as well as on temp., pressure and the pH of the soln.

FACTORS AFFECTING SOLUBILITY 

1.Temperature:- The solubility of a given solute in a given solvent typically depends on temp. As the temp. of a liquid increases, the solubilities of gases in that liquid decrease and vice versa.

2.Pressure :- For condensed phases (solids and liquids), the pressure dependence of solubility is typically weak and usually neglected in practice.

SOLUBILITY OF GASES 

Henry's law is used to quantify the solubility of gases in solvents. The solubility of a gas in a solvent is directly proportional to the partial pressure of that gas above the solvent. This relationship is written as:

RAOULT’S

LAW

According to this law, the partial pressure of any volatile constituent of a soln. at a const. temp. is equal to the vapour pressure of pure constituent multiplied by the mole fraction of that constituent in the soln.

Let a mixture (soln.) be prepared by mixing nA moles of liquid A and nB moles of liquid B. Let pA and pB be the partial pressures of two constituents A and B in soln. and pA

0 and pB0 the

vapour pressures in pure state respectively.

Thus, according to Raoult’s law,      pA = nA/nA+nB pA

0 = mole fraction of A × pA

0 = XApA0

And pB = nB/nA+nB pA0 = mole

fraction of B × pB0 = XBpB

0

If the total pressure be P, thenP = pA + pB

        = nA/nA+nB pA0 + nB/nA+nB pA

0

= XAPA0 + XBPB

0

P = XAPA0 + XBPB

0

IDEAL AND

NON-IDEAL

SOLUTIONS

Ideal solutions Non-ideal solutions

Positive deviation from Raoult’s law

Negative deviation from Raoult’s law

1. Obey Raoult’s law at every

range of concentration.

1. Do not obey Raoult’s law.

1. Do not obey Raoult’s law.

2. Δ Hmix = 0; is neither evolved

nor absorbed during dissolution.

2. Δ Hmix>0. Endothermic dissolution; heat is

absorbed.

2. Δ Hmix<0. Exothermic

dissolution; heat is evolved.

3. ΔVmix = 0; total volume of

solution is equal to sum of

volumes of the components.

3. Δ Vmix > 0. Volume is increased

after dissolution.

3. Δ Vmix <0. Volume is

decreased during dissolution.

4. A—A, A—B, B—B interactions

should be same

4. A—B attractive force should be

weaker than A—A and B—B

attractive forces.

4. A—B attractive force should be greater than A—

A and B—B attractive forces.

Examples:Dil. solns;C6H6 + toluene;n-C6H14 + n-C7H16;chlorobenzene + bromobenzene;n-butyl chloride + n-butyl bromide.

Examples:oacetone + C2H5OH;oacetone + CS2;oH2O + CH3OH;oH2O + C2H5OH;oCCl4 + toluene;oCCl4 + CHCl3;oacetone + C6H6;oCCl4 + CH3OH;oCyclohexane + C2H5OH

Examples:oacetone + aniline;oacetone + CCl3;oCH3OH + CH3COOH;oH2O + HNO3;oCCl3 + diethyl ether,oH2O + HCl;oCH3COOH+ pyridine;oCCl3 + C6H6

VAPOUR

PRESSURE

OF LIQUID

SOLUTIONS

1.  VAPOUR PRESSURE OF LIQUID-LIQUID SOLUTIONS 

If  P  = total vapour pressure pA  = partial vapour pressure of component 1.

pB  = partial vapour pressure of component 2.

The quantitative relationship b/w the partial pressures (  pA  ,  pB  ) and the corresponding mole fractions (XA, XB ) of the two components 1 & 2 respectively, was given by “Marte Raoult” known as Raoult’s law

FROM DALTON’S LAW OF PARTIAL PRESSURE

The plot shows that the graph of pA &,pB verses the mole fractions XA &XB for a soln. is linear hence pA &,pB are directly proportional to XA &XB respectively.

In Figure, the dashed lines I & II represent the partial pressures of components, while the marked Line III represents the total vapour pressure of the soln.

2.  VAPOUR PRESSURE OF SOLUTIONS OF SOLIDS IN LIQUIDS

In a pure liquid the entire surface is filled by only liquid molecules, but when we add a non-volatile solute to a liquid soln., a part of the liquid surface is also occupied by these solutes. This reduces the space to the liquid molecules to evaporate out resulting the lowering in vapour pressure because only solvent part of the soln. can vaporise and creates vapour pressure, but not the solute part being non-volatile.

1.Relative lowering of vapour pressure

2.Elevation of boiling point3.Depression of freezing

point4.Osmotic pressure.

COLLIGATIVE

PROPERTIES

The word Colligative is derived from the Latin ’Colligatus’ meaning bound together. 

This no. can be related to the various units for conc. of solns.Colligative properties are mostly studied for dil. soln.s, whose behaviour may often be approximated as that of an ideal soln.

In chemistry, Colligative Properties are properties of soln.s that depend upon the ratio of the no. of solute particles to the no. of solvent molecules in a soln. We will only consider those properties which result because of dissolution of non -volatile solute in a volatile liquid solvent. They are independent of the nature of the solute particles, and are due to the dil.n of the solvent by the solute.

1. RELATIVE LOWERING OF VAPOUR PRESSURE

When a non-volatile solute is added to a solvent, the vapour pressure is lowered due to the foll. reason:

Percentage surface area occupied by the solvent decreases. Thus the rate of evaporation and vapour pressure decreases. The solute molecules occupy the surface, and so the per cent surface area occupied by the solvent decreases.

Derivation Of Raoult’s Law For A Dilute Solution

It states that the relative lowering in vapour pressure of a dil. soln. is equal to mole fraction of the solute present in the soln.

If n moles of solute be dissolved in N moles of the solvent, the mole fraction of the solute will be n/n+N.

According to Raoult’s law, [p0–ps/p0] = n/n+N.

This is the mathematical expression for modified form of Raoult’s law. The above r/l can be written as,

      p0/P0–ps = n+N/n = 1 + N/n 

or p0/p0–ps – 1 = N/n or p0/p0–ps= N/n

or p0–ps/ps = n/N = wA/mA × mB/wB

2. ELEVATION OF BOILING POINT

The b. pt. of a liquid is the temp. at which its vapour pressure is equal to the atmospheric pressure.

The vapour pressure of a liquid is lowered when a non-volatile solute is added to it. Hence, the b. pt. of the solvent is elevated by the addition of non-volatile solute.

The difference in the b. pt. of the soln. & the b. pt. of the pure solvent is termed elevation of b. pt.

Elevation of b. pt., (ΔT) = B. pt. of the soln. – B. pt. of pure solvent

P0C = atmospheric pressure T0, T1 and T2= b. pts. of pure solvent,

soln. I and soln. II respectively P0, P1 and P2 = vapour pressure of pure

solvent, soln. I and soln. II at temp. T0  respectively

T2–T0/T1–T0 = P0–P2/P0–P1

or ΔT2/ΔT1 = ΔP2/ΔP1

  or ΔT ∝ ΔP   From Raoult’s law for dil. soln.

       p0–ps/p0 = wA/mA.mB/wB  

 (ps = vapour pressure of soln.)

  or p0–ps = wA/mA.mB/wB . p0

For the pure solvent, P0 (its vapour pressure at the b. pt.) and mB (its molecular mass) are const.  

Therefore,             p0 – ps ∝ wA/mAwB

or Δp ∝ ΔT ∝ wA/mAwB

orΔT =K.wA/mAwB  …(i)

where K is a const., c/a as elevation const. 

If wA/mA and wB = 1000 g.

        ΔT = K/1000 = Kb

Kb is c/a molal elevation const. It is defined as the elevation in b. pt. produced when 1 mol of the solute is dissolved in 1000 g of the solvent.

Thus, K = 1000 Kb

Putting this value in Eq. (i),         ΔT° = 1000 Kb wA/mAwB

  …(ii) or      ΔT = Molality × Kb

    (since wA/mA×wB × 1000 = molality).

3. DEPRESSION OF FREEZING POINT

Freezing point of a substance is defined as the temp. at which the vapour pressure of its liquid is equal to the vapour pressure of the corresponding solid.

Since the addition of a non-volatile solute always lowers the vapour pressure of solvent, therefore, it will be in equilibrium with solid phase at a lower pressure and hence at a lower temp.

The difference b/w the f. pts. of the pure solvent and its soln, is c/a depression of f. pt.

Depression of f. pt. (ΔT) = F. pt. of the solvent – F. pt. of the soln.

CFB is a curve for solid solvent. The solvent, soln.I and soln.II vapour

pressure curves meet CFB curve at pts.B, F and C respectively.

Thus, T0, T1 and T2 for solvent, soln. I and soln. II are, thus, P0, P1 and P2 respectively.

T0–T2/T0–T1 = P0–P2/P0–P1

or ΔT2/ΔT1 = ΔP1/ΔP2           or             ΔT ∝ ΔP

From Raoult’s law for dil. soln.s,  p0–ps/p0 = wA/mA . mB/wB 

or   p0–ps = wA/mA . mB/wB . P0

For the pure solvent, p0 and mB are const. Therefore,

p0 – ps ∝ wA/mAwB 

or Δp ∝ wA/mAwB

or Δp ∝ ΔT ∝ wA/mAwB

or ΔT =K. wA/mAwB   …(i)

where K is a const., c/a depression const.

When wA/mA (or mole of solute) and wB = 1 g

  ΔT = KIf wA/mA and wB = 100 g,

 ΔT = K/100 = K'Thus, K = 100 K’ Putting this value in Eq. (i), ΔT =

100K'×wA/mA×wB                    …(ii)

If wA/mA and wB = 1000 g

   ΔT = K/1000 = Kf 

Kf is c/a molal depression const. It is defined as the depression of f. pt. produced when 1 mole of solute is dissolved in 1000 g of the solvent.

Thus, K = 1000Kf

Putting this value in Eq. (i), ΔT° = 1000 Kf wA/mAwB               

                     …(iii)  or  ΔT = molality × Kf

4. OSMOTIC PRESSURE.

OSMOSIS: When a semi-permeable membrane is placed b/w a soln. and a solvent, it is observed that solvent molecules enter the soln. and the volume of soln. increases.

It is also observed that if two soln.s of unequal conc.s are separated by a semi permeable membrane, the solvent molecules from a soln. of lower conc. move towards a soln. of higher conc.

Osmotic Pressure is the pressure which needs to be applied to a soln. to prevent the inward flow of water across a semi permeable membrane.

The phenomenon of osmotic pressure arises from the tendency of a pure solvent to move through a semi-permeable membrane and into a soln. containing a solute to which the membrane is impermeable.

By Van't Hoff's law, Osmotic Pressure Formula can be calculated as:

Where ,i = number of ions formed by dissociation of a solute molecule

R = ideal gas const.T = absolute temp. (measured in K)

c = molal or molar conc.

The expression which relates osmotic pressure to conc. & temp. is similar to the ideal gas eqn:

The Osmotic Pressure Formula of an ideal soln. can be calculated as follows:

Where ,P = osmotic pressure in mm

Hgn = no. of solute particlesc/M = molar conc. of the

substance in mol/kgR = universal gas const. T = absolute temp.

TYPES OF SOLUTIONS BASED ON SOLUTE CONCENTRATION

The two soln.s are equal in their solute concs. We say that they are isotonic to each other.

The soln. in the bag contains less solute than the soln. in the beaker. The soln. in the bag is hypotonic (lower solute conc.) to the soln. in the beaker. The soln. in the beaker is hypertonic  (higher solute conc.) to the one in the bag. Water will move from the hypotonic soln. into the hypertonic soln.

REVERSE OSMOSIS

Reverse osmosis (RO) is a water purification technology that uses a semi permeable membrane.

In RO, an applied pressure is used to overcome osmotic pressure, that is driven by chemical potential. RO can remove many types of molecules and ions from solutions and is used in both industrial processes and to produce potable water.

The result is that the solute is retained on the pressurized side of the membrane and the pure solvent is allowed to pass to the other side.

To be "selective," this membrane should not allow large molecules or ions through the pores (holes), but should allow smaller components of the solution (such as the solvent) to pass freely.

Reverse osmosis is most commonly known for its use in drinking water purification from seawater, removing the salt and other effluent materials from the water molecules.