2. Solutions A solution is a homogeneous mixture of two or more substance A solution is generally composed of two substances Solute: substance that is being dissolved and is in smaller quantity Solvent: A substance that dissolves the solute and is in larger quantity

3. Properties of Solution Solutions are homogeneous mixtures. The properties are uniform throughout the mixture. In a solution, proportions of solute and solvent can be varied. That is, the amount of solute and solvent can be changed. The solute is not visible to the unaided eye. This is because the solute breaks down to ionic or molecular size. The solute does not settle out of solution. The solute can not be filtered out of solution. The solute can be separated from solution by changing temperature. That is, by freezing or boiling, or evaporation

4. Factors That Affect Solubility What determines if a solute will dissolve in a given solvent? The nature of solute and solvent. " LIKE DISSOLVES LIKE". Water dissolves salts, but not oil or gasoline. Temperature: the higher the temperature, more solute will dissolve in a given solvent. Pressure: pressure has little or no effect on the solubility of a solid or liquid, but the solubility of gases is greatly affected by pressure (Henry's Law).

5. Henry's Law - the solubility of a gas in a liquid is directly proportional to the applied pressure. This means that higher the pressure, the more gas dissolves in a liquid. Carbonated soft drinks are examples of a gas (carbon dioxide) dissolved in a liquid (water) at high pressure. When a soft drink can is opened, the dissolved carbon dioxide escapes ( bubbles). If this can is left open for some time, the drink goes "flat". All of the carbon dioxide has escaped.

6. Factors That Affect the Rate of Dissolution  Particle size: the smaller the particle size, the faster the rate of dissolution. Powders dissolve faster than large lumps.  Temperature: the higher the temperature, the faster is the rate of dissolution. It is easier to dissolve sugar in a glass of hot tea than it is in a glass of iced tea.  Rate of stirring: stirring or agitation of the solute in a solvent causes it to dissolve faster.  Concentration: if a solution has some solute dissolved in it, the rate of dissolving additional solute will be slower.

7. Relative Terms for Expressing Solute Concentration Solubility is the amount of solute that can be dissolved in a given amount of solvent. Non-quantitaive terms: this means it does not contain any numbers but is based on relative strengths. Concentrated solution: the solution contains more solute than a dilute solution. Dilute solution: the solution contains less solute than concentrated solution. Unsaturated solution: the solution can dissolve more solute. Saturated solution: the solution contains the maximum amount of solute the solvent can dissolve Supersaturated solution: the solution contains more solute than a saturated solution. This is very unstable condition and slight disturbance causes the excess solute to settle out.

8. Solutions are made up of solute and solvent. Concentration terms are basically a relationship between the amount of solute present and the amount of solution. Terms like percent, parts per million (ppm), molarity and normality each describe the concentration of a solute in a solution. The particular concentration term to use is dependent on the eventual use of the solution. Molarity(moles of solute per liter of solution) and Normality (equivalents of solute per liter of solution) are generally used for calculations involving chemical reactions Percent, parts per thousand (ppt), parts per million (ppm) and parts per billion (ppb) are useful for describing concentrations on a mass or volume basis. Concentrated acids and bases are given in percent concentration Quantitative Terms for Expressing Solute Concentration

9. Molarity Defined as no of moles of solute/ litre of solution Molarity is often used when more precise measurements of the concentration of solutions are needed and the solutions are being use in reactions and volumetric measurements Molarity= no of moles/litre of solution Moles= grams of solute/ formula mass of solute  Grams of any solute required to prepare a required molar solution could also be calculated from above formula  Example: 120grams of NaCl is diluted to 750ml.what would be its molarity. 1. First write molarity formulas  Molarity= no of moles/litre  No of moles=grams of solute/formula mass of solute

10. 2. Next make data of all what you need Molarity= Moles= Litre= Grams=120g Formula mass 3. Next convert ml to litres(750ml/1000=litres) 4. Next determine the formula mass of NaoH Na=23, o=16, H=1 NaOH=40 5. Next calculate the moles of NaOH Moles of NaOH=120g/40amu =3.0moles 6. Finally calculate Molarity Molarity= 3 moles/0.75litre=4.0M

11. Normality Normality is often used when more precise measurements of the concentration of solutions are needed and the solutions are being use in reactions of acids and bases. Normality= Equivalents of solute/litre of solution Equivalents= grams of solute/equivalent mass of solute Equivalent mass = Formula mass of solute/no of replaceable H+ or OH+ ions Equivalents can be defined in several ways No of replaceable H ions No of replaceable OH- ions No of moles of electrons transferred in a chemical reaction (redox)

12. Example 296.4 g of Ca(OH)2 is diluted to 600ml. What would be normality of this solution Steps to follow 1. First write normality formulas Normality=equivalents of solute litres of solution Equivalents=grams of solute equivalent mass of solute Equivalent mass of solute=formula mass of solute no of replaceable H or OH- ions

13. 2. Next make a table of all data needed for formula 3. Next convert 600ml to litre(600/l000=litres) 4.Next determine the formula mass of Ca(OH)2 Ca= 40x1=40 0= 16x2=32 H= 1.0x2=2 Ca(OH)2= 74.1 amu 5. Next calculate the no of replaceable OH Ca(OH)2 has 2 replaceable OH 6. Next calculate equivalent mass of Ca(OH)2 Equivalent mass= 74.1/2=37.05

14. 7. Calculate equivalents of Ca(OH)2 Eq=296.4g/37.05=8 8.Fially calculate the normality of Ca(OH)2 Normality= Equivalents/litres of solution =8/0.6litres =13.3N

15. Percentage solutions Percentage of solution can be expressed as percentage of solute based on entire solution The concentration can be expressed as percent by mass % mass(w/w)=mass of solute x 100 mass of solution %mass (w/W)= mass of solute x100 Mass of solute+ mass of solvent  When dealing with two solutions, the concentration could also be expressed as % by volume % by volume (v/v)= Volume of solute X 100 Volume of solution % by volume (v/v)= Volume of solute X 100 Volume of solvent+ volume of solute

16.  Sometimes the concentration is expressed as combination of mass of solute and volume of solution % (m/V)= mass of solute X 100 Volume of solution  Concentration could also be expressed as parts of solute per million parts of solution  ppm= mass of solute X 10 ^6 Mass of solution

17. Dilution formula The dilution formula is used to dilute concentrated stock solutions to desired concentrations C1V1=C2V2 C1= initial concentration V1= the final concentration C2= Final concentration V1= final volume

18. The END