1. Solutions

2. Definitions n A solution is a homogeneous mixture n Solvent is the liquid in which the solute is dissolved n an aqueous solution has water as solvent n A saturated solution is one where the concentration is at a maximum - no more solute is able to dissolve. n A solution is composed of a solvent which is the dissolving medium and a solute which is the substance dissolved. n In a solution there is an even distribution of the molecules or ions of the solute throughout the solvent.

3. n The concentration of a solution can be expressed in a variety of ways (qualitatively and quantitatively). n Formerly, the concentration of a solution can be expressed in four ways: –Molarity(M): moles solute / Liter solution –Mass percent: (mass solute / mass of solution) * 100 –Molality (m) - moles solute / Kg solvent –Mole Fraction( c A ) - moles solute / total moles solution

4. Qualitative Expressions of Concentration A solution can be qualitatively described as n dilute: a solution that contains a small proportion of solute relative to solvent, or n concentrated: a solution that contains a large proportion of solute relative to solvent.

5. Semi-Quantitative Expressions of Concentration n A solution can be semi-quantitatively described as n Unsaturated: a solution in which more solute will dissolve, or n Saturated: a solution in which no more solute will dissolve. n The solubility of a solute is the amount of solute that will dissolve in a given amount of solvent to produce a saturated solution.

6. Percent Composition (by mass) n We can consider percent by mass (or weight percent, as it is sometimes called) n We need two pieces of information to calculate the percent by mass of a solute in a solution: n The mass of the solute in the solution. n The mass of the solution. n Use the following equation to calculate percent by mass:

7. Molality n Molality, m, tells us the number of moles of solute dissolved in exactly one kilogram of solvent. (Note that molality is spelled with two "l"'s and represented by a lower case m.) n We need two pieces of information to calculate the molality of a solute in a solution: n The moles of solute present in the solution. n The mass of solvent (in kilograms) in the solution. n To calculate molality we use the equation: n

8. Molarity n Molarity tells us the number of moles of solute in exactly one liter of a solution. (Note that molarity is spelled with an "r" and is represented by a capital M.) n We need two pieces of information to calculate the molarity of a solute in a solution: n The moles of solute present in the solution. n The volume of solution (in liters) containing the solute. n To calculate molarity we use the equation:

9. Mole per liter solutions n A mole per litre (mol/l) solution contains one mole of a solute dissolved in, and made up to 1 litre with solvent.

10. Mole: n A counting unit n Similar to a dozen, except instead of 12, it’s 602 billion trillion 602,000,000,000,000,000,000,000 n 6.02 X 10 23 (in scientific notation) n A Mole of Particles Contains 6.02 x 10 23 particles n 1 mole C= 6.02 x 10 23 C atoms n 1 mole H 2 O= 6.02 x 10 23 H 2 O molecules n 1 mole NaCl = 6.02 x 10 23 NaCl “molecules” n The mole is the SI base unit that measures an amount of substance. n One mole contains Avogadro's number (approximately 6.023×10 23 ) (number of atoms or molecules).

11. Preparation of mol/l solution n To preparation a mol/l solution, use the following formula: Number of grams to be dissolved in 1 litre of solution= Required mol/l solution X Molecular mass of substance

12. Example n Make a solution of 50 ml of sodium chloride, 0.15 mol/l: n Required mol/l concentration= 0.15 n Molecular mass of NaCl= 58.44 n Therefore 50 ml NaCl, 0.15 mol/l contains: 0.15X 58.44 X 50 = 0.438 g of the chemical. 1000 substance dissolved in 50 ml of solvent.

13. Make liter of sodium chloride (NaCl), 1 mol/l n To make 1 litre of sodium chloride (NaCl), 1 mol/l: n Required mol/l concentration= 1 n Molecular mass of NaCl= 58.44 n Therefore 1 lit NaCl, 1 mol/l contains: n 1 X 58.44= 58.44 g of the chemical dissolved in 1 litre of solvent.

14. To make 1 litre of sodium chloride, 0.15 mol/l (physiological saline)

15. Conversion a percentage solution into a mol/l solution: By the following formula: n Mol/l solution = g% (w/v) solution X 10 n Molecular mass of the substance Examples: n To convert a 4% w/v NaOH solution into a mol/l solution: n Gram % solution = 4 n Molecular mass of NaOH= 40 n Conversion to mol/l= 4X10 = 1 n 40 n Therefore 4% w/v NaOH is equivalent to NaOH, 1 mol/l solution.

16. Convert a 0.9% w/v NaCl solution into a mol/l solution

17. Conversion of a normal solution into a mol/l solution: n By the following formula: mol/l solution= Normality of solution Valence of substance n Examples: Convert 0.1 N (N/10) HCl into a mol/l solution: Normality of solution= 0.1 Valence of HCl= 1 Conversion to mol/l= 0.1/1= 0.1 n Therefore 0.1 N HCl is equivalent to HCl, 0.1mol/l solution.

18. To convert 1 N Na 2 CO 3 into a mol/l solution (valence =2)

19. How to dilute solutions and body fluids In the laboratory it is frequently necessary to dilute solutions and body fluids to reduce its concentrations. n Diluting solutions: n A weaker solution can be made from a stronger solution by using the following formula: n Volume (ml) of stronger solution required= R X V n O n Where: R= concentration of solution required n V= volume of solution required. n O= strength of original solution.

20. n Examples: n To make 500 ml of NaOH, 0.25 mol/l from a 0.4 mol/l solution: C= 0.25 mol/l, V= 500 ml, S=0.4 mol/l ml of stronger solution required: 0.25 x 500 = 312.5 ml 0.4 n Therefore, measure 312.5 ml NaOH, 0.4 mol/l and make up to 500 ml with distilled water.

21. Example 1 n Make 100 ml glucose, 3 mmol/l in 1 g/l benzoic acid from glucose 100 mmol/l solution : n C= 3 mmol/l V= 100 ml S= 100 mmol/l n ml of stronger solution required= 3 x 100 = 3 100 n Therefore, measure 3 ml of glucose, 100 mmol/l and make up to 100 ml with 1 g/l benzoic acid.

22. Example 2 n To make 500 ml H 2 SO 4, 0.33 mol/l from concentrated H 2 SO 4 which has an approximate concentration of 18 mol/l: C= 0.33 mol/l, V= 500 ml S= 18 mol/l n ml of stronger solution required= 0.33 x 500 = 9.2 18 n Therefore, measure 9.2 ml conc. H 2 SO 4, and slowly add it to about 250 ml of distilled water in a volumetric flask. Make up to 500 ml with DW.

23. Make 1 litre of HCl, 0.01 mol/l from a 1.0 mol/l solution:

24. Diluting body fluids and calculating dilutions n To prepare a dilution or series of dilutions of a body fluid: Examples: To make 8 ml of 1 in 20 dilution of blood: n Volume of blood required= 8/20 = 0.4 ml Therefore, to prepare 8 ml of a 1in 20 dilution, add 0.4 ml of blood to 7.6 ml of diluting fluid. 2- To make 4 ml of a 1 in 2 dilution of serum in physiological saline: Volume of serum required= 4/2 = 2 ml Therefore, to prepare 4 ml of a 1in 2 dilution, add 2 ml of serum to 2 ml of physiological saline.

25. To calculate the dilution of a body fluid Examples: 1- Calculate the dilution of blood when using 50 µl of blood and 950 µl of diluting fluid: n Total volume of body fluid and diluting fluid= 50+950=1000 µl n Therefore, dilution of blood: 1000/50= 20 n i. e. 1 in 20 dilution. 2- Calculate the dilution of urine using 0.5 ml of urine and 8.5 ml of diluting fluid (physiological saline): n Total volume of urine and diluting fluid= 8.5+0.5=9 ml n Therefore, dilution of urine: 9.0/0.5= 18 n i. e. 1 in 18 dilution.