2. Copyright Sautter 2003

3. SOLUTIONS & CONCENTRATIONS WHAT IS A SOLUTION ? WHAT IS CONCENTRATION & HOW IS IT MEASURED ?

4. IS DISSOLVING A CHEMICAL OF A PHYSICAL PROCESS ? IF SUGAR IS DISSOLVED IN WATER IS ITS CHEMICAL NATURE CHANGED ? NO !! IF THE SOLUTION IS DRIED, A WHITE, SWEET CRYSTALLINE SUBSTANCE (SUGAR) IS OBSERVED. THEREFORE DISSOLVING IS A PHYSICAL CHANGE. THE CHEMICAL PROPERTIES OF A SUBSTANCE MUST BE CHANGED IN ORDER FOR A CHEMICAL CHANGE TO OCCUR !

5. A SOLUTION THEN IS THE RESULT OF THE PHYSICAL CHANGE CALLED DISSOLVING, BUT PRECISELY, WHAT IS A SOLUTION? THE OPPPOSITE OF A SOLUTION IS A SUSPENSION. MUDDY WATER IS A SUSPENSION. SUGAR DISSOLVED IN WATER IS A SOLUTION. CAN YOU NAME SOME DIFFERENTS BETWEEN THEM?

6. SOME OBSERVED DIFFERENCES DIFFERENCES IN CLARITY ! DIFFERENCES IN UNIFORMITY ! DIFFERENCES IN SETTLING ! DIFFERENCES IN FILTRATION !

7. CLARITY SOLUTIONS ARE TRANSPARENT * SUSPENSIONS ARE OPAQUE * * ALTHOUGH SOLUTIONS ARE TRANSPARENT (YOU CAN SEE THROUGH THEM) THEY MAY BE COLORED * OPAQUE MEANS CLOUDLY

8. UNIFORMITY UNFORMITY OR HOMOGENOUS MEANS THE SAME THROUGHOUT SOLUTIONS ARE HOMOGENOUS ( THE SAME AMOUNT OF DISSOLVED SUBSTANCE FOR THE SAME AMOUNT OF SOLUTION EVERYWHERE IN THE SYSTEM) SUSPENSIONS ARE HETEROGENOUS (THE AMOUNT OF SUSPENSED MATERIAL DIFFERS FROM PLACE TO PLACE) FOR EXAMPLE, THE AMOUNT OF MUD IS MORE CONCENTRATED AT THE BOTTOM THAN AT THE TOP

9. SETTLING OUT SOLUTIONS NEVER SETTLE. THE DISSOLVED MATERIAL WILL NOT FALL OUT OF THE SOLUTION (UNLESS THE TEMPERATURE IS CHANGED) IN SUSPENSIONS, EVENTUALLY THE SUSPENDED MATERIAL WILL FALL TO THE BOTTOM OF THE CONTAINER !

10. FILTRATION THE DISSOLVED MATERIAL IN A SOLUTION CANNOT BE FILTERED OUT BY ORDINARY MEANS (FOR EXAMPLE THE SALT CANNOT BE FILTERED FROM SEA WATER WITH FILTER PAPER) IN A SUSPENSION SUCH A MUDDY WATER, THE SUSPENDED SUBSTANCE CAN BE FILTERED LEAVING A CLEAR LIQIUD

11. SOLUTIONS VS. SUPSENSIONS SOLUTIONS (1) TRANSPARENT (2) HOMOGENOUS (3) WILL NOT SETTLE OUT (4) CANNOT BE FILTERED SUSPENSIONS (1) OPAQUE (2) HETEROGENOUS (3) WILL SETTLE OUT (4) CAN BE FILTERED FOG COLLOIDAL SUSPENSION

12. WHAT IS A SOLUTION ? DEFINITION: A SOLUTE DISSOLVED IN A SOLVENT (FOR EXAMPLE A SUGAR / WATER SOLUTION) SOLUTE IS THE DISSOLVED SUBSTANCE (THE SUGAR) SOLVENT IS THE DISSOLVING MEDIUM IN WHICH THE SOLUTE IS DISSOLVED (THE WATER)

13. WHAT COMBINATION OF PHASES CAN FORM SOLUTIONS ? THINK OF SOME DIFFERENT KINDS OF SOLUTIONS. THEY MUST HAVE THE PROPERTIES OF SOLUTIONS TO BE CONSIDERED. SOME COMMON COMBINATIONS: SOLID SOLUTE / LIQUID SOLVENT (SUGAR DISSOLVED IN WATER) LIQUID SOLUTE / LIQUID SOLVENT (ANTIFREEZE DISSOLVED IN WATER) GAS SOLUTE / LIQUID SOLVENT (SODA WATER – CARBON DIOXIDE DISSOLVED IN WATER) GAS SOLUTE / GAS SOLVENT ( AIR – OXYGEN DISSOLVED IN NITROGEN)

14. WHAT ARE THE GENERAL CLASSIFICATIONS OF SOLUTIONS ? THERE ARE GENERALLY TWO TYPES: ELECTROLYTES AND NON ELECTROLYTES ELECTROLYTIC SOLUTIONS ARE ELECTRICALLY CONDUCTIVE. THEY CONSIST OF IONIC SOLUTES DISSOLVED IN POLAR SOLVENTS NON ELECTROLYTIC SOLUTIONS ARE NON CONDUCTIVE AND THEY CONSIST OF MOLECULAR SOLUTES DISSOLVED IN NON POLAR SOLVENTS.

16. TYPES OF ELECTROLYTIC SOLUTIONS STRONG ELECTROLYTES SOLUTIONS IN WHICH ALL OF THE DISSOLVED SOLUTE FORMS IONS WEAK ELECTROLYTES SOLUTIONS IN WHICH ONLY A PERCENTAGE OF THE DISSOLVED SOLUTE FORMS IONS

17. OTHER WAYS TO CLASSIFY SOLUTIONS SATURATED SOLUTIONS NO MORE SOLUTE CAN BE DISSOLVED (SOLUBILITY LIMIT HAS BEEN REACHED) UNSATURATED SOLUTIONS ADDITIONAL SOLUTE CAN STILL BE DISSOLVED

18. WHAT DETERMINES THE SATURATION POINT OF A SOLUTION ? (1) THE TYPE OF SOLUTE AND SOLVENT USED (2) THE TEMPERATURE OF THE SOLUTION (Generally solids dissolve better at higher temperatures while gases dissolve more poorly) (3) GAS PRESSURE WHEN A GAS IS THE SOLUTE (HENRY’S LAW)* *THE SOLUBILITY OF A GAS IS DIRECTLY RELATED TO THE PRESSURE OF THAT GAS ABOVE THE SOLUTION Solubility = a constant x Pressure of the gas

19. How Temperature Effects Solubility Note that all substances (even solids), do not dissolve better at higher temperatures although most do.

20. HOW CAN THE SOLUTIONS COMPOSED OF THE SAME SUBSTANCES BE DIFFERENT ? FOR EXAMPLE, HOW CAN ONE AQUEOUS* SUGAR SOLUTION BE DIFFERENTIATED FROM ANOTHER ? DIFFERENT SOLUTIONS CONSISTING OF THE SAME SOLUTE / SOLVENT COMBINATIONS MAY BE DIFFERENT IN CONCENTRATION ! WHAT DOES CONCENTRATION MEAN ?? * AQUEOUS MEANS THAT WATER IS THE SOLVENT MEDIUM IN THE SOLUTION

21. CONCENTRATION CONCENTRATION REFERS TO A RATIO OF SOLUTE AMOUNT TO SOLVENT OR SOLUTION AMOUNT. FOR EXAMPLE, CONCENTRATED ORANGE JUICE MEANS THAT THE SOLUTE (THE ORANGE COMPONENT) IS PRESENT IN LARGE QUANTITY RELATIVE TO THE SOLVENT (THE WATER COMPONENT).

22. METHODS OF MEASURING SOLUTION CONCENTRATION DEPENDING ON THE UNITS OF MEASURE AND THE WHETHER SOLVENT OR SOLUTION QUANTITIES ARE MEASURED, CONCENTRATION CALCULATIONS VARY. WHEN THE SOLUTE QUANTITY IS MEASURED IN MOLES AND THE SOLUTION VOLUME IS MEASURED IN LITERS, THE CONCENTRATION IS EXPRESSED AS MOLARITY OR MOLES PER LITER OF SOLUTION. MOLARITY = MOLES OF SOLUTE / LITER OF SOLUTION

23. MOLARITY CALCULATIONS WHAT IS THE MOLARITY OF A SOLUTION WITH A VOLUME OF 2.0 LITERS AND CONTAINING 90.0 GRAMS OF GLUCOSE (C 6 H 12 O 6 ) ? THE DEFINITION OF MOLARITY IS MOLES OF SOLUTE PER LITER OF SOLUTION ! STEP I – FIND THE NUMBER OF MOLES OF SOLUTE PRESENT 90.0 GRAMS / 180 GRAMS PER MOLE OF C 6 H 12 O 6 GIVES 0.50 MOLES OF GLUCOSE STEP II – MOLARITY = MOLES / LITERS 0.50 MOLES / 2.0 LITERS = 0.25 M (CAPITAL M = MOLARITY)

24. MOLARITY CALCULATIONS (CONTINUED) HOW MANY MOLES OF ZINC CHLORIDE ARE CONTAINED IN 500 ML OF A 0.20 M SOLUTION? SINCE M = MOLES / LITERS, MOLES = M X LITERS 500 MLS = 0.500 LITERS MOLES = 0.20 M X 0.500 L = 0.100 MOLES

25. MOLARITY CALCULATIONS (CONTINUED) HOW MANY GRAMS OF SODIUM CHLORIDE (NaCl) ARE CONTAINED IN 250 MLS OF A 0.50 M SOLUTION ? STEP I – CALCULATION MOLES AS IN THE PREVIOUS PROBLEM MOLES = MOLARITY X LITERS MOLES = 0.50 M X 0.250 L = 0.125 MOLES STEP II – CONVERT MOLES TO GRAMS NaCl IS 58.5 GRAMS PER MOLE 0.125 MOLES X 58.5 GRAMS PER MOLE = 7.31 GRAMS OF NaCl ARE CONTAINED IN THE SOLUTION

26. MOLARITY CALCULATIONS (CONTINUED) HOW MANY MILLILITERS OF A 0.40 M SOLUTION ARE NEEDED TO OBTAIN 45.0 GRAMS OF GLUCOSE ? MOLARITY = MOLES / LITERS LITERS = MOLES / MOLARITY 45.0 GRAMS / 180 GRAMS PER MOLE = 0.25 MOLES LITERS = 0.25 MOLES / 0.40 M = 0.625 LITERS 0.625 LITERS = 625 MILLILITERS

27. SOLUTIONS AND DILUTION WHAT DOES DILUTION MEAN ? DILUTE MEANS LESS CONCENTRATED HOW CAN A SOLUTION BE DILUTED? BY THE ADDITION OF MORE SOLVENT (MOST OFTEN WATER) WHAT HAPPENS TO THE ORIGINAL CONCENTRATION OF THE SOLUTION ? IT IS REDUCED !

28. STARTING VOLUME V 1 STARTING MOLARITY M 1 FINAL VOLUME V 2 (ORIGINAL VOLUME PLUS ADDED WATER) FINAL MOLARITY M 2 M 1 V 1 = M 2 V 2

29. DILUTION CALCULATIONS THE DILUTION FORMULA: M 1 V 1 = M 2 V 2 M 1 = ORIGINAL CONCENTRATION OF SOLUTION V 1 = ORIGINAL VOLUME OF SOLUTION (ML OR L) M 2 = CONCENTRATION AFTER DILUTION V 2 = VOLUME OF SOLUTION OF AFTER DILUTION (ML OR L) (ORIGINAL VOLUME + VOLUME OF WATER ADDED)

30. DILUTION CALCULATIONS WHAT IS THE CONCENTRATION OF 200 ML OF A 0.50 M SUGAR SOLUTION AFTER 100 ML OF WATER HAVE BEEN ADDED? M 1 = 0.50 M, V 1 = 200 ML V 2 = 200 ML + 100 ML = 300 ML, M 2 = ? M 1 V 1 = M 2 V 2 (0.50) x (200) = M 2 x (300) M 2 = 0.33 M

31. DILUTION CALCULATIONS (continued) HOW MUCH WATER MUCH BE ADDED TO 500 ML OF A 2.0 M SOLUTION OF GLUCOSE TO DILUTE IT TO A CONCENTRATION OF 0.50 M ? M 1 = 2.0 M, V 1 = 500 ML V 2 = ?, M 2 = 0.50 M M 1 V 1 = M 2 V 2 (2.0) x (500) = (0.50) x V 2 V 2 = 2000 ML NEW VOL OF SOLUTION – ORIGINAL VOL = VOL OF WATER ADDED 2000 ML - 500 ML = 1500 ML ADDED

32. OTHER CONCENTRATION MEASURES MOLALITY The concentration of a solution can be measured in molality(m) units as well as molarity (M). Molality is defined as moles of solute divided by kilograms of solvent. Notice the differences between molality and molarity. Molarity uses a volume measure in liters, molality uses mass measures in kilograms. Molarity is based on the quantity of solution, molality is based on the quantity of solvent. Both however use moles of solute as the numerator term. Molality = moles of solute / kilograms of solvent Remember grams must be divided by 1000 to obtain kilograms!

33. Molality Problem: What is the molality of a solution containing 20.0 grams of glucose (C 6 H 12 O 6 ) in 100 grams of water? Solution: m = mole / Kg Moles of glucose = 20.0 / 180 = 0.111 m = 0.111 moles / 0.100 Kg = 1.11 molal Molality is common used in calculations involving the freezing and boiling points of solutions.

34. Mole Fractions Measure concentrations using mole fractions is used in the determination of the vapor pressure of solutions. It is the decimal percent of a component in mole terms. For example, a solution in which half the molecules are water and half are alcohol would contain a mole fraction of water equal to 0.50 and a mole fraction of alcohol equal to 0.50. The sum of the mole fractions present in a solution must equal 1.00. Mole fraction of X = moles of X / total moles present Capital X is used to represent mole fraction. Mole fraction contains no unit values but is simply a number value.

35. Mole Fractions Problem: A solution contains 25.0 grams of ethyl alcohol (C 2 H 5 OH) and 150.0 grams of water. What is the mole fraction of each component? Solution: X = moles / total moles For C 2 H 5 OH, moles = 25.0 / 46.0 =0.543 For H 2 O, moles = 150.0 / 18.0 =8.33 X C2H5OH = 0.543 / (0.543 + 8.33) = 0.0612 X H2O = 8.33 / (0.543 + 8.33) = 0.939 Check: 0.0612 + 0.939 = 1.000

36. Percent Solution by Mass Percent solution by mass is percent of the solution which is the solute. It follows the usual percent calculation of parts divided by the whole times 100 % % solution = (gram of solute/gram of solution) x 100% Problem: What is the percent solution of the solution in the previous problem? (25.0 grams of alcohol in 150.0 grams of water) Solution: % = (g solute / total grams) x 100 % % alcohol = (25.0 g / (25.0 + 150.0 g)) x 100 % % alcohol = 14.3 %

37. Normality Often, when dealing with acids and bases especially, concentrations are expressed in normality terms. Normality, for acids and bases is the molarity of the hydrogen or hydroxide ion in solution. One mole of hydrogen or hydroxide ions is called an equivalent. The weight of acid or base which contained one mole of hydrogen or hydroxide ions is called the equivalent weight of that substance. Examples: A 1.0 molar solution of HCl is a 1.0 normal solution A 1.0 molar solution of H 2 SO 4 is a 2.0 normal solution because each sulfuric acid releases two hydrogen ions. A 1.0 molar solution of H 3 PO 4 is a 3.0 normal solution because each phosphoric acid releases three hydrogen ions.

38. Normality A 1.0 molar solution of NaOH is a 1.0 normal solution A 1.0 molar solution of Ca(OH) 2 is a 2.0 normal solution because each calcium hydroxide releases two hydroxide ions. A 1.0 molar solution of Al(OH) 3 is a 3.0 normal solution because each aluminum hydroxide releases three hydroxide ions. The equivalent weight of HCL is 36.5 grams (its molecular weight) The equivalent weight of H 2 SO 4 is 49 grams (98 / 2 or half of its molecular weight) The equivalent weight of H 3 PO 4 is 32.7 grams (98 / 3 or one third of its molecular weight) Normality = equivalents / liter Normality of acids = moles of H + ions / liter Normality of bases = moles OH - / liter

39. Concentration Unit Conversions Percent Solution to Molality Problem: What is the molality of 10.0 % NaCl solution? Solution: A 10.0 % solution contains 10.0 grams of solute in 100.0 grams of solution. The solvent must have a mass of 90.0 grams (100.0 g – 10.0 g) or 0.090 kg Molality = moles / kg Moles NaCl = 10.0 / (23.0 + 35.5) = 0.171 m = 0.171 moles / 0.090 kg = 1.89 molal

40. Concentration Unit Conversions Percent solution to molarity Problem: Concentration HCl is 36.0 %. The density of the solution is 1.19 grams per milliliter. Find its molarity. Solution: Find the mass of 1.0 liter of solution. Density x volume = mass (1.19 g/ml) x 1000 ml = 1190 g Find the mass of solute 36.0 % of 1190 = 428.4 grams is HCl, the rest is water. Convert mass of solute to moles Moles of HCl = 428.4 g / (1 + 35.5) =11.7 There are 11.7 moles of HCl in one liter of solution therefore the solution is 11.7M

41. Concentration Unit Conversions Molality to mole fraction Problem: Find the mole fractions of each component in a 4.57 molal solution of CaCl 2. Solution: A 4.57 molal solution of CaCl 2 contains 4.57 moles of calcium chloride in 1.0 kg (1000 grams) of water Moles of water = 1000 g / 18.0 g per mole = 55.5 X = moles / total moles X CaCl2 = 4.57 / (4.57 + 55.5) = 0.0761 X water = 55.5 / (4.57 + 55.5) = 0.924 Check: 0.0761 + 0.924 = 1.000

42. Concentration Unit Conversions Molarity to Molality Problem: A solution of HCl has a molarity of 11.8 M and a density of 1.19 g per ml. Find its molality. Solution: Find the mass of 1.0 liter Mass = density x volume Mass = 1.19 g per ml x 1000 ml = 1190 grams Find the mass of solute in one liter Mass = mole x molar mass Mass = 11.8 x (1 + 35.5) = 431 grams HCl, the rest is water Mass of water = mass of solution – mass of solute Mass of water = 1190 g – 431 g = 569 grams Molality = moles / kg = 11.8 moles / 0.569 kg = 20.7 molal

43. Concentration Unit Conversions Molality to molarity Problem: A solute with molar mass of 72.0 grams per mole is dissolved in water to make a 2.00 m solution. The density of the solution is 1.25 g / ml. Find its molarity. Solution: Find the mass of 1.0 liters of solution Mass = density x volume Mass = (1.25 g /ml) x 1000 ml = 1250 g Find the mass of solution containing 2.00 moles of solute A 2.00 molal solution consists of 142.0 grams of solute (2.00 x 72) plus 1000 g of solvent = 1420 g Set up a ratio to find the moles in 1.0 liter or 1250 grams of solution (2.00 moles / 1420 grams solution) = x moles / 1250 grams solution Solving the proportion gives 1.76 moles in 1.0 liter or 1.76 M

44. SOLUTION VOLUME SOLVENT MASS MOLES OF H + OR OH - THE NUMERATORS IN ALL OF THE EQUATIONS REFER TO THE SOLUTE

45. Summary (1) Molarity (M) = moles solute / liters of solution (2) M 1 V 1 = M 2 V 2 (Dilution formula) (3) Molality (m) = moles solute / kg of solvent (4) Mole Fraction (X A ) = (moles A / total moles) (5) Percent Sol’n = (grams solute / total grams) x 100 % (6) Normality = equivalents / liter (7) Normality of acids = moles of H + ions / liter (8) Normality of bases = moles OH - / liter