1. Solutions Notes

2. Words to Know Solution – homogenous mixture Solvent – substance present in the largest amount Solutes – substance present in the smallest amount Aqueous solution – solutions with water as the solvent Concentration – the amount of solute in a given volume of solution Concentrated – large amount of solute dissolved in solvent Dilute – small amount of solute dissolved in solvent

3. Saturated – a solution that contains as much solute as will dissolve at that temperature Unsaturated – a solution that hasn’t reached that limit of solute that will dissolve Supersaturated - a solution that contains more solute than should dissolve at that temperature

4. Effect of Temperature on Solubility Increasing the temperature of a solution, increases the amount of solute that can be dissolved Decreasing the temperature of a solution, causes the solute to recrystallize

5. Learning Check 1.How many grams of NaCl will dissolve in 100 g of H 2 O at 90°C? 2.50 g of KCl is dissolved in 100 g of water at 50°C. Is the solution saturated, unsaturated or supersaturated?

6. Effect of Pressure on Solubility Pressure has a major effect on the solubility of gas- liquid systems An increase in pressure increases the solubility of a gas in the liquid

7. Polar molecules dissolve other polar molecules and ionic compounds. Nonpolar molecules dissolve other nonpolar molecules. Alcohols, which have characteristics of both polar & nonpolar, tend to dissolve in both types of solvents, but will not dissolve ionic solids. “Like dissolves like” – a solvent usually dissolves solutes that have polarities similar to itself SOLUTES SOLVENTS WaterCCl 4 Alcohol NaCl I2 I2 C 3 H 7 OH benzene (nonpolar) Br 2 KNO 3 toluene (polar) Ca(OH) 2 methanol NH 3 CO 2 polar non polar alcohol Alcohols are organic, covalent molecules with an –OH group. Alcohol names end with “-ol.” ionic polar non polar ionic alcohol and alcohols and other alcohols

8. Colligative properties Colligative properties - the physical changes that result from adding solute to a solvent. Colligative Properties depend on how many solute particles are present as well as the solvent amount, but they do NOT depend on the type of solute particles. Boiling Point Elevation Freezing Point Depression Osmotic Pressure Increasing Vapor Pressure Lowering Conductivity Increasing More particles/ions = greater change

9. Learning Check 1. Which substance will provide the greatest change in freezing point of water? A.NaClB. CaCl 2 C. C 6 H 12 O 6 D. H 2 O 2. Which of the following reflect colligative properties? (I) A 0.5 m NaBr solution has a higher vapor pressure than a 0.5 m BaCl 2 solution. (II) A 0.5 m NaOH solution freezes at a lower temperature than pure water (III) Pure water freezes at a higher temperature than pure methanol. A. only I B. only II C. only III D. I and II E. I and III 2 ions 3 ions 1 particle no change in H 2 O 2 ions 3 ions = vapor pressure lowering (more ions lower pressure) 2 ions 0 ions = freezing point depression Freezing point is a physical property, not a colligative property

10. 3. A student measured the conductivity in water, of unlabeled liquids, after each added drop. The following graph was produced... a. Identify the line that represents : –aluminum chloride –water –magnesium chloride –sugar –sodium chloride b.Which line could also represent potassium iodide? # of Drops Conductivity (µs/cm) AlCl 3, 4 ions H 2 O, no change in H 2 O MgCl 2, 3 ions C 6 H 12 O 6, 1 particle NaCl, 2 ions H2OH2O C 6 H 12 O 6 NaCl MgCl 2 AlCl 3 KI, 2 ions = same as NaCl

11. Solution Composition - Mass Percent Mass percent – describes a solution’s composition expresses the mass of solute present in a given mass of solution Mass Percent = mass of solute x 100% mass of solution* * mass of solution = mass of solute + mass of solvent Example – A solution is prepared by mixing 1.00g of C 2 H 5 OH, with 100.0g of H 2 O. Calculate the mass percent of ethanol. Given mass of solute = 1.00 g mass of solution = 100.0 g + 1.00 g = 101.0 g Mass Percent = mass of solute x 100% mass of solution Mass % = 1.00 g x 100 % 101.0 g Mass % = 0.990 %

12. Solution Composition – Molarity Molarity – measure of concentration - number of moles of solute per volume of solution in liters Molarity = moles of solute = mol = M L of solution L Example – Calculate the molarity of a solution prepared by dissolving 11.5 g NaOH in enough water to make 1.50L solution. = x __________ 1 1.50 L NaOH x ___________ mol NaOH 40.00 1 g NaOH 11.5 g NaOH 0.192 M

13. Ex: Calculate the mass of solid AgCl formed when 1.50L of a 0.100M AgNO 3 solution is reacted with excess NaCl. NaCl + AgNO 3  AgCl + NaNO 3 1.50 L 0.100 M ? g 1.50 L AgNO 3 x ______________ L AgNO 3 mol AgNO 3 0.100 1 x ____________ mol AgNO 3 mol AgCl 1 1 g AgCl 1 x ____________ 143.32 = 21.5 g

14. Example – How many moles of Ag + ions are present in 25mL of a 0.75M Ag 2 SO 4 solution? Ag 2 SO 4  2 Ag +1 + SO 4 -2 25 mL Ag 2 SO 4 L Ag 2 SO 4 mL Ag 2 SO 4 1000 1 x _____________ x ______________ L Ag 2 SO 4 mol Ag 2 SO 4 1 0.75 x ______________ mol Ag 2 SO 4 mol Ag +1 2 1 = 0.038 mol Ag +1

15. Learning check Calculate the molarity of a solution prepared by dissolving 25.6 g NaC 2 H 3 O 2 in enough water to make 200.0 mL solution.

16. Standard Solution Standard Solution – a solution whose concentration is accurately known Example – A chemist needs 1.0 L of a 0.200M K 2 Cr 2 O 7 solution. How much solid K 2 Cr 2 O 7 must be weighed out to make this solution? x ______________ g K 2 Cr 2 O 7 mol K 2 Cr 2 O 7 294.20 1 x ________________ mol K 2 Cr 2 O 7 1 0.200 L K 2 Cr 2 O 7 1.0 L K 2 Cr 2 O 7 = 59 g K 2 Cr 2 O 7

18. Dilution Dilution – process of adding more solvent to a solution Moles of solute before dilution = Moles of solute after dilution M 1 V 1 = M 2 V 2 Example: What volume of 16M H 2 SO 4 must be used to prepare 1.5L of a 0.10M H 2 SO 4 solution? Given V 1 = ? M 1 = 16 M V 2 = 1.5 L M 2 = 0.10 M V 1 = M 2 V 2 M 1 _____ V 1 = (0.10 M)(1.5 L) ____________ 16 M V 1 = 0.0094 L

19. Given V 1 = 500.0 mL M 1 = 1.00 M M 2 = 17.5 M V 2 = ? V 2 = M 1 V 1 M 2 _____ V 2 = (1.00 M)(500.0 mL) _______________ 17.5 M V 2 = 28.6 mL Example: Prepare 500.0mL of 1.00 M HC 2 H 3 O 2 from a 17.5 M stock solution. What volume of the stock solution is required? Learning Check

20. Notes- Acids and Bases

21. Acids and Bases Arrhenius ACIDS – produces hydrogen ions in aqueous solutions, sour taste, low pH, and the fact that they turn litmus paper red HCl (aq)  H + (aq) + Cl - (aq) Arrhenius BASES – produces hydroxide ions in aqueous solutions, bitter taste, slippery feel, high pH, and the fact that they turn litmus paper blue NaOH (aq)  Na + (aq) + OH - (aq) Arrhenius definition – limits the concept of a base

22. Bronsted – Lowry definition – gives a broader definition of a base Bronsted – Lowry ACID – a proton (H + ) donor Bronsted – Lowry BASE – a proton (H + ) acceptor General Reaction – HA (aq) + H 2 O (l)  H 3 O + (aq) + A - (aq) Acid Base Conjugate Conjugate Acid Base Conjugate Base – everything that remains of the acid molecule after a proton is lost Conjugate Acid – the base with the transferred proton (H+) Conjugate Acid – Base Pair – two substances related to each other by the donating and accepting of a single proton proton donor proton acceptor

23. Examples: Finish each equation and identify each member of the conjugate acid –base pair. H 2 SO 4 (aq) + H 2 O (l)  HSO 4 -1 (aq) + H 3 O + (aq) AcidBase Conjugate Base Conjugate Acid CO 3 2- (aq) + H 2 O (l)  HCO 3 -1 (aq) + OH - (aq) BaseAcid Conjugate Acid Conjugate Base The hydronium ion, H 3 O +, forms when water behaves as a base. This happens when the two unshared pairs of electrons on O bond covalently with the H +.

24. Learning check Write the conjugate ACID a.NH 3 b.HCO 3 -1 Write the conjugate BASE a.H 3 PO 4 b.HBr Finish each equation and identify each member of the conjugate acid –base pair. a. H 2 SO 3 (aq) + H 2 O (l)  b. SO 4 -2 (aq) + H 2 O (l) 

25. Water as an Acid and a Base Amphoteric – a substance that can behave as either an acid or a base - water is the most common amphoteric substance Ionization of Water – H 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH - (aq) In the shorthand form: H 2 O (l)  H + (aq) + OH - (aq)

26. Ion-product constant – K w refers to the ionization of water K w = [H + ][OH - ] At 25  C, Kw = [H + ][OH - ] = [1.0 x 10 -7 ] [1.0 x 10 -7 ] = 1.0 x 10 -14 If [H + ] increases, the [OH - ] decreases, so the products of the two is still 1.0 x 10 -14. There are three possible situations – 1.A neutral solution, where [H + ] = [OH - ] 2.An acidic solution, where [H + ]  [OH - ] 3.A basic solution, where [H + ]  [OH - ] [ ] = concentration [H + ] = hydrogen ion concentration in M [OH - ] = hydroxide ion concentration in M

27. Example: Calculate [H+] or [OH-] as required for each of the following solutions at 25  C, for each solution state whether it is neutral, acidic, or basic. a. 1.0 x 10 -5 M OH - b. 10.0 M H + K w = [H + ][OH - ] 1 x 10 -14 = [H + ][1.0 x 10 -5 M] [H + ] = 1.0 x 10 -9 M BASIC K w = [H + ][OH - ] 1 x 10 -14 = [10.0 M][OH - ] [OH - ] = 1.00 x 10 -15 M ACIDIC

28. pH scale pH scale – because the [H+] in an aqueous solution is typically small, logarithms are used to express solution acidity pH = -log [H+]pOH = -log [OH-] Graphing calculatorNon graphing calculator 1. Press the +/- key1. Enter the [H+]2. Press the log key 3. Enter the [H+]3. Press the +/- key Significant Figure Rule – The number of places to the right of the decimal for a log must be equal to the number of significant figures in the original number.

29. pH Scale

30. Example – Calculate the pH or pOH a. [H + ] = 5.9 x 10 -9 M b. [OH - ] = 2.4 x 10 -6 M pH = - log [H + ] pH = - log (5.9 x 10 -9 M) pH = 8.23 pOH = - log [OH - ] pOH = - log (2.4 x 10 -6 M) pOH = 5.62

31. Since K w = [H + ][OH - ] = 1.0 x 10 -14, pH + pOH = 14.00 Example - The pH of blood is about 7.4. What is the pOH of blood? pH + pOH =14.00 7.4 + pOH = 14.00 pOH = 6.6

32. In order to calculate the concentration from the pH or pOH, [H + ] = 10 -pH [OH - ] = 10 -pOH Graphing calculator Non-graphing calculator 1.Press the 2nd 1. Enter the pH function, then log2. Press the +/- key 2. Press the +/- key3. Press the inverse 3. Enter the pH log key

33. Example - The pH of a human blood sample was measured to be 7.41. What is the [H + ] in blood? [H + ] = 10 -pH [H + ] = 10 -7.41 [H + ] = 3.9 x 10 -8 M

34. Example – The pOH of the water in a fish tank is found to be 6.59. What is the [H + ] for this water? [OH - ] = 10 -pOH [OH - ] = 10 -6.59 [OH - ] = 2.6 x 10 -7 M K w = [H + ][OH - ] 1 x 10 -14 = [H + ][2.6 x 10 -7 M] [H + ] = 3.8 x 10 -8 M

35. Learning check Determine the pH of a solution with a hydrogen ion concentration of 3.2 x10 -12 M. What is the [OH - ] concentration of a solution with a hydrogen ion concentration of 8.9x10 -4 M? What is the pH of a solution with a hydroxide ion concentration of 5.7x10 -10 M?

36. How Do We Measure pH? For less accurate measurements, one can use –Litmus paper “Red” paper turns blue above ~pH = 8 “Blue” paper turns red below ~pH = 5 –An indicator

37. How Do We Measure pH? For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

38. Strong Acids seven strong acids are HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 3, and HClO 4. These are, by definition, strong electrolytes and exist totally as ions in aqueous solution.

39. Strong Bases Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca 2+, Sr 2+, and Ba 2+ ). Again, these substances dissociate completely in aqueous solution, strong electrolytes

40. Strong, Weak, or Nonelectrolyte Electrolytes are substances which, when dissolved in water, break up into cations (plus-charged ions) and anions (minus-charged ions). We say they ionize. Strong electrolytes ionize completely (100%), while weak electrolytes ionize only partially (usually on the order of 1–10%). The ions in an electrolyte can be used to complete an electric circuit and power a bulb. Strong electrolytes fall into three categories: strong acids, strong bases, and soluble salts. The weak electrolytes include weak acids, weak bases and insoluble salts. Molecules are nonelectrolytes. Substance Classification - Strong acid, weak acid, strong base, weak base, soluble salt, insoluble salt, molecule Strong electrolyte, weak electrolyte, nonelectrolyte sodium hydroxide acetic acid potassium nitrate hydrobromic acid silver chloride Carbon dioxide strong base strong electrolyte weak acidweak electrolyte soluble saltstrong electrolyte strong acid strong electrolyte insoluble saltweak electrolyte moleculenonelectrolyte

41. Learning check Substance Classification - Strong acid, weak acid, strong base, weak base, soluble salt, insoluble salt, molecule Strong electrolyte, weak electrolyte, nonelectrolyte chloric acid barium carbonate nitric acid sulfurous acid strontium sulfate ethanol octane (gasoline)

42. Titration A known concentration of base (or acid) is slowly added to a solution of acid (or base).

43. Titration A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

44. Titration of a Strong Acid with a Strong Base From the start of the titration to near the equivalence point, the pH goes up slowly.

45. Titration of a Strong Acid with a Strong Base Just before and after the equivalence point, the pH increases rapidly.

46. Titration of a Strong Acid with a Strong Base At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.

47. Titration of a Strong Acid with a Strong Base As more base is added, the increase in pH again levels off.

48. Neutralization Neutralization Reaction = Acid + Base  Salt + Water Salt – ionic compound containing a positive ion other than H + and a negative ion other than OH -

49. Buffered solutions – resists a change in its pH even when a strong acid or base is added to it - A solution is buffered in the presence of a weak acid and its conjugate base

50. pH and pOH Calculations