Solutions

Solution Definitions Solution:Solution: homogeneous mixture, evenly mixed at the particle level (like salt water). Solvent: Solvent: present in greater amount that can dissolve the solute Solute: Solute: substance being dissolved

Classes of Solutions aqueous solution: water = “the universal solvent” solvent = water amalgam:solvent = Hg e.g., dental amalgam tincture:solvent = alcohol e.g., tincture of iodine (for cuts) organic solution:solvent contains carbon e.g., gasoline, benzene, toluene, hexane

Solution Definitions alloy: solvent: the substance that dissolves the solute watersalt a solid solution of metals -- e.g., bronze = Cu + Sn; brass = Cu + Zn “will dissolve in” refers to two gases or two liquids that form a solution; more specific than the term “soluble” miscible: soluble:

Solubility Solubility - Solubility - maximum grams of solute that will dissolve in 100 g of solvent at a given temperature varies with temperature

Solubility SATURATED SOLUTION no more solute dissolves UNSATURATED SOLUTION more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form increasing concentration

Temp. ( o C) Solubility (g/100 g H 2 O) KNO 3 (s) KCl (s) HCl (g) SOLUBILITY CURVE Solubility  how much solute dissolves in a given amt. of solvent at a given temp. below unsaturated:solution could hold more solute; below line on saturated:solution has “just right” amt. of solute; on line supersaturated:solution has “too much” solute dissolved in it; above the line

Solubility Table LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World, 1996, page Temperature (°C) Solubility vs. Temperature for Solids Solubility (grams of solute/100 g H 2 O) KI KCl NaNO 3 KNO 3 HClNH 4 Cl NH 3 NaCl KClO 3 SO 2 shows the dependence of solubility on temperature gases solids

per 100 g H 2 O Classify as unsaturated, saturated, or supersaturated. 80 g NaNO 30 o Cunsaturated 60 g 60 o Csaturated 50 g NH 10 o Cunsaturated 70 g NH 4 70 o Csupersaturated So sat. 40 o C for 500 g H 2 O = 5 x 66 g = 330 g 120 g < 330 gunsaturated saturation 40 o C for 100 g H 2 O = 66 g KNO 3 Per 500 g H 2 O, 120 g KNO 40 o C

Solubility  Solids are more soluble at... high temperatures.  Gases are more soluble at... low temperatures

(A) Per 100 g H 2 O, 100 g Unsaturated; all solute NaNO 50 o C. dissolves; clear solution. (B) Cool solution (A) very Supersaturated; extra slowly to 10 o C. solute remains in solution; still clear. Describe each situation below. (C) Quench solution (A) in Saturated; extra solute an ice bath to 10 o C. (20 g) can’t remain in solution, becomes visible.

Non-Solution Definitions insoluble:“will NOT dissolve in” e.g., sand and water immiscible:refers to two gases or two liquids that will NOT form a solution e.g., water and oil suspension:appears uniform while being stirred, but settles over time

Water HOT Solubility A B Before Water COLD Add 1 drop of red food coloring Miscible – “mixable” two gases or two liquids that mix evenly Experiment 1: Water HOT AFTER Water COLD A B

Solubility Water Oil T 30 sec AFTER Before Add oil to water and shake Immiscible – “does not mix” two liquids or two gases that DO NOT MIX Experiment 2: T 0 sec

Solutions What the solute and the solvent are determines: –whether a substance will dissolve. –how much will dissolve.

As size, rate As T, rate 3. mixing Factors Affecting the Rate of Dissolution 1. temperature 2. particle size 4. nature of solvent or solute More mixing, rate

Solid Solubility Timberlake, Chemistry 7 th Edition, page 297 KI NaNO 3 KNO 3 Na 3 PO 4 NaCl Temperature ( o C) Solubility (g solute / 100 g H 2 O)

Gas Solubility CH 4 O2O2 CO He Temperature ( o C) Solubility (mM) Higher Temperature …Gas is LESS Soluble

Particle Size and Mixing In order to dissolve - the solvent molecules must come in contact with the solute. Stirring moves fresh solvent next to the solute. The solvent touches the surface of the solute. Smaller pieces increase the amount of surface of the solute.

Molecular Polarity Nonpolar molecules: electrons are shared equally –Molecule has no charge e.g. Ethane

O 2- H+H+ H+H+  ++ Polar Molecules Electrons are not equally shared Molecule has slight charge “Like dissolves like” (Polarity)

Solubility Rules 1.Most nitrate (NO 3 1- ) salts are soluble. 2.Most salts of Na +, K +, and NH 4 + are soluble. 3.Most chloride salts are soluble. Notable exceptions are AgCl, PbCl 2, and Hg 2 Cl 2. 4.Most sulfate salts are soluble. Notable exceptions are BaSO 4, PbSO 4, and CaSO 4. 5.Most hydroxide compounds are only slightly soluble.* The important exceptions are NaOH and KOH, Ba(OH) 2 and Ca(OH) 2 are only moderately soluble. 6.Most sulfide (S 2- ), carbonate (CO 3 2- ), and phos- phate (PO 4 3- ) salts are only slightly soluble. *The terms insoluble and slightly soluble really mean the same thing. Such a tiny amount dissolves that it is not possible to detect it with the naked eye. General Rules for Solubility of Ionic Compounds (Salts) in Water at 25 o C Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 218

Oil and Water Don’t Mix Oil is nonpolar Water is polar “Like dissolves like” polar + polar = solution nonpolar + nonpolar = solution polar + nonpolar = suspension (won ’ t mix evenly) Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 470

Solvation  Soap / Detergent polar “head” with long nonpolar “tail” dissolves nonpolar grease in polar water micelle

Solubility of Common Compounds NO 3 - salts Na +, K +, NH 4 + salts Soluble compounds Cl -, Br -, I - salts S 2-, CO 3 2-, PO 4 3- salts OH 1- salts SO 4 2- salts Except for those containing Ag +, Hg 2 2+, Pb 2+ Except for those containing Ba 2+, Pb 2+, Ca 2+ Insoluble compounds Except for those containing Na +, K +, Ca 2+ Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 218

Making Molar Solutions …from liquids (More accurately, from stock solutions)

Concentration…a measure of solute-to-solvent ratio concentrated vs. dilute “lots of solute” “not much solute” “watery” Add water to dilute a solution; boil water off to concentrate it.

Concentration “The amount of solute in a solution” mol L M A. mass % = mass of solute mass of sol’n B. parts per million (ppm)  also, ppb and ppt – commonly used for minerals or contaminants in water supplies C. molarity (M) = moles of solute L of sol’n – used most often in this class D. molality (m) = moles of solute kg of solvent M = mol L % by mass – medicated creams % by volume – rubbing alcohol

V = 1000 mL n = 2 moles Concentration = 2 molar V = 1000 mL n = 4 moles [ ] = 4 molar V = 5000 mL n = 20 moles [ ] = 4 molar Molarity (M) = # of moles volume (L) V = 250 mL n = 8 moles [ ] = 32 molar

Glassware

precise; expensive; good for making solutions Range: Glassware – Precision and Cost beakervs.volumetric flask When filled to 1000 mL line, how much liquid is present? beaker 5% of 1000 mL = 50 mL volumetric flask 1000 mL mL 950 mL – 1050 mL mL– mL imprecise; cheap Range:

How to mix a Standard Solution Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 480 Wash bottle Volume marker (calibration mark) Weighed amount of solute

How to mix solid chemicals How many grams of sodium hydroxide would we need to make 100. mL of 3.0 M NaOH? M = mol L 3 M = ? mol 1 L How much will this weigh? 1 23g/mol + 16g/mol g/mol MM NaOH = 40g/mol 40.0 g NaOH 1 mol NaOH X g NaOH = 3.0 mol NaOH = To mix this: add 120 g NaOH into 1L volumetric flask with~750 mL cold H 2 O. Mix, allow to return to room temperature – bring volume to 1 L. 120 g NaOH

40.0 g NaOH 1 mol NaOH How many moles of solute are required to make 1.35 L of 2.50 M solution? mol = M L B. What mass of magnesium phosphate is this? A. What mass of sodium hydroxide is this? mol L M 3.38 mol = 2.50 M (1.35 L) = X g NaOH = 3.38 mol NaOH = 135 g NaOH g Mg 3 (PO 4 ) 2 1 mol Mg 3 (PO 4 ) 2 X g Mg 3 (PO 4 ) 2 = 3.38 mol Mg 3 (PO 4 ) 2 = 889 g Mg 3 (PO 4 ) 2

0.342 mol 5.65 L Find molarity if 58.6 g of barium hydroxide are in 5.65 L solution g Ba(OH) 2 1 mol Ba(OH) 2 Step 2). What is the molarity of a 5.65 L solution containing mol solute? Step 1). How many moles barium hydroxide is this? M Ba(OH) 2 = X mol Ba(OH) 2 = 58.6 g Ba(OH) 2 = mol Ba(OH) 2 M = mol L M =

You have 10.8 g potassium nitrate. How many mL of solution will make this a 0.14 M solution? convert to mL

Molality mass of solvent only 1 kg water = 1 L water

Molality  Find the molality of a solution containing 75 g of MgCl 2 in 250 mL of water. 75 g MgCl 2 1 mol MgCl g MgCl 2 = 3.2 m MgCl kg water

Molality  How many grams of NaCl are required to make a 1.54 m solution using kg of water? kg water1.54 mol NaCl 1 kg water = 45.0 g NaCl g NaCl 1 mol NaCl

500 mL volumetric flask Preparing Solutions 500 mL of 1.54M NaCl 500 mL water 45.0 g NaCl –mass 45.0 g of NaCl –add water until total volume is 500 mL –mass 45.0 g of NaCl –add kg of water (volume will be just over 500 total mL) 500 mL mark 1.54m NaCl in kg of water molality molarity

Dilution Preparation of a desired solution by adding water to a concentrate. Moles of solute remain the same.

Dilution What volume of 15.8M HNO 3 is required to make 250 mL of a 6.0M solution? GIVEN: M 1 = 15.8M V 1 = ? M 2 = 6.0M V 2 = 250 mL WORK: M 1 V 1 = M 2 V 2 (15.8M) V 1 = (6.0M)(250mL) V 1 = 95 mL of 15.8M HNO 3

occurs when neutral combinations of particles separate into ions while in aqueous solution. sodium chloride sodium hydroxide hydrochloric acid sulfuric acid acetic acid Dissociation: NaCl  Na 1+ + Cl 1– NaOH  Na 1+ + OH 1– HCl  H 1+ + Cl 1– H 2 SO 4  2 H 1+ + SO 4 2– CH 3 COOH  CH 3 COO 1– + H 1+

NaCl Na 1+ + Cl 1– CH 3 COOH CH 3 COO 1– + H 1+ Weak electrolytes exhibit little dissociation. “Strong” or “weak” is a property of the substance. We can’t change one into the other. Strong electrolytes exhibit nearly 100% dissociation. NOT in water: in aq. solution: NOT in water: in aq. solution:

electrolytes: solutes that dissociate in solution -- conduct electric current because of free-moving ions e.g., acids, bases, most ionic compounds -- are crucial for many cellular processes -- obtained in a healthy diet -- For sustained exercise or a bout of the flu, sports drinks ensure adequate electrolytes. nonelectrolytes: solutes that DO NOT dissociate -- DO NOT conduct electric current (not enough ions) e.g., any type of sugar

…normal boiling point (NBP)…higher BP FREEZING PT. DEPRESSION BOILING PT. ELEVATION Colligative Properties  depend on concentration of a solution Compared to solvent‘s: a solution w/that solvent has a: …normal freezing point (NFP)…lower FP

1. salting roads in winter FP BP water0 o C (NFP) 100 o C (NBP) 2. antifreeze (AF) /coolant FP BP water0 o C (NFP) 100 o C (NBP) water + a little AF–10 o C110 o C 50% water + 50% AF–35 o C130 o C water + a little salt water + more salt –11 o C 103 o C –18 o C105 o C Applications (NOTE: Data are fictitious.)

3. law enforcement white powder starts melting at… finishes melting at… penalty, if convicted A109 o C175 o Ccomm. service B150 o C180 o C2 years C194 o C196 o C20 years

Calculations for Colligative Properties The change in FP or BP is found using…  T x = K x m i  T x = change in T o (below NFP or above NBP) K x = constant depending on… (A) solvent (B) freezing or boiling m = molality of solute = mol solute / kg solvent i = integer that accounts for any solute dissociation any sugar (all nonelectrolytes)……………...i = 1 table salt, NaCl  Na 1+ + Cl 1– ………………i = 2 barium bromide, BaBr 2  Ba Br 1– ……i = 3

Freezing Point DepressionBoiling Point Elevation  T f = K f m i  T b = K b m i Then use these in conjunction with the NFP and NBP to find the FP and BP of the mixture. (K b = ebullioscopic constant, which is 0.51 K kg/mol for the boiling point of water) (K f = cryoscopic constant, which is 1.86 K kg/mol for the freezing point of water)

(NONELECTROLYTE) 168 g glucose, C 6 H 12 O 6, (a sugar, nonelectrolyte) are mixed w /2.50 kg H 2 O. Find BP and FP of mixture. For water, K b = 0.512, K f = –1.86. i = 1  T b = K b m i = (0.373) (1) = 0.19 o C BP = ( ) o C = o C  T f = K f m i = –1.86 (0.373) (1) = –0.694 o C FP = (0 + –0.69) o C = –0.694 o C

168 g cesium bromide are mixed w /2.50 kg H 2 O. Find BP and FP of mixture. For H 2 O, K b = 0.512, K f = –1.86.  T b = K b m i = (0.316) (2) = 0.32 o C BP = ( ) o C = o C  T f = K f m i = –1.86 (0.316) (2) = –1.18 o C FP = (0 + –1.18) o C = –1.18 o C Cs 1+ Br 1– i = 2 CsBr  Cs 1+ + Br 1–

1 Pb(NO 3 ) 2 (aq) + KI (aq)  PbI 2 (s) + KNO 3 (aq) __ 212 Molarity and Stoichiometry M M V V P P mol M L M = mol L mol = M L What volume of 4.0 M KI solution is required to yield 89 g PbI 2 ?

461 g PbI 2 Strategy: 1 Pb(NO 3 ) 2 (aq) + 2 KI (aq)  1 PbI 2 (s) + 2 KNO 3 (aq) X mol KI = 89 g PbI 2 1 mol PbI 2 2 mol KI = 0.39 mol KI (1) Find mol KI needed to yield 89 g PbI 2. (2) Based on (1), find volume of 4.0 M KI solution. What volume of 4.0 M KI solution is required to yield 89 g PbI 2 ? 89 g ? L 4.0 M M = mol L L = mol M = 0.39 mol KI 4.0 M KI = L of 4.0 M KI

= mol CuSO 4 How many mL of a M CuSO 4 solution will react w /excess Al to produce 11.0 g Cu? __CuSO 4 (aq) + __Al (s)  Al 3+ SO 4 2– CuSO 4 (aq) + Al (s)  Cu(s) + Al 2 (SO 4 ) 3 (aq)3231 x mol11 g __Cu(s) + __Al 2 (SO 4 ) 3 (aq) X mol CuSO 4 = 11 g Cu 1 mol Cu 63.5 g Cu 3 mol CuSO 4 3 mol Cu M = mol L L = mol M mol CuSO M CuSO 4 = L L 1000 mL 1 L = 346 mL

Limiting Reactants 79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP? Zn + 2HCl  ZnCl 2 + H g ? L 0.90 L 2.5M Courtesy Christy Johannesson

Limiting Reactants 79.1 g Zn 1 mol Zn g Zn = 27 L H 2 1 mol H 2 1 mol Zn 22.4 L H 2 1 mol H 2 Zn + 2HCl  ZnCl 2 + H g ? L 0.90 L 2.5M Courtesy Christy Johannesson

Limiting Reactants 22.4 L H 2 1 mol H L 2.5 mol HCl 1 L = 25 L H 2 1 mol H 2 2 mol HCl Zn + 2HCl  ZnCl 2 + H g ? L 0.90 L 2.5M Courtesy Christy Johannesson

Limiting Reactants Zn: 27 L H 2 HCl: 25 L H 2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25 L H 2 left over zinc Courtesy Christy Johannesson