Factors Affecting the Rate of Dissolution 1. temperature As T o, rate. 2. particle size As size, rate. 3. mixing More mixing, rate

Transcription

1 Unit 10: Solutions Solution Definitions solution: a homogeneous mixture -- evenly mixed at the particle level -- e.g., salt water alloy: a solid solution of metals -- e.g., bronze = Cu + Sn; brass = Cu + Zn solvent: the substance that dissolves the solute water salt soluble: will dissolve in miscible: refers to two gases or two liquids that form a solution; more specific than soluble -- e.g., food coloring and water Factors Affecting the Rate of Dissolution 1. temperature As T o, rate. particle size As size, rate 3. mixing ore mixing, rate 4. nature of solvent or solute

2 Classes of Solutions aqueous solution: solvent = water water = the universal solvent 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 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

3 olecular Polarity nonpolar molecules: e.g., fats and oils polar molecules: e.g., water Like dissolves like. -- e are shared equally -- tend to be symmetric -- e NOT shared equally H H O H H C H H C H H C H H C H H polar + polar = solution nonpolar + nonpolar = solution polar + nonpolar = suspension (won t mix evenly) Using Solubility Principles Chemicals used by body obey solubility principles. -- water-soluble vitamins: e.g., vit. C -- fat-soluble vitamins: e.g., vits. A, D Dry cleaning employs nonpolar liquids. -- polar liquids damage wool, silk -- also, dry clean for stubborn stains (ink, rust, grease) -- tetrachloroethylene is in common use Cl Cl C=C Cl Cl

4 emulsifying agent (emulsifier): -- molecules w /both a polar AND a nonpolar end -- allows polar and nonpolar substances to mix e.g., soap detergent lecithin eggs ODEL OF A SOAP OLECULE Na 1+ POLAR HEAD NONPOLAR HYDROCARBON TAIL soap vs. detergent -- made from animal and -- made from petroleum vegetable fats -- works better in hard water Hard water contains minerals w /ions like Ca +, g +, and Fe 3+ that replace Na 1+ at polar end of soap molecule. Soap is changed into an insoluble precipitate (i.e., soap scum). micelle: a liquid droplet covered w /soap or detergent molecules

5 Solubility SOLUBILITY CURVE how much solute dissolves in a given amt. of solvent at a given temp. KNO 3 (s) Solubility (g/100 g H O) KCl (s) HCl (g) Temp. ( o C) unsaturated: sol n could hold more solute; below line saturated: sol n has just right amt. of solute; on line supersaturated: sol n has too much solute dissolved in it; above the line Solids dissolved in liquids Gases dissolved in liquids Sol. Sol. T o T o As T o, solubility As T o, solubility

6 Classify as unsaturated, saturated, or supersaturated. per 100 g H O 80 g NaNO 30 o C unsaturated 45 g 60 o C saturated 50 g NH 10 o C unsaturated 70 g NH 4 70 o C supersaturated Per 500 g H O, 10 g KNO 40 o C saturation 40 o C for 100 g H O = 66 g KNO 3 So sat. 40 o C for 500 g H O = 5 x 66 g = 330 g 10 g < 330 g unsaturated Describe each situation below. (A) Per 100 g H O, 100 g NaNO 50 o C. (B) Cool sol n (A) very slowly to 10 o C. (C) Quench sol n (A) in an ice bath to 10 o C. Unsaturated; all solute dissolves; clear sol n. Supersaturated; extra solute remains in sol n; still clear. Saturated; extra solute (0 g) can t remain in sol n, becomes visible.

7 Glassware Precision and Cost beaker vs. volumetric flask 1000 ml + 5% 1000 ml ml When filled to 1000 ml line, how much liquid is present? beaker volumetric flask 5% of 1000 ml = 50 ml Range: ml Range: 950 ml 1050 ml ml imprecise; cheap precise; expensive water in grad. cyl. mercury in grad. cyl. ~ ~ ~ ~ easure to part of meniscus w /zero slope.

8 Concentration a measure of solute-to-solvent ratio concentrated dilute lots of solute not much solute watery Add water to dilute a sol n; boil water off to concentrate it. Selected units: 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 () = moles of solute L of sol n -- used most often in this class mol L mol L

9 D. molality (m) = moles of solute kg of solvent 7.85 kg KCl are dissolved in.38 L of sol n. Find molality. m kg solute L sol'n 7.85 kg.38 L 3.30 m KCl 4.8 g table sugar (i.e., sucrose, C 1 H O 11 ) are mixed into 450 g water. Find molality. m kg solute L sol'n kg L m C 1HO11 What mass of CaF must be added to 1,000 L of water so that fluoride atoms are present at a conc. of 1.5 ppm? X m' cule H O 1000L 1000 ml 1L 1g 1mL 1mol 18 g x 10 m' cule 1mol X X g CaF = 3.34 x 10 8 m cules H O 1.5 atom F 1,000,000 m' cule H 5.01x 10 O at.f times X atoms F x 10 m' cule H 1m' c CaF at.f 1mol 6.0 x 10 O.505 x g 1mol m' c CaF.505 x 10 m' c 3.5 g CaF 3 m' c

10 1: How many mol solute are req d to make 1.35 L of.50 sol n? mol = L =.50 (1.35 L) = 3.38 mol A. What mass sodium hydroxide is this? X gnaoh 3.38 mol 40.0 g 1mol 135 gnaoh B. What mass magnesium phosphate is this? X gg 6.9 g 1mol 3 (PO4 ) 3.38 mol 889 gg3(po 4 ) : Find molarity if 58.6 g barium hydroxide are in 5.65 L sol n. 1mol Ba(OH) 58.6 g Ba(OH) gba(oh) X Ba(OH) 5.65 L 3: You have 10.8 g potassium nitrate. How many ml of sol n will make this a 0.14 sol n? X L 1mol 10.8 gkno g L 1000 ml 1L 763 ml convert to ml

11 olarity and Stoichiometry V mol mol V P L L P Pb(NO 3 ) (aq) + KI (aq) PbI (s) + KNO 3 (aq) 1 Pb(NO 3 ) (aq) + KI (aq) 1 PbI (s) + KNO 3 (aq) What volume of 4.0 KI sol n is req d to yield 89 g PbI? Strategy: (1) Find mol KI needed to yield 89 g PbI. () Based on (1), find volume of 4.0 KI sol n. X mol KI 89 gpbi 1mol PbI 461g PbI mol KI 1mol PbI 0.39 mol KI mol L L mol 0.39 mol KI 4.0 KI L of 4.0 KI

12 How many ml of a CuSO 4 sol n will react w /excess Al to produce 11.0 g Cu? Al 3+ SO 4 CuSO 4 (aq) + Al (s) Cu(s) + Al (SO 4 ) 3 (aq) 3 CuSO 4 (aq) + Al (s) 3 Cu(s) + 1 Al (SO 4 ) 3 (aq) 1mol Cu 3 mol CuSO4 X mol CuSO4 11g Cu 63.5 g Cu 3 mol Cu = mol CuSO 4 mol L L mol mol CuSO CuSO L L 1000 ml 1L 346 ml

13 Dilutions of Solutions Acids (and sometimes bases) are purchased in concentrated form ( concentrate ) and are easily diluted to any desired concentration. **Safety Tip: When diluting, add acid or base to water. Dilution Equation: C VC D VD C = conc. D = dilute Conc. H 3 PO 4 is What volume of concentrate is req d to make 5.00 L of H 3 PO 4? C V C D V D 14.8 (V C ) (5.00 L) V C = L = 845 ml How would you mix the above sol n? 1. easure out L of conc. H 3 PO 4.. In separate container, obtain ~0 L of cold H O. 3. In fume hood, slowly pour H 3 PO 4 into cold H O. 4. Add enough H O until 5.00 L of sol n is obtained. You have 75 ml of conc. HF (8.9 ); you need 15.0 L of HF. Do you have enough to do the experiment? C V C D V D 8.9 (0.075 L) (15.0 L) Yes; we re OK mol HAVE > 1.50 mol NEED

14 Dissociation occurs when neutral combinations of particles separate into ions while in aqueous solution. sodium chloride NaCl Na 1+ + Cl 1 sodium hydroxide NaOH Na 1+ + OH 1 hydrochloric acid HCl H 1+ + Cl 1 sulfuric acid H SO 4 H 1+ + SO 4 acetic acid CH 3 COOH CH 3 COO 1 + H 1+ In general, acids yield hydrogen (H 1+ ) ions in aqueous solution; bases yield hydroxide (OH 1 ) ions. Strong electrolytes exhibit nearly 100% dissociation. NaCl Na 1+ + Cl 1 NOT in water: in aq. sol n: Weak electrolytes exhibit little dissociation. CH 3 COOH CH 3 COO 1 + H 1+ NOT in water: in aq. sol n: Strong or weak is a property of the substance. We can t change one into the other.

15 electrolytes: solutes that dissociate in sol n -- conduct elec. 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 elec. current (not enough ions) -- e.g., any type of sugar

16 Colligative Properties depend on conc. of a sol n Compared to solvent s normal freezing point (NFP) a sol n w/that solvent has a lower FP FREEZING PT. DEPRESSION normal boiling point (NBP) higher BP BOILING PT. ELEVATION Applications (NOTE: Data are fictitious.) 1. salting roads in winter FP BP water 0 o C (NFP) 100 o C (NBP) water + a little salt 11 o C 103 o C water + more salt 18 o C 105 o C. antifreeze (AF) /coolant FP BP water 0 o C (NFP) 100 o C (NBP) water + a little AF 10 o C 110 o C 50% water + 50% AF 35 o C 130 o C 3. law enforcement white powder starts melting at finishes melting at penalty, if convicted A 10 o C 150 o C comm. service B 130 o C 140 o C years C 134 o C 136 o C 0 years

17 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 = barium bromide, BaBr Ba + + Br 1 i = 3 Freezing Point Depression T f = K f m i Boiling Point Elevation 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.

18 168 g glucose (C 6 H 1 O 6 ) are mixed w /.50 kg H O. Find BP and FP of mixture. For H O, K b = 0.51, K f = i = 1 (NONELECTROLYTE) m mol C6 H1O kg H O g 180 g.50 kg m T b = K b m i = 0.51 (0.373) (1) = 0.19 o C BP = ( ) o C = o C T f = K f m i = 1.86 (0.373) (1) = 0.69 o C FP = ( ) o C = 0.69 o C 168 g cesium bromide are mixed w /.50 kg H O. Find BP and FP of mixture. For H O, K b = 0.51, K f = Cs 1+ Br 1 CsBr Cs 1+ + Br 1 i = m mol CsBr kg H O 168 g 1.8 g.50 kg m T b = K b m i = 0.51 (0.316) () = 0.3 o C BP = ( ) o C = o C T f = K f m i = 1.86 (0.316) () = 1.18 o C FP = ( ) o C = 1.18 o C