Chemistry Chapter 14: 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 e.g., water e.g., salt

soluble: “will dissolve in” miscible: refers to two liquids that mix evenly in all proportions -- e.g., food coloring and water

Factors Affecting the Rate of Dissolution 1. temperature As temp. , rate As size , rate 2. particle size 3. mixing With more mixing, rate We can’t control this factor. 4. nature of solvent or solute

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

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

-- e– are shared equally H–C–H H Molecular Polarity nonpolar molecules: -- e– are shared equally -- tend to be symmetric e.g., fats and oils polar molecules: -- e– NOT shared equally H O e.g., water “Like dissolves like.” polar + polar = solution nonpolar + nonpolar = solution polar + nonpolar = suspension (won’t mix evenly)

unsaturated: sol’n could hold more solute; below the line Temp. (oC) Solubility (g/100 g H2O) KNO3 (s) KCl (s) HCl (g) SOLUBILITY CURVE sudden stress causes this much ppt Solubility how much solute dissolves in a given amt. of solvent at a given temp. unsaturated: sol’n could hold more solute; below the line saturated: sol’n has “just right” amt. of solute; on the line supersaturated: sol’n has “too much” solute dissolved in it; above the line

Solids dissolved Gases dissolved in liquids in liquids To Sol. To Sol. [O2] As To , solubility ___ As To , solubility ___

Using an available solubility curve, classify as unsaturated, saturated, or supersaturated. per 100 g H2O 80 g NaNO3 @ 30oC unsaturated 45 g KCl @ 60oC saturated 30 g KClO3 @ 30oC supersaturated 70 g Pb(NO3)2 @ 60oC unsaturated

Glassware – Precision and Cost beaker vs. volumetric flask 1000 mL + 5% 1000 mL + 0.30 mL When filled to 1000 mL line, how much liquid is present? WE DON’T KNOW. 5% of 1000 mL = 50 mL min: max: Range: min: max: Range: 950 mL 999.70 mL 1050 mL 1000.30 mL imprecise; cheap precise; expensive

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

Selected units of concentration mass % = mass of solute x 100 mass of sol’n B. parts per million (ppm) = mass of solute x 106 mass of sol’n  also, ppb and ppt (Use 109 or 1012 here) mol -- commonly used for minerals or contaminants in water supplies M L C. molarity (M) = moles of solute L of sol’n -- used most often in this class

mol L M How many mol solute are req’d to make 1.35 L of 2.50 M sol’n? mol = M L = 2.50 M(mol/L) (1.35 L ) = 3.38 mol A. What mass sodium hydroxide is this? Na1+ OH1– NaOH 3.38 mol = 135 g NaOH B. What mass magnesium phosphate is this? Mg2+ PO43– Mg3(PO4)2 3.38 mol = 886 g Mg3(PO4)2

Find molarity if 58.6 g barium hydroxide are in 5.65 L sol’n. mol OH1– Ba(OH)2 58.6 g = 0.342 mol = 0.061 M Ba(OH)2 You have 10.8 g potassium nitrate. How many mL of sol’n will make this a 0.14 M sol’n? K1+ NO31– KNO3 10.8 g = 0.1068 mol (convert to mL) = 0.763 L = 763 mL

Molarity and Stoichiometry V P mol M L V of gases at STP PbI2 V of sol’ns KI 1 2 1 2 __Pb(NO3)2(aq) + __KI (aq)  __PbI2(s) + __KNO3(aq) What volume of 4.0 M KI sol’n is req’d to yield 89 g PbI2? Strategy: (1) (2) Find mol KI needed to yield 89 g PbI2. Based on (1), find volume of 4.0 M KI sol’n.

1 2 1 2 89 g PbI2 = 0.39 mol KI mol L M = 0.098 L of 4.0 M KI __Pb(NO3)2(aq) + __KI (aq)  __PbI2(s) + __KNO3(aq) What volume of 4.0 M KI sol’n is req’d to yield 89 g PbI2? Strategy: (1) (2) Find mol KI needed to yield 89 g PbI2. PbI2 KI M L Based on (1), find volume of 4.0 M KI sol’n. 89 g PbI2 = 0.39 mol KI mol L M = 0.098 L of 4.0 M KI

Cu CuSO4 How many mL of a 0.500 M CuSO4 sol’n will react w/excess Al to produce 11.0 g Cu? M L Al3+ SO42– __CuSO4(aq) + __Al(s)  3 2 __Cu(s) 3 + __Al2(SO4)3(aq) 1 11.0 g Cu = 0.173 mol CuSO4 mol L M = 0.346 L = 346 mL of 0.500 M CuSO4

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. C = conc. D = dilute Dilution Equation: MC VC = MD VD

Conc. H3PO4 is 14.8 M. What volume of concentrate is req’d to make 25.00 L of 0.500 M H3PO4? MC VC = MD VD 14.8 (VC) = 0.500 (25) 14.8 VC = 0.845 L How would you mix the above sol’n? 1. Measure out _____ L of conc. H3PO4. 0.845 2. In separate container, obtain ____ L of cold H2O. ~20 3. In fume hood, slowly pour H3PO4 into cold H2O. 4. Add enough H2O until 25.00 L of sol’n is obtained.

Cost Analysis with Dilutions 2.5 L of 12 M HCl (i.e., “concentrate”) 0.500 L of 0.15 M HCl Cost: $25.71 Cost: $6.35 How many 0.500 L-samples of 0.15 m HCl can be made from the bottle of concentrate? MC VC = MD VD (Expensive water!) 12 M (2.5 L) = 0.500 M (VD) Moral: Buy the concentrate and mix it yourself to any desired concentration. VD = 200 L 400 samples @ $6.35 ea. = $2,540

Calc. how much conc. you need… You have 75 mL of conc. HF (28.9 M); you need 15.0 L of 0.100 M HF. Do you have enough to do the experiment? MC VC = MD VD 28.9 M (0.075 L) = 0.100 M (15 L) Calc. how much conc. you need… 28.9 M (VC) = 0.100 M (15 L) VC = 0.052 L = 52 mL needed Yes; we’re OK.