SOLUTIONS HW = Read 15.5 & 15.6, finish Molarity POGIL Tomorrow we’ll be in the lab 

Definitions Solution - homogeneous mixture Solute - substance being dissolved Solvent - present in greater amount, substance that dissolves the solute

SATURATION Unsaturated – A solution that has NOT reached the limit of how much solute can be dissolved Saturated – A solution that contains as much solute as can dissolve at that temperature Supersaturated – A solution that contains more dissolved solute than is theoretically possible at that temperature

CONCENTRATION Concentrated – A solution that contains a relatively large amount of solute Dilute – A solution that contains relatively little solute Standard – A solution whose concentration is known

Solubility vs. Temperature An aqueous solution containing 90g of potassium nitrate and 200g of water at 60oC is ... 200 180 160 140 120 100 80 60 40 20 KI KNO3 Solubility (g solute / 100 g H2O) NaNO3 Na3PO4 The general rule of thumb is that solubility of solids increases with increases in temperature. Maximum amount of a solute that can dissolve in a solvent at a specified temperature and pressure is its solubility. – Solubility is expressed as the mass of solute per volume (g/L) or mass of solute per mass of solvent (g/g) or as the moles of solute per volume (mol/L). – Solubility of a substance depends on energetic factors and on the temperature and, for gases, the pressure. • A solution that contains the maximum possible amount of solute is saturated. • If a solution contains less than the maximum amount of solute, it is unsaturated. When a solution is saturated and excess solute is present, the rate of dissolution is equal to the rate of crystallization. • Solubility increases with increasing temperature — a saturated solution that was prepared at a higher temperature contains more dissolved solute than it would contain at a lower temperature, when the solution is cooled, it can become supersaturated. Solubility of a substance generally increases with increasing temperature No relationship between the structure of a substance and the temperature dependence of its solubility Solubility may increase or decrease with temperature; the magnitude of this temperature dependence varies widely among compounds This variation of solubility with temperature is used to separate the components of a mixture by fractional crystallization, the separation of compounds based on their solubilities in a given solvent Fractional crystallization is a common technique for purifying compounds; the compound of interest must be more soluble at high temperature than at low temperature, so that lowering the temperature causes it to crystallize out of solution Solubility of gases in liquids decreases with increasing temperature Attractive intermolecular interactions in the gas phase are essentially zero for most substances When a gas dissolves, its molecules interact with solvent molecules and heat is released when these new attractive interactions form, therefore, dissolving most gases in liquids is an exothermic process (Hsoln < 0) Adding heat to the solution provides thermal energy that overcomes the attractive forces between the gas and the solvent molecules, thereby decreasing the solubility of the gas NaCl 20 40 60 80 100 Temperature (oC) a) Unsaturated (b)saturated (c) supersaturated Timberlake, Chemistry 7th Edition, page 297

Pure water does not conduct an electric current Source of electric power Pure water Pure water doesn’t conduct electricity because it contains no ions. Ions (cations have (+) charges) carry electrons in solution. The flow of electrons is called electricity. Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 215

Ionic Solutions conduct a Current Source of electric power Free ions present in water To make a solution that conducts electricity (an electrolyte) – an ionic salt is added. An alternative method is to add acid. In truth, most acids are oxysalts dissolved in water. Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 215

Factors Affecting the Rate of Dissolution 1. temperature As To , rate 2. particle size As size , rate 3. mixing More mixing, rate Formation of a solution from a solute and a solvent is a physical process, not a chemical one. Both solute and solvent can be recovered in chemically unchanged form using appropriate separation methods. Dissolution of a solute in a solvent to form a solution does not involve a chemical transformation. Substances that form a single homogeneous phase in all proportions are said to be completely miscible in one another. If two substances are essentially insoluble in each other, they are immiscible. 4. nature of solvent or solute

Concentration…a measure of solute-to-solvent ratio concentrated vs. dilute “lots of solute” “not much solute” “watery” Concentration of a solution describes the quantity of a solute that is contained in a particular quantity of solvent or solution Knowing the concentration of solutes is important in controlling the stoichiometry of reactant for reactions that occur in solution A concentrated solution contains a large amount of solute in a given amount of solution. A 10 mol/L solution would be called concentrated. A dilute solution contains a small amount of solute in a given amount of solution. A 0.01 mol/L solution would be called dilute. Add water to dilute a solution; boil water off to concentrate it.

Concentration “The amount of solute in a solution” 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 % by mass – medicated creams % by volume – rubbing alcohol MOLARITY - Most common unit of concentration Most useful for calculations involving the stoichiometry of reactions in solution Molarity of a solution is the number of moles of solute present in exactly 1 L of solution: moles of solute molarity = liters of solution Units of molarity — moles per liter of solution (mol/L), abbreviated as M Relationship among volume, molarity, and moles is expressed as VL M Mol/L = L (mol) = moles (L) There are several different ways to quantitatively describe the concentration of a solution, which is the amount of solute in a given quantity of solution. 1. Molarity – Useful way to describe solution concentrations for reactions that are carried out in solution or for titrations – Molarity is the number of moles of solute divided by the olume of the solution Molarity = moles of solute = mol/L liter of solution – Volume of a solution depends on its density, which is a function of temperature 2. Molality – Concentration of a solution can also be described by its molality (m), the number of moles of solute per kilogram of solvent – Molality = moles of solute kilogram solvent – Depends on the masses of the solute and solvent, which are independent of temperature – Used in determining how colligative properties vary with solute concentrations 3. Mole fraction – Used to describe gas concentrations and to determine the vapor pressures of mixtures of similar liquids – Mole fraction () = moles of component total moles in the solution – Depends on only the masses of the solute and solvent and is temperature independent 4. Mass percentage (%) – The ratio of the mass of the solute to the total mass of the solution – Result can be expressed as mass percentage, parts per million (ppm), or parts per billion (ppb) mass percentage = mass of solute  100% mass of solution parts per million (ppm) = mass of solute  106 parts per billion (ppb) = mass of solute  109 – Parts per million (ppm) and parts per billion (ppb) are used to describe concentrations of highly dilute solutions, and these measurements correspond to milligrams (mg) and micrograms (g) of solute per kilogram of solution, respectively – Mass percentage and parts per million or billion can express the concentrations of substances even if their molecular mass is unknown because these are simply different ways of expressing the ratios of the mass of a solute to the mass of the solution M = mol L D. molality (m) = moles of solute kg of solvent

Glassware

Glassware – Precision and Cost beaker vs. volumetric flask When filled to 1000 mL line, how much liquid is present? beaker 5% of 1000 mL = volumetric flask 50 mL 1000 mL + 0.1 mL Range: 950 mL – 1050 mL Range: 999.9 mL– 1000.1 mL imprecise; cheap precise; expensive

How to mix a standard solution from a solid solute An aqueous solution consists of at least two components, the solvent (water) and the solute (the stuff dissolved in the water). Usually one wants to keep track of the amount of the solute dissolved in the solution. We call this the concentrations. One could do by keeping track of the concentration by determining the mass of each component, but it is usually easier to measure liquids by volume instead of mass. To do this measure called molarity is commonly used. Molarity (M) is defined as the number of moles of solute (n) divided by the volume (V) of the solution in liters. It is important to note that the molarity is defined as moles of solute per liter of solution, not moles of solute per liter of solvent. This is because when you add a substance, perhaps a salt, to some volume of water, the volume of the resulting solution will be different than the original volume in some unpredictable way. To get around this problem chemists commonly make up their solutions in volumetric flasks. These are flasks that have a long neck with an etched line indicating the volume. The solute (perhaps a salt) is added to the flask first and then water is added until the solution reaches the mark. The flasks have very good calibration so volumes are commonly known to at least four significant figures. http://www.chem.ucla.edu/~gchemlab/soln_conc_web.htm http://www.chem.ucla.edu/~gchemlab/soln_conc_web.htm

How to mix a standard solution from a solid solute Wash bottle Volume marker (calibration mark) Weighed amount of solute Use a VOLUMETRIC FLASK to make a standard solution of known concentration Step 1> add the weighed amount of solute in the volumetric flask Step 2> add distilled water (about half of final volume) Step 3> cap volumetric flask, and shake to dissolve solute completely Step 4> add distilled water to volume marker (calibration mark) The solution process may be exothermic (release heat). This may cause the liquid to show a larger volume than is real. Allow the solution to return to ambient (room) temperature and check volume again. Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 480

Preparing Solutions How to prepare 500 mL of 1.54 M NaCl solution mass 45.0 g of NaCl add water until total volume is 500 mL 500 mL volumetric flask 500 mL mark 45.0 g NaCl solute

Making a Dilute Solution remove sample moles of solute initial solution same number of moles of solute in a larger volume mix Making a Dilute Solution diluted solution Timberlake, Chemistry 7th Edition, page 344

Process of Making a Standard Solution from Liquids Solutions can be made using liquids or solids (or gases). To make a 5% solution v/v (volume to volume) This means to add 5 mL of solute in 95 mL of solvent. The total is 5 mL / 100 mL or 5%. For the diagram add 25 mL of liquid solute and add water to bring volume to 500 mL (about 475 mL water). SAFETY NOTE: Always add acid concentrate to water…never add water to concentrated acid. If you’ve seen what happens when water or ice crystals hit hot oil…a similar phenomenon occurs when water is added to concentrated acid. The addition of water to concentrated dissipates a large amount of heat. This heat rapidly boils the acid and causes it to spatter. If however, you start with a large volume of water and slowly add acid, the same amount of heat is generated. This time, the large volume of water is capable of absorbing the heat. The solution will not splatter. Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 483

How to mix a dilute solution from a concentrated stock solution A solution of a desired concentration can be prepared by diluting a small volume of a more-concentrated solution, a stock solution, with additional solvent. – Calculate the number of moles of solute desired in the final volume of the more-dilute solution and then calculate the volume of the stock solution that contains the amount of solute. – Diluting a given quantity of stock solution with solvent does not change the number of moles of solute present. – The relationship between the volume and concentration of the stock solution and the volume and concentration of the desired diluted solution is (Vs) (M s) = moles of solute = (Vd) (M d). Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.

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

Dilution What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution? GIVEN: M1 = 15.8M V1 = ? M2 = 6.0M V2 = 250 mL WORK: M1 V1 = M2 V2 V1 = (6.0M)(250mL) (15.8M) V1 = 95 mL of 15.8M HNO3

Dilution Example Problem 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? M1V1 = M2V2 28.9 M (0.075 L) = 0.100 M (15.0 L) Yes; we’re OK. 2.1675 mol HAVE > 1.50 mol NEED

STOICHIOMETRY MAP (22.4 L/mol) LITERS OF GAS AT STP Molar Volume *only at STP! (22.4 L/mol) MOLES MASS IN GRAMS NUMBER OF PARTICLES Molar Mass (g/mol) 6.022  1023 particles/mol Molarity M = mol/L SOLUTIONS

SOLUTION STOICHIOMETRY (new) 2 Al (s) + 3 CuCl2(aq) → 2AlCl3 (aq) + 3Cu (s) ???? 20.0 mL 0.21g ?M Trial # Original Mass Al (g) Final Mass Al (g) Mass of Al Consumed in rxn (g) Volume of CuCl2 Solution Used (mL) Calculated Molarity (M) 1 0.59 0.38 0.21 20.0 0.58 ????   0.21 g Al 1 mol Al 3 mol CuCl2 = 0.01168 mol CuCl2 26.98 g Al 2 mol Al M = mol = 0.01168 mol CuCl2 = 0.58 M CuCl2 L 0.020 L

Solution Stoichiometry Problem (w/limiting reactants too!) Calculate the mass of chalk formed when 25.0 mL of a 0.100 M calcium nitrate solution is mixed with 20.0 mL of a 1.50 M sodium carbonate solution. Ca(NO3)2 (aq) + Na2CO3 (aq)  CaCO3 (s) + 2NaNO3 (aq) 25.0mL 0.100M 20.0mL 1.50M ? g 0.00250 mol .030 mol X 1 mol CaCO3 1 mol Ca(NO3)2 X 100.09 g CaCO3 1 mol CaCO3 = 0.250 g CaCO3 0.0025 mol Ca(NO3)2 X 1 mol CaCO3 1 mol Na2CO3 X 100.09 g CaCO3 1 mol CaCO3 0.030 mol Na2CO3 = 3.00 g CaCO3

Remember? Standard Molar Volume 1 mol of a gas = 22.4 L at STP Standard Temperature & Pressure 0°C and 1 atm

Gases Stoichiometry Review 2 H2O2 (l) ---> 2 H2O (g) + O2 (g) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP? Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

Gases Stoichiometry Review 2 H2O2 (l) ---> 2 H2O (g) + O2 (g) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP? Solution x 1 mol H2O2 34 g H2O2 x 1 mol O2 2 mol H2O2 x 22.4 L O2 1 mol O2 1.1g H2O2 = 0.36 L O2 at STP