Chemistry Ch 15 (Solutions) Study Guide Introduction

Chemistry Ch 15 (Solutions) Study Guide Introduction Name: Note: a word marked (?) is a vocabulary word you should know the meaning of. A homogeneous (?) mixture, or, is a mixture in which the individual components are uniformly distributed. In other words, if you were to sample a solution at various locations, the compositions would be the same. A solution can be a solid, liquid or. Making a solution: The solute (?) is the component that changes percentage., or is present in the The solvent (?) is the component into which the solutes dissolve. In the above diagram, identify the solute (?) solvent (?) If a small amount of carbon monoxide gas were released into the air, would the CO be the solvent or a solute? Special solutions (?) Solutions in which water is the solvent are called solutions. are solutions that have alcohol as the solvent dental fillings. are solutions in which mercury is the solvent. These were formerly used in (?) Solid solutions of one metal dissolved into another are. Brass is a special alloy of and, while bronze is copper and zinc with. Another alloy you may know is pewter, made of tin and. 1

Solubility (?) the mass (grams) of solute that will dissolve in each 100 grams of water, at a specified temperature. When an ionic substance dissolves in water it breaks up into its component. Example: Sodium chloride (NaCl) dissolving in water Additionally, when salts (ionic solids) dissolve in water, the resulting solutions will conduct an electric current. When covalent substances dissolve in water, the molecules do not separate. Solutions of covalent substances do not conduct electric currents well. Acids are exceptions to this rule. Acids are covalent compounds, but behave like ionic compounds, losing their H s (one at a time, if they have more than one H.) Practice problems: Identify the ions that are produced when each of the following dissolve in water: cation anion NaNO 3 : CaCO 3 : NH 4 Cl: K 3 PO 4 : HCl: H 2 SO 4 : *** It would be to your benefit to review the solubility rules (ch. 7) *** 2

What factors determine how much solute will dissolve in a solvent? 1) Explain the phrase like dissolves like : 2) Temperature. Usually, more of a solid solute (like sugar) will dissolve when the solvent s temperature is. This relationship is different for, however. The graph at right indicates that as the temperature of water increases, what happens to the solubilities of common gases? Why would fish prefer cold water over warm water? 3) Pressure. As the pressure of the gas over a liquid increases, what happens to the solubility of the gas in the liquid? If you pop the tab on a soda and release the pressure what happens to the carbonation? What will quickly happen to a soda when the pressure is released and the temperature is warm? What gas is dissolved in pop? In other words, what is carbonation? The solubility of a substance at various temperatures is shown on a graph called a solubility curve. See the next page. 3

Name 1) Which salt (a synonym for ionic compound ) on the chart is most affected by changes in temperature? Solubility Curves 2) Which salt is least affected by changes in temperature? 3) Which salt on the chart is most soluble at 10 o C? 4) At what temperature is the solubility of KNO 3 the same as the solubility of KCl? 5) What is the solubility of potassium nitrate at 50 o C? (Another way of saying this is how many grams of KNO 3 will dissolve in 100 g of water at 50 o C? ) 6) How many grams of KNO 3 could be dissolved in 500 grams of water at 5 o C? Set up a proportion; show your work. 7) How many grams of NaNO 3 can be dissolved in 100 grams of water at 30 o C? 8) How much NH 4 Cl could be dissolved in 200 grams of water at 20 o C? Show your work. 9) At what temperature is the solubility of HCl 60 grams/100 grams of water? 10) Based on their decreasing solubility as temperature increases, NH 3, HCl, and SO 2 must all be. 11) How many grams of KClO 3 can be dissolved in 50 grams of water at 25 o C? Show your work. 12) How many grams of sodium nitrate can be dissolved in 180 grams of water at 15 o C? Show your work. 4

13) How many grams of KCl are in 250 g of water that is saturated with KCl at 25 o C? Show your work. 14) How many grams of sodium nitrate are needed to saturate 40 grams of water at 65 o C? Show your work. 15) A solution of 20 grams of KNO 3 in 100 grams of water at 20 o C is ( unsaturated / saturated )? 16) A solution of 47 grams of NaNO 3 in 50 grams of water would be saturated at what temperature? 17) If 285 grams of water at 15 o C that is saturated with NaNO 3 were allowed to evaporate, how many grams of NaNO 3 would be left behind in the beaker? Show your work. 5

Solution Concentration For every substance there is a limit to how much solute can dissolve into a given amount of solvent. For example as you add table salt to water it disappears relatively quickly. However, as you continue to add salt to the water, the salt dissolves more slowly, ultimately reaching a point where it stops dissolving and begins to accumulate at the bottom of the container. If more solute will dissolve into the solution, the solution is said to be. If no more solute will dissolve into the solution, at that temperature, the solution is said to be. Solutions of certain compounds (e.g. sodium acetate) can also be supersaturated, having more solute dissolved than should be possible at a particular temperature. As long as a supersaturated solution is left untouched, it will just sit there. But if disturbed in any way, the excess solute precipitates rapidly. For a good look at a supersaturated solution visit this website: http://www.youtube.com/watch?v=nvhrxr5jajg also search hot ice, supersaturated solutions, and sodium acetate If we look at the terms saturated, unsaturated and super saturated in reference to a solubility curve, saturated solutions would be at temperature-mass combinations on the curve, unsaturated solutions would be at temperature-mass combinations the curve and super saturated solutions would be at temperature-mass combinations the curve. Solutions can also be discussed in the even more generic terms concentrated and dilute. Concentrated solutions have a relatively amount of solute whereas dilute solutions have a amount of solute dissolved in the solution. While these generic terms are of use in an everyday setting you should be able to recognize that they are not specific enough to be useful in a scientific setting, especially chemistry. We need terms that give actual quantitative (number-based) descriptions of solution concentration: mass percent and molarity. ( At this point, just be aware that there are other measures of concentration also that we will not discuss in class: (1) molality, (2) normality, (3) volume %, and (4) mole fraction.) 6

Mass Percent Mass Percent = grams of solute x 100 total grams of solution Example: 3.00g of KCl are dissolved in 17.00g of water [3.00g / (3.00g + 17.00g)] x 100 = 15% The KCl is the solute (3.00g) and the combination of KCl (3.00g) and water (17.00g) is the solution. Example: How much water must be added to 25 grams of NaCl to have a 30.0% solution? 30% = 0.30 = [25 g NaCl / (25 g NaCl + X g H 2 O) Solve for X. 0.30(25 + X) = 25 7.5 + 0.30X = 25 0.30X = 17.5 X = 58.3 grams Sample problem: Calculate the mass percent of a solution containing 1.60 g of barium chloride dissolved in 50.00 g of water. Molarity As we have seen throughout the year, chemical reactions work in terms of Therefore, being able to calculate the molarity (M) of a solution is useful. instead of grams. M = moles of solute liters of solution units: mol/l M (molar) Definition: A 1.00 M (1.00 Molar) solution contains 1 mole of solute dissolved in enough water to make exactly 1 L of solution. Example: What is the molarity of 6.50g of NaOH dissolved into 2.00L of solution? 6.50g / 40g/mol = 0.1625mol NaOH (turn grams into moles by dividing by the molar mass) 0.1625mol NaOH / 2.00L = 0.0813M NaOH solution Sample problem Calculate the molarity of a Barium hydroxide solution made by dissolving 2.60g of barium hydroxide in water to a total volume of 2500. mls 7

Converting Between Mass % and Molarity Mass % can be converted to Molarity because grams can be converted to moles. The molar mass of the solute and the density of the solution must both be known. Example: Formalin is a 37.0% solution of formaldehyde (CH 2 O) in water. The density of formalin is 1080 g/l. What is the molarity of formaldehyde in a formalin solution? (The molar mass of CH 2 O is 30 g/mol.) Let s assume that we have 1.00 L of solution. 1.00 L x 1080 g/l = 1080 grams of solution. 37.0% of 1080 g = 399.6 grams (don t worry about sig. figs. yet) 399.6 grams / 30 g/mol = 13.32 moles. 13.32 moles dissolved in 1.00 L = 13.3 M Dilutions In the laboratory it is often necessary to make solutions that are less concentrated than the concentrated or stock solutions sold by chemical companies. In order to make a weaker (more dilute) solution, more solvent is added in a process known as. By adding more water, the moles of solute remains the same but the volume of the solution increases, making the molarity of the new solution than the molarity of the original solution. Because molarity and volume are inversely proportional, the equation for calculating the new molarity takes the same form as the (hopefully) now familiar Boyle s Law. Example: Dilution Equation: M 1 V 1 = M 2 V 2 If 0.100L of 6.00 M HCl is diluted by adding enough water make the volume 2.50 L, what is the new molarity? M 1 V 1 =M 2 V 2 (6.00 M)(0.100 L) = M 2 (2.50 L) M 2 = 0.24 M Sample problem 350ml of 8.0M HNO 3 is diluted to a new volume of 500ml. What is the new molarity? 8

Dissociation As you saw yesterday, when an ionic compound or an acid is dissolved in water, the solute dissociates (or ionizes) into its component ions. For example: NaCl (s) Na + + Cl -. One mole of NaCl dissolved in water will produce 1 mole of Na + ions and 1 mole of Cl - ions. However, 1 mole of K 3 PO 4 dissolved in water will produce 3 moles of K + ions and 1 mole of PO -3 4 ions. Therefore, the molarity of K +, sometimes written as [K + ], will be 3 times the molarity of K 3 PO 4. Example: What is the molarity of nitrate ion, if 10.00 g of Mg(NO 3 ) 2 is dissolved in 1.00 L of solution? 10.00 g of Mg(NO 3 ) 2 / 148.3 g/mol = 0.06743 moles of Mg(NO 3 ) 2 0.06743 moles / 1.00 L = 0.0674 M Mg(NO 3 ) 2 However, since there are 2 NO 3 - ions in each mag. nitrate molecule, the M of nitrate ion is 2 x 0.0674 = 0.135 M. Sample problem Calculate the molarity of ammonium ion in 0.350 M ammonium phosphate. 9

Molality (The following material is not in the book!) Just like molarity, molality (m) is based on the number of of solute; this time in relation to the mass (in kilograms) of solvent. Molality is most often used when the solvent isn t water (and so doesn t have a density of 1.00 g/ml), or for calculating certain chemical properties called colligative properties. m = moles of solute mass (in kg) of solvent units: mol/kg m (molal) Example What is the molality of a solution made by dissolving 0.50mol of NaCl into 1000ml of water? 1000ml of water = 1000g of water 1.000kg of water m = 0.50mol / 1.00kg = 0.50m NaCl solution Sample problem 4.3g of FeCl 3 is dissolved in 2,000g of water. What is the molality of the resulting solution? Normality The final unit of concentration we will discuss is normality (N). Normality is used primarily when discussing acids and bases, which we will examine in chapter 17. Normality is defined as the number of per liter of solution. Hopefully, you are now wondering what is an equivalent? if not, the next explanation will come as a complete surprise. An equivalent is the mass (in grams) of the solute divided by the gram equivalent weight (gew) of the solute. At this point you should be wondering what the heck is the gram equivalent weight? If not, I must admit to being a bit surprised, and frankly a bit concerned! As you should recall, from 6 sentences ago, you learned that Normality is primarily used with acids and bases. So the gram equivalent weight is going to be defined in terms of H + (acid) and OH - (base) ions. The gram equivalent weight is the molar mass of the solute divided by the number of hydrogens (from H + or OH - ) in the solute. HCl hydrochloric acid 1 H + H 2 SO 4 sulfuric acid 2H + N = number of equivalents liters of solution units: equiv./l N (normal) equivalents = mass of solute gew gew = molar mass of solute number of Hydrogens 10

Example What is the normality of a 4.5L solution made by dissolving 1.00g of H 2 SO 4 in water? 1.) Determine the gram equivalent weight (gew) [(1.0g/mol)2]+32g/mol+[(16g/mol)4] = 98g/mol 98g/mol / 2 mol H = 49g/equiv 2.) Determine the number of equivalents 1.00g / 49g/equiv = 0.020408equiv. 3.) Calculate the normality Sample problem N = 0.020408equiv / 4.5L = 0.0045N H 2 SO 4 Determine the normality of a solution made by dissolving 3.50g of HNO 3 in enough water to reach a total volume of 1500ml. Converting Between Molarity and Normality: M x n H = N (Molarity x moles of Hydrogen = Normality) Complete the following problems as indicated, show all work be sure to report your answers using the correct number of significant figures and the proper units. 11

What is the mass percent of camphor in a solution that contains 10.00g of camphor dissolved in 120.00g of toluene? (Hint: how many total grams are present?) An alloy is made by dissolving 5.31g Cu and 4.03g Zn in 145g of Fe. Calculate the mass percent of all 3 components of this alloy. What is the molarity of a solution that contains 0.250g of sodium chromate in 100.0ml of solution? How many grams of MgCl 2 are needed to prepare 300.0ml of a 0.400M solution? What is the molarity of a solution that contains 25.0ml of ethanol ( = 0.89g/ml) in 50.0ml of solution? What is the molality of the solution in problem #5 if the solvent was water? What is the molality of a solution containing 60.0mg of Ni(NO 3 ) 2 dissolved into 45.00ml of water? 12

1. Indicate if the following are Unsaturated, Saturated or Supersaturated: a.) A cup of tea into which you could dissolve more sugar b.) A saltwater solution that is beginning to form salt crystals at the bottom of the glass as you stir the solution c.) 100.ml of water at 40 o C that contains 20.g of dissolved KClO 3 (Hint: consult the solubility curve) Chapter 15 problem set pp. 462 466 #1, 9 15, 19, 25, 29, 32, 33, 43, 53 55, 57, 65, 67, 79 & 80 Colligative Properties: (Supplemental to book) A colligative property is a physical property that depends on concentration, but not the type of solute particle. Example: Vapor Pressure Raoult s Law: P = X solvent * P o P vapor pressure of the solvent in a solution P o vapor pressure of the pure solvent X solvent mole fraction of the solvent [x = n solvent / n total ] 13

Boiling Point Elevation and Freezing Point Depression: When you add a solute to a solvent, the vapor pressure of the solution ends up being lower than that of the pure solvent. As a result the boiling point of the solution increases and the melting point decreases. Everyday example antifreeze (ethylene glycol, C 2 H 6 O 2 ) Explain why the boiling point elevation and freezing point depression caused by adding antifreeze to your car radiator is beneficial. Formulas: T BP = k b m m molality T FP = -k f m k b molal boiling point elevation constant k f molal freezing point depression constant 14

Solution Stoichiometry Once again, the concept of stoichiometry has made its way into the study of chemistry (and it will at least one more time this year). The steps are basically the same as they were in chapters 10 and 13: 1. the chemical equation 2. Convert information about the given substances (molarity, volume) into. 3. Moles given x mole ratio =. 4. Convert the moles target back into the units required by the problem (grams, M, liters) 5. Overall: info given moles given x mole ratio moles target unit required Most reactions are done in solution, not with dry crystals. We need a different first step to find out how many moles of starting material there are: Example: How many grams of NaCl would be needed to precipitate out all the Ag + in 500.0 mls of 0.100M AgNO 3? Chemical equation: NaCl (aq) + AgNO 3 (aq) AgCl (ppt) + NaNO 3 (aq) M x V = moles, so 0.100 M x 0.500 L = 0.0500 moles of AgNO 3 (= 0.0500 moles of Ag + ). Use mole ratio: 1 NaCl used : 1 AgNO 3 used. Therefore 0.0500 moles of NaCl are used also. 0.0500 moles NaCl x 58.5 g/mol = 2.92 g of NaCl. Equation: (M given x V given ) x mole ratio x molar mass target = grams target. Practice problem: Vinegar is a 0.85M solution of acetic acid in water. How many grams of baking soda, sodium bicarbonate, would be needed to neutralize 50.0 mls of vinegar? 3.57 g HC 2 H 3 O 2 + NaHCO 3 CO 2 + H 2 O + NaC 2 H 3 O 2 How about a reaction where both the material given & the material sought are solutions? Now, we would also need to change the 3 rd step in our stoichiometry equation. Example: How many mls of 0.250 M Ba(NO 3 ) 2 solution are needed to exactly react with 35.0 mls of 0.360 M K 2 CrO 4 solution? K 2 CrO 4 is the given, and Ba(NO 3 ) 2 the target. Chemical equation: Ba(NO 3 ) 2 (aq) + K 2 CrO 4 (aq) BaCrO 4 (ppt) + 2KNO 3 (aq) (0.360 M x 0.0350L) x (1/1) x (1 / 0.250 M) = 0.0504 L (or 50.4 mls of Ba(NO 3 ) 2 solution) for 1:1 mole ratio reactions, M 1 V 1 = M 2 V 2 for non 1:1 mole ratio reactions, M 1 V 1 (mole ratio) = M 2 V 2 Equation: (M given x V given ) x mole ratio x (1/M target ) = V target Practice problem: How many mls of 0.150 M NaOH solution will exactly react with 25.00 mls of 0.475 M HCl solution? 79.17 mls. 15

A limiting reactant problem: Example: 50 mls of 0.250M Na 2 SO 4 solution is mixed with 35 mls of 0.450 M SrCl 2 solution. What is the maximum mass of SrSO 4 precipitate that you can make? Chemical equation: Na 2 SO 4 (aq) + SrCl 2 (aq) SrSO 4 (ppt) + 2NaCl (aq) As always for a limiting reactant problem, we do stoichiometry twice, starting with both reactants and ending with the precipitate, and keep the smaller amount of product: (0.250 M x 0.050L) x 1/1 x (184 g/mol) = 2.30 g of SrSO 4 (0.450 M x 0.035L) x 1/1 x (184 g/mol) = 2.90 g of SrSO 4 Practice problem: How much lead(ii) sulfate is precipitated when 125 mls of 0.0500 M lead(ii) nitrate reacts with 200. mls of 0.0250 M sodium sulfate? 1.52 grams of PbSO 4 Pb(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) 2 NaNO 3 (aq) + PbSO 4 (ppt) 331 g/mol 142 g/mol 85 g/mol 303 g/mol 16