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Solutions, Solubility, and Reaction Types. Brown, LeMay Ch 4 AP Chemistry. 4.1: Solutions & Expressing Concentration. Solution: homogeneous mixture of 2 or more substances solute: component present in lesser quantity solvent: component present in greater quantity.

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Solutions, Solubility, and Reaction Types

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    1. Solutions, Solubility,and Reaction Types Brown, LeMay Ch 4 AP Chemistry

    2. 4.1: Solutions & Expressing Concentration Solution: homogeneous mixture of 2 or more substances • solute:component present in lesser quantity • solvent:component present in greater quantity • Molarity (M)= moles solute = molvolume of solution L • Since volume of a solution changes with temperature, M of a solution changes with temp. too. M is a good unit in measuring the conc. of a solution under constant temp conditions like during an experiment in the lab. Ex. Titration • Molality (m)= moles solute = molkg of solvent kg • Since mass of a solution is temp. independent, m does not change with change in temp, hence m is a good unit to use for solution conc. when temp is changing. Ex. B.P. Elevation, F.P. Depression

    3. Which is more concentrated- 1M aq NaCl or I m aq NaCl ?

    4. 4.1: Solutions & Expressing Concentration • *Normality (N):associated with acid & base strength. Normality = molarity x n (where n = the number of protons exchanged in a reaction). = (moles solute)(# of acid/base equivalents) = volume of solution mol/L • Ex. N of 1M H2SO4 will be 2N. • N was used earlier to express concentration of acids/bases but is obsolete now.

    5. Mole Fraction • XA= nA/ nA + nB ….. • Practice Problem: What is the mole fraction of each component in a solution in which 3.57 g of sodium chloride, NaCl, is dissolved in 25.0 g of water? Ans: X H2O 0.0421, X NaCl 0.958 • Note: Sum total of mole fractions of all components is always 1 • Mole fraction is a good unit but not so easy to compute, hence has limited use.

    6. Mass Percent Mass percent is an easy way to express the concentration of solution and is most commonly used. Mass percent can be written in as mass/mass (m/m) or mass/volume (m/v) or volume/volume (v/v). Let us look at each of these carefully. Mass % = mass of solute X100 mass of solution Mass % = mass of solute X100 volume of solution Mass %= volume of solute X 100 volume of solution

    7. Practice Problem on Mass % • What does 2% (m/m) aq NaCl mean? 2% m/V? 2% v/v • What mass % unit are you most likely to find vinegar’s conc. expressed in? Salt solution’s? • Calculate the mole fraction of HCl in a solution of hydrochloric acid containing 36% HCl by weight. 0.22 • What is the molality of a solution made by dissolving 5.0g of toluene (C7H8) in 225g of benzene (C6H6)? 0.24m

    8. Titration movie on Titration -It is a method to determine the molarity of unknown acid or base • In titration, an acid or base of unknown molarity is titrated against a standard solution (whose M is known) of acid or base. • The end point in a titration is indicated by a color change by the indicator. Indicators are weak acids or bases and are added in small quantity (1-3 drops) to indicate the end point. • At equivalence point (which should be close to end point), moles of H+ = moles of OH- M1V1= M2 V2 (sometimes used to get moles , M= moles/L , so moles= M XV) -What other ways can you get the moles- for a solid acid or base? For a gas? -color change by indicator indicates end point -end point and equivalence points should be close. Equivalence point is defined when moles of acid and base become equal in a titration and end point is where the indicator changes color.

    9. Dilution:(Minitial)(Vinitial) = (Mfinal)(Vfinal) Ex: How much water must be added to a solution that contains 100.0 g NaOH in 500. mL of solution to make the final concentration 2.00 M? Minitial = (100.0 g NaOH)/(40.00 g/mol NaOH) = 5.00 M 0.500 L (Mi)(Vi) = (Mf)(Vf) = (5.00 M)(0.500 L) = (2.00 M)(Vf) Vf = 1.25 L, so 0.75 L water must be added.

    10. Practice Problems: M, m and Titrations • What is the molality of lead nitrate in 0.726 M Pb(NO3)2 is: (The density of the solution is 1.202 g/ml) Ans. 0.755m • A solution is prepared by dissolving 15.0g of NH3 in 250g of water. The density of the resulting solution is 0.974 g/ml. What is the mole fraction of NH3 in the solution. Ans. 0.060 • How many mL of a 3M NaOH solution are required to completely neutralize 20.0 mL of 1.5M H2SO4? (Start by writing a balanced equation!) Ans. 20.0 mL • How many g of NaOH is required to completely react with 100. mL of 1M HCl?

    11. 4.2: Properties of Solutes • Ionic compounds (salts):dissociate in water by solvation, in which polar H2O molecules surround and separate ions from ionic crystal lattice • Molecular compounds:do not ionize because intramolecular bonds are stronger than bonds that form between H2O and molecule • Acids may ionize or not, depending on their strength. Bases dissociate based upon their strength. Refer to Naming Notes for further details.

    12. Electrolyte:substance which, in aqueous solution, ionizes and thus conducts electricity. Ex: salt in water.(salt dissolving in water) Non-electrolyte:substance which, in aqueous solution, does not dissociate and thus does not conduct electricity Strong & weak electrolytes:conductivity depends on degree of dissociation and equilibrium position: HA (aq) ↔ H+ (aq) + A- (aq)Strong = nearly completely dissociated Weak = partially dissociated • ( electrolytes animation) • (Conductivity of different substances)

    13. 4.3: Acids & Bases • Acids: H+ donor • Bases: H+ acceptor • Be (OH)2 and Mg(OH)2 are weak bases. Why? Group 1A’s: LiOH, NaOH, KOH, RbOH, CsOH Heavy 2A’s: Ca(OH)2, Sr(OH)2, Ba(OH)2 HBr, HCl, HI HNO3, H2SO4 HClO4, HClO3

    14. 4.4: Ionic Equations • Molecular equation: shows complete chemical equation with states of matter, undissociated BaCl2 (aq) + Na2SO4 (aq) → 2 NaCl (aq) + BaSO4 (s) • Complete ionic equation: shows complete chemical equation with states of matter, dissociated if appropriate Ba2+(aq) + 2 Cl-(aq) + 2 Na+(aq) + SO42-(aq) → 2 Cl-(aq) + 2 Na+(aq) + BaSO4 (s) • Spectator ions:present in reaction but do not “participate”; depend on solubility rulesCl- (aq) and Na+ (aq) ( Precipitation Animation) 3. Net ionic equation:shows chemical equation without spectator ionsBa2+ (aq) + SO42- (aq) → BaSO4 (s) Good Practice for writing equations

    15. Solubility Rules (memorize them!) • Also, compounds with Group 1A and NH4+ cations

    16. Other anions (e.g., BrO3- or Cr2O72-) are generally insoluble except with the soluble cations (Group 1A and NH4+) • Precipitation Reaction Animation

    17. AP Exam: Predicting Reaction Products Reactants: • Write all soluble ionic compounds, strong acids, and strong bases in aqueous solution as their constituent ions. Never write NaCl or Ca(NO3)2, HCl or HNO3, NaOH or Ba(OH)2, etc. Products: • Assume a reaction occurs. • Inspect all of the ions to see if any precipitates will form. • Cross out any spectator ions. • Balance the reaction (and states of matter are not needed.) • For reaction prediction, please go to podcasts and WS on the website.