Presentation on theme: "Daniel L. Reger Scott R. Goode David W. Ball Lecture 04 (Chapter 4) Chemical Reactions in Solution."— Presentation transcript:
Daniel L. Reger Scott R. Goode David W. Ball http://academic.cengage.com/chemistry/reger Lecture 04 (Chapter 4) Chemical Reactions in Solution
Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7 th Edition, 2011; http://en.wikipedia.org/; http://www.medfinity.com/index.php?cPath=704_737http://en.wikipedia.org/ Solutions: Homogeneous mixtures of two or more substances Particles uniformly distributed and transparent (clear) Particles in constant motion (KE), don’t settle under influence of gravity Solution = Solvent + Dissolved Solute Solvent is most abundant substance comprising solution Solute(s) is/are other substance(s) that are not the solvent Solutions usually liquids but may be gases or solids Aqueous solution: water is the solvent. Strong electrolyte: compound that separates completely into ions in water. Weak electrolyte: molecule that only partially ionizes when dissolved in water. Solutions
Solubility Soluble substances: Dissolve completely in solvent Insoluble substances: Do not dissolve in solvent Immiscible: Liquid solute that does not dissolve in a liquid solvent The extent of solubility can vary Isopropyl alcohol and water are completely soluble in any proportion Limit on how much sugar will dissolve Oil and water form distinct, separate layers Saturated solution: Solution with max amount of dissolved solute Supersaturated solution: Solutions where amount of solute dissolved is greater than solute solubility Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7 th Edition, 2011
Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7 th Edition, 2011 When solid, ionic compounds placed in water, polar water molecules orient themselves by charge along surface of solid Produces a shielding effect, allows water to remove ions from matrix (i.e., ions become hydrated) Hydrated ions become evenly distributed throughout solution Formation of Solutions Process continues until number of ions in solution results in saturation. Ions begin to reform solid equilibrium. Similar results are observed with polar, non-ionic solids (e.g., sugar)
Two reasons why solute won’t dissolve in solvent: 1.Forces between solute particles stronger than solvent particles 2.Solvent particles more attractive to each other than solute particles “Like dissolves like” Polar molecules dissolve in polar solvents (e.g., water) Non-polar molecules dissolve in non-polar solvents (e.g., CCl 4 ) Formation of a Solution
Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7 th Edition, 2011 Predict the solubility of the following: 1.NH 3 in water 2.O 2 in water 3.Ca(NO 3 ) 2 in water 4.Mg 3 (PO 4 ) 2 in water 5.Paraffin wax (nonpolar) in CCl 4 6.BaCO 3 in water 7.Li 3 PO 4 in water 8.(NH 4 ) 2 SO 4 9.PbCl 2 Predicting Solubility 1.Soluble, NH 3 is polar 2.Insoluble, O 2 is nonpolar 3.Soluble 4.Insoluble 5.Soluble 6.Insoluble 7.Soluble 8.Soluble 9.Insoluble
A precipitation reaction involves the formation of an insoluble product or products from the reaction of soluble reactants. Example: Mixing AgNO 3 and LiCl, both of which are soluble, produces insoluble AgCl. Precipitation Reactions AgNO 3 (aq) + LiCl(aq) AgCl(s) + LiNO 3 (aq)
A. Build a table of the reactants and possible products; label each as soluble or insoluble. PbSO 4 is insoluble and the equation is: Pb(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) PbSO 4 (s) + 2NaNO 3 (aq) Q. What insoluble compound, if any, will form when solutions of Pb(NO 3 ) 2 and Na 2 SO 4 are mixed? Write the chemical equation. Precipitation Reactions
Ionic Equations Ionic compounds and some polar covalent compounds may dissociate in water H +, Na +, K +, Mg 2+, Ca 2+, Fe 2+, Fe 3+, Ag 1+, Pb 2+, F -, Cl -, Br -, Br -, OH -, NH 4 +, SO 4 2-, SO 3 2-, PO 4 3-, NO 2 -, NO 3 -, CO 3 2- Reactions can be represented by complete ionic equation or net ionic equation Net Ionic Equation: The actual chemistry that happens Complete Ionic Equation: Shows all strong electrolytes as ions in solution HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O( l ) Molecular equation H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq) Na + (aq) + Cl - (aq) + H 2 O( l ) Complete ionic equation Net ionic equation H + (aq) + OH - (aq) H 2 O( l ) In the total ionic equation, Na+ and Cl- appear as both reactants and products These are spectator ions Spectator ions: Do not participate in reaction excluded from net ionic equation
Overall equation: Pb(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) PbSO 4 (s) + 2NaNO 3 (aq) Complete ionic equation: Pb 2+ (aq) + 2NO 3 - (aq) + 2Na + (aq) + SO 4 2- (aq) PbSO 4 (s) + 2Na + (aq) + 2NO 3 - (aq) Net ionic equation: Pb 2+ (aq) + SO 4 2- (aq) PbSO 4 (s) Ionic Equations (Example 02)
Ionic Equations: Example 01 Na 2 SO 4 (aq) + BaCl 2 (aq) BaSO 4 (s) + 2NaCl(aq) Molecular equation 2Na + (aq) + SO 4 2- (aq) + Ba 2+ (aq) + 2Cl - (aq) BaSO 4 (s) + 2Na + (aq) + 2Cl - (aq) Total ionic equation Net ionic equation SO 4 2- (aq) + Ba 2+ (aq) BaSO 4 (s)
Ionic Equations: Example 02 CaCl 2 (aq) + Na 2 CO 3 (aq) 2NaCl(aq) + CaCO 3 (s) Molecular equation Ca 2+ (aq) + 2Cl - (aq) + 2Na + (aq) + CO 3 2- (aq) CaCO 3 (s) + 2Cl - (aq) + 2Na + (aq) Total ionic equation Net ionic equation Ca 2+ (aq) + CO 3 2- (aq) CaCO 3 (s)
2.00 L of solution contains 1.50 mol of solute. Calculating Solution Concentrations (Molarity) Molarity 150 mL of solution contains 0.210 mol of solute. 315 mL of solution contains 10.3 g of C 3 H 7 OH. Note the units
Molarity of Ions One mole of K 2 SO 4 dissolves in water to form two moles of K + ions and one mole of SO 4 2- ions. K 2 SO 4 (s) 2K + (aq) + SO 4 2- (aq)
Solution Preparation by Mixture Describe how to prepare 1.00 L of 1.50 M CoCl 2 solution Describe how to prepare 1.50 L of 0.50 M CoCl 2 solution Describe how to prepare 200 mL of 0.200 M CoCl 2 solution Describe how to prepare 500 mL of 1.6 × 10 -4 M Pb(NO 3 ) 2 solution
Solution Preparation by Dilution Solutions of lower concentration can be prepared by dilution of more concentrated solutions of known molarity. Volume (L) of dilute solution Moles of solute Molarity of dilute solution Molarity of concentrated solution Volume (L) of concen- trated solution
Solution Preparation - Dilution This equation works for any form of concentration based on volume Describe how to prepare 250 mL of 0.100 M NaCl solution from 2.00 M NaCl Describe how to prepare 500 mL of 0.250 M NaOH solution from 6.00 M NaOH
Test Your Skills (a) Calculate the volume of 4.00 M K 2 SO 4 that is needed to prepare 600 mL of a 0.0200 M solution of K 2 SO 4. (b) Calculate the molar concentration of K + ions in the 0.0200 M solution. Answers: (a) 8.33 mL (b) 0.0400 M K +
In stoichiometric calculations, molarity is used to calculate moles from volume of solution analogous to using molar mass from mass of a solid. Solution Stoichiometry Calculations
Solution Stoichiometry Stoichiometry deals with relative quantities of reactants and products in balanced chemical equation We can now apply everything learned about mass, moles, concentrations and balancing equations to solutions Ex. 01: What volume of 0.200 M NaOH solution is needed to exactly react with 0.150 moles of HCl? We need 0.150 moles of NaOH for 0.150 moles of HCl (1:1 stoichiometry) NaOH solution is 0.200 M (moles/L)
Solution Stoichiometry Ex. 01: What volume of 0.185 M NaOH solution is needed to exactly react with 25.0 mL of 0.255 M HCl? We need to figure out moles in 25.0 mL of 0.255 M HCl Now we can find volume of of 0.185 M NaOH from moles HCl and stoichiometry
Calculate the mass of lead(II) sulfate formed in the reaction of 145 mL of 0.123 M lead(II) nitrate and excess sodium sulfate. Pb(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) PbSO 4 (s) + 2NaNO 3 (aq) Solution Stoichiometry Calculations
Test Your Skills Calculate the mass of magnesium hydroxide need to react completely with 356 mL of 6.92 M H 2 SO 4.
Test Your Skills Calculate the mass of magnesium hydroxide need to react completely with 356 mL of 6.92 M H 2 SO 4. Answer: 287 g Mg(OH) 2
Titrations In a titration, the concentration and volume of a solution of known concentration is used to determine the concentration of an unknown solution. Equivalence point: the point in a titration where stoichiometrically equivalent amounts of the two reactants have been added. An indicator is a compound that changes color as an acidic solution becomes basic or basic solution becomes acidic. The indicator changes color at the end point – the end point of the indicator should match the equivalence point.
Titration with Phenolphthalein Indicator Left: acidic solution with indicator added Center: end point - very slight pink color Right: pink color after excess base added
Example: Titrations Calculate the molarity of an HCl solution if 26.4 mL of the solution neutralizes 30.0 mL of 0.120 M Ba(OH) 2.
Test Your Skills Calculate the molarity of an NaOH solution if 33.4 mL of the solution is neutralized by 16.0 mL of a 0.220 M solution of H 2 SO 4.
Test Your Skills Calculate the molarity of an NaOH solution if 33.4 mL of the solution is neutralized by 16.0 mL of a 0.0550 M solution of H 2 SO 4. Answer: 0.0527 M
Gravimetric Analysis Calculate the molarity of Cl - ions in a 250 mL solution if addition of excess silver nitrate yielded 1.34 g of silver chloride.
Test Your Skill What mass of lead(II) chloride forms in the reaction of 24.3 mL of 1.34 M lead(II) nitrate and 38.1 mL of 1.22 M sodium chloride? Hint: Limiting reactant?
Test Your Skill What mass of lead(II) chloride forms in the reaction of 24.3 mL of 1.34 M lead(II) nitrate and 38.1 mL of 1.22 M sodium chloride?