Presentation on theme: "Daniel L. Reger Scott R. Goode David W. Ball Chapter 4 Chemical Reactions in Solution."— Presentation transcript:
Daniel L. Reger Scott R. Goode David W. Ball http://academic.cengage.com/chemistry/reger Chapter 4 Chemical Reactions in Solution
Solvent: compound that has the same physical state as the solution - frequently a liquid. Solute: substance being dissolved. 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 of Ionic Compounds Predict the solubility of (a) (NH 4 ) 2 SO 4 and (b) PbCl 2.
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. AgNO 3 (aq) + LiCl(aq) AgCl(s) + LiNO 3 (aq) Precipitation Reactions
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
Build a table of the reactants and possible products; label each as soluble or insoluble. Cations Anions Na + Pb 2+ NO 3 - soluble NaNO 3 soluble Pb(NO 3 ) 2 SO 4 2- soluble Na 2 SO 4 insoluble PbSO 4 PbSO 4 is insoluble and the equation is: Pb(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) PbSO 4 (s) + 2NaNO 3 (aq) Precipitation Reactions
Complete ionic equation shows all strong electrolytes as ions in solution 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) Complete Ionic Equation
Net ionic equation shows only those species in the solution that actually undergo a chemical change. Overall equation: Pb(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) PbSO 4 (s) + 2NaNO 3 (aq) Net ionic equation: Pb 2+ (aq) + SO 4 2- (aq) PbSO 4 (s) Spectator ions are those that do not participate in the chemical reaction. Net Ionic Equation
Test Your Skill Write the net ionic equation for the reaction of HCl(aq) and KOH(aq).
Molarity (M) is the number of moles of solute in one liter of solution. moles of solute Molarity = liters of solution Molarity
Example: Molarity of Solution What is the molar concentration of NaF in a solution prepared by dissolving 2.51 g of NaF in enough water to form 200 mL of solution?
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)
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
Dilution In a dilution problem: moles of solute in dilute solution = moles of solute in the concentrated solution molarity(conc) x volume(conc) = molarity(dil) x volume(dil) Use this formula only for dilution problems, not for problems involving equations.
Example: Dilution Calculate the volume of 6.00 M H 2 SO 4 that is needed to prepare 2.00 L of a 0.200 M solution of H 2 SO 4.
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.
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
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.
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.
Titrations 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.
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?