Chemistry 101 : Chap. 4 Aqueous Reactions and Solution Stoichiometry (1) General Properties of Aqueous Solutions (2) Precipitation Reactions (3) Acid-Base Reactions (4) Concentration of Solutions (5) Solution Stoichiometry and Chemical Analysis

Aqueous Solutions  Solution : A homogeneous mixture of two or more substance  Solvent : The substance present in the largest amount  Solute(s) : The other substance in the solution Aqueous Solutions : Solutions in which water is the dissolving medium (or solvent)

Ionic Compounds in Water When ionic compounds dissolve in water, they dissociate completely into ions. The ions are surrounded by water molecules H 2 O NaCl (s) NaCl (aq) or Na + (aq) + Cl - (aq) [ Solvation of Na + and Cl - ]

Molecular Compounds in Water When molecular compounds dissolve in water, they do not dissociate into ions in general. The molecules remain intact and are surrounded by water molecules + H 2 O CH 3 OH (l) CH 3 OH (aq)

Molecular Compounds in Water Some molecular compounds, when dissolved in water, dissociate (ionoized) into ions.  Strong Acid, such as hydrochloric acid, dissociate completely HCl (aq)  H + (aq) +Cl - (aq)  Weak Acid, such as acetic acid, dissociate only partially CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) NOTE: All three species are present in solutions

Electrolyte vs. Nonelectrolyte  Electrolyte : A substance whose aqueous solutions contains ions  Ionic compounds  Strong acids  Weak acids strong electrolyte weak electrolyte  Nonelectrolyte : A substance that does not form ions in solution NOTE: Don’t confuse the extent to which an electrolyte dissolves with whether it is strong or weak Ex. CH 3 COOH is extremely soluble, but it is a weak electrolyte. Ba(OH) 2 is not very soluble, but it is a strong electrolyte.  the dissolved Ba(OH) 2 completely dissociate

Electrolyte Solutions Electrolyte solutions conduct electricity !!  Electrical conductivity is the flow of charged particles through substance, such as the flow of electrons through a wire or the flow of ions through a solution.  Testing electrical conductivity of electrolyte solutions

Electrolyte Solutions

Precipitation Reactions  Recall : Three types of chemical reactions studied in Ch. 3  Combination Reaction : A + B  C  Decomposition Reaction : A  B + C  Combustion Reaction : C x H y + O 2  CO 2 + H 2 O  Exchange Reaction (Metathesis Reaction) AB + CD  AD + CB Reactants “exchange partners” to form the products

Precipitation Reactions  Precipitation Reactions : Reactions that result in the formation of insoluble products Pb(NO 3 ) 2 (aq) + 2KI (aq)  PbI 2 (s) + 2KNO 3 (aq) yellow ppt AgNO 3 (aq) + NaCl (aq)  AgCl (s) + NaNO 3 (aq) white ppt Mg(NO 3 ) 2 (aq) + 2NaOH (aq)  Mg(OH) 2 (s) + 2NaNO 3 (aq) white ppt

Precipitation Reactions What is happening in molecular level (nanoscale view)? AgNO 3 (aq) + NaI (aq)  AgI (s) + NaNO 3 (aq) brown ppt + NOTE: Water molecules are omitted in these pictures

Solubility Guidelines  Question: How can you tell if a precipitation will form during an exchange reaction?

Solubility Guidelines  Example: Which of the following compounds is likely soluble in water? Fe(NO 3 ) 3 AgCl CaCO 3 (NH 4 ) 2 SO 4 CuSO 4 Cu(OH) 2

Solubility Guidelines  Empirical Rule : Salts containing following ions are always soluble  Group 1A ion (alkali metal): Li +, Na +, K +, …  Ammonium ion : NH 4 +  Nitrate ion : NO 3 -  Acetate : CH 3 COO -  Insoluble Ionic compounds : Ag +, Hg 2 +, Pb 2+ + Halogen anions (Cl -, Br -, I - )

Predicting Precipitation Reactions  Example : Complete the following equation and determine whether or not a precipitation reaction (exchange reaction) will occur. If a reaction will occur, balance the equation Pb(NO 3 ) 2 (aq) + KI (aq)  (1) Determine the products of exchange reaction: (2) Determine if there is any insoluble product (3) Balance the chemical equation

Predicting Precipitation Reactions  Example : Complete the following equation and determine whether or not a precipitation reaction (exchange reaction) will occur. If a reaction will occur, balance the equation Mg(NO 3 ) 2 (aq) + KCl (aq)  (1) Determine the products of exchange reaction: (2) Determine if there is any insoluble product (3) Balance the chemical equation (?)

Net Ionic Equation Pb(NO 3 ) 2 (aq) + 2 KI (aq)  PbI 2 (s) +2 KNO 3 (aq)  Molecular equation :  (complete) Ionic equation: Pb 2+ (aq) + 2NO 3 - (aq) + 2K + (aq) + 2I - (aq)  PbI 2 (s) + 2K + (aq) + 2NO 3 - (aq) spectator ions  Net ionic equation : Pb 2+ (aq) + 2I - (aq)  PbI 2 (s) NOTE: Mg(NO 3 ) 2 (aq) + 2 KCl (aq)  All ions are spectators ! No net reaction !!

Net Ionic Equation  Summary:  Spectator Ions : Ions present in solution before and after the reaction.  Net Ionic Equation : Spectators are omitted and only the ions involved in forming the precipitation are shown NOTE: Net ionic equation must be mass balanced and charge balanced.  Writing net ionic equation : (1) Write a balanced molecular equation for the reaction (2) Write the complete ionic equation [precipitation or not] (3) Identify and remove the spectator ions

Net Ionic Equation  Example : Write the net ionic equation for the precipitation reaction that occurs when solutions of nickel nitrate and sodium hydroxide are mixed. (1) Write the balanced molecular equation (2) Write the complete ionic equation (3) Write the net ionic equation

Net Ionic Equation  Example : Write the net ionic equation for the precipitation reaction that may occur when solutions of calcium nitrate and sodium acetate are mixed. (1) Write the balanced molecular equation (2) Check the solubility of products (3) Net ionic equation:

Acids and Bases  Properties of Acids  Acids ionize to produce protons: HCl (aq)  H + (aq) + Cl - (aq)  Turn blue litmus red  Taste sour  Aqueous solution has a pH less than 7  Properties of Bases  Bases produce OH - in water: NaOH (aq)  Na + (aq) + OH - (aq)  Turn red litmus blue  Bitter taste  Aqueous solution has a pH more than 7  React with metal to produce H 2 gas

Acids and Bases  Strong acids and bases Strong acids and bases ionize completely in aqueous solution. Strong acids and bases are strong electrolytes. HCl (aq)  H + (aq) + Cl - (aq) 100% ionized NaOH (aq)  Na + (aq) + OH - (aq) 100% ionized HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 4, HClO 3 strong acids group 1A hydroxide [LiOH, NaOH, KOH] group 2A hydroxide [Ca(OH) 2, Sr(OH) 2, Ba(OH) 2 ] strong bases

Acids and Bases  Weak acids and bases Weak acids and bases ionize partially in aqueous solution. Weak acids and bases are weak electrolytes. HF (aq)  H + (aq) + F - (aq) ~10% ionized HF, CH 3 COOH, HNO 2, … weak acids NH 3, …weak bases NOTE: NH 3 do not have OH. However, they produce OH - by accepting a proton from H 2 O. NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH - (aq)

Acid-Base Reactions  acid-base neutralization reactions: An exchange reaction in which a molecular compound (H 2 O) is formed. molecular equation: HCl(aq) + NaOH(aq)  H 2 O(l) + NaCl(aq) complete ionic equation : H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq)  H 2 O(l) + Na + (aq) + Cl - (aq) net ionic equation : H + (aq) + OH - (aq)  H 2 O(l)

Acid-Base Reactions NOTE: (1) The products of simple acid-base reactions are water and ionic compounds called “salts”. HNO 3 (aq) + NaOH (aq)  H 2 O(l) + NaNO 3 (aq) acid base water salt (2) The formation of water is what drives the acid-base neutralization reaction H + (aq) + OH - (aq) H 2 O(l) possible, but much less favorable than the other way

Acid-Base Reactions  Example : Complete and balance the following equation for an acid-base neutralization reaction: H 3 PO 4 (aq) + KOH(aq)  What is the net ionic equation?

Acid-Base Reactions  Example : Complete and balance the following equation for an acid-base neutralization reaction: Mg(OH) 2 (s) + HCl(aq)  What is the net ionic equation?

Acid-Base Reactions  acid-base reactions with gas formation : Acids reacting with certain weak bases produce a salt (ionic compound) and gas (molecular compound). The most common reaction of this type involves carbonates (CO 3 2- ). 2 HCl(aq) + Na 2 CO 3 (aq)  H 2 CO 3 (aq) + 2NaCl(aq) unstable H 2 CO 3 (aq)  H 2 O (l) + CO 2 (g) The overall reaction will be 2 HCl(aq) + Na 2 CO 3 (aq)  H 2 O(l) + 2NaCl(aq) + CO 2 (g)

Acid-Base Reactions  Example : What is the net ionic equation for the reaction between Na 2 CO 3 (aq) and HCl(aq)?

Acid-Base Reactions  Example: Complete and balance the following equation. HCl(aq) + CaCO 3 (s)  What is the net ionic equation?

Concentrations of Solutions  Concentration: The amount of solute dissolved in a given amount of solvent or solution  Molarity: The most common way of expressing concentration in chemistry A 1.00 molar solution (= 1.00M) contains 1.00 mol of solute in every liter of solution.

Concentrations of Solutions 1 mol 1 liter 1M 1 mol ? liter 1M

Concentrations of Solutions  Example : Calculate the molarity of a solution made by dissolving 23.4 g of Na 2 SO 4 (MW= 142) in enough water to form 125 mL of solution.

Concentrations of Electrolyte 1 mole of Na 2 SO 4 (s) 1M solution of Na 2 SO 4 1L 2Na + (aq) + SO 4 2- (aq) Each formula unit of Na 2 SO 4 that dissolve produces two Na + ions and one SO 4 2- ion.  The concentration of Na + (2M) will be twice the SO 4 2- concentration (1M).

Concentrations  Example: How many grams of Na 2 SO 4 (MW=142) are required to make 0.350L of 0.500M Na + (aq)?

Dilutions add solvent number of solute molecules before dilution number of solute molecules after dilution = number of moles of solute before dilution number of moles of solute after dilution = M conc, V conc M dilute, V dilute M conc  V conc = M dilute  V dilute

Dilutions  Example : What is the molarity of the resulting solution when 15 mL of 0.65M solution is diluted to 315mL?

Solution Stoichiometry Example : What volume of a M HNO 3 solution is needed to completely neutralize 0.1 g of Ca(OH) 2 ? Stratagy: (1) What is the chemical reaction involving HNO 3 (aq) and Ca(OH) 2 (s)? (2) How many moles of HNO 3 is needed to neutralize 0.1 g of Ca(OH) 2 ? (3) What is the volume of 0.100M HNO 3 solution neutralizing that many moles of Ca(OH) 2 ?

Solution Stoichiometry aA + bB  cC + dD