LECTURE ELEVEN CHM 151 ©slg Topics: 1. Precipitation Reactions 2. Acid / Base Reactions

Acids: Mostly water soluble, commercially available in water solution Salts and Bases: if both cation and anion are large in size and small in charge, ( +1,- 1), it is probably soluble in H 2 O. Checkout following tables... SOLUBILITY OF “Strong Electrolytes” IN WATER To predict when a precipitate will form, we need to know some solubility guidelines:

The Electrolyte is Usually Water Soluble if: THE CATION IS: Na + K + NH 4 + OR THE ANION IS*: Cl -, Br -, I - ClO 4 -, ClO 3 - NO 3 - SO 4 2- C 2 H 3 O 2 -

THE COMPOUND IS PROBABLY INSOLUBLE IF: THE CATION IS NOT: Na + K + NH 4 + THE ANION IS: CO 3 2- PO 4 3- S 2- O 2- OH -

There are a few notable exceptions to the solubility guide on the last slide, principally the ones noted below, which you should be aware of: Insoluble in Water: AgCl, AgBr, AgI PbCl 2, PbBr 2, PbI 2 BaSO 4, PbSO 4

Reactions in Aqueous Solutions #1 Precipitation Reactions This type of reaction goes to completion if and only if any recombination of the reactant ions produces an insoluble precipitate. Let’s Consider two possible reaction sequences: K 2 SO 4 (aq) + (NH 4 ) 3 PO 4 (aq)---> ? CuSO 4 (aq) + (NH 4 ) 3 PO 4 (aq)---> ?

K 2 SO 4 (aq) + (NH 4 ) 3 PO 4 (aq)---> ? Step One: Write out the four reactant ions, decide if any combination is insoluble: Cations: Anions: K + (no ppt) SO 4 2- (usually soluble) NH 4 + (no ppt) PO 4 3- (ppt???) Decision: No cation to precipitate with the phosphate ion, therefore, no reaction! K 2 SO 4 (aq) + (NH 4 ) 3 PO 4 (aq) ---> NR, no reaction

CuSO 4 (aq) + (NH 4 ) 3 PO 4 (aq)---> ? Step One. Write out each reactant ion. Decide if any combination of ions will produce a precipitate: Cations: Anions: Cu 2+ (ppt???) SO 4 2- (usually soluble) NH 4 + (no ppt) PO 4 3- (ppt???) Decision: The copper(II) ion will form an insoluble precipitate with the phosphate ion and therefore reaction will occur.

Step Two: since reaction will occur to form insoluble copper(II) phosphate, determine the correct formula for the products. Cu 3 (PO 4 ) 2 (NH 4 ) 2 SO 4 Charge per ion: Total Charge: Cu 2+ (PO 4 ) > Cu 3 (PO 4 ) 2 (NH 4 ) + + (SO 4 ) > (NH 4 ) 2 SO 4 SMART ACTION: DOUBLE CHECK YOUR FORMULAS!

[3 Cu 2+ (aq) + 3 SO 4 2- (aq) ] + [6 NH 4 + (aq) + 2 PO 4 3- (aq) ] ---> Cu 3 (PO 4 ) 2 (s) + [6 NH 4 + (aq) + 3 SO 4 2- (aq) ] Step Three: Do a total balanced equation: 3CuSO 4 (aq) + 2(NH 4 ) 3 PO 4 (aq) ---> Cu 3 (PO 4 ) 2 (s) + 3 (NH 4 ) 2 SO 4 (aq) Step Four: Do a total ionic equation, showing the actual species involved in solution:

Step Five: Prepare a “net ionic equation” for this reaction, by eliminating any ion which appears on both sides of the equation, A “SPECTATOR ION” : 3Cu 2+ (aq) + 3SO 4 2- (aq) + 6NH 4 + (aq) + 2 PO 4 3- (aq) ---> Cu 3 (PO 4 ) 2 (s) + 6NH 4 + (aq) + 3SO 4 2- (aq) NET IONIC EQUATION 3Cu 2+ (aq) + 2 PO 4 3- (aq) ---> Cu 3 (PO 4 ) 2 (s)

Group Work: a) Na 2 CO 3 (aq) + Al(NO 3 ) 3 ( aq) ---> ? b) K 2 CO 3 (aq) + NH 4 NO 3 (aq) ---> ? c) CrCl 3 (aq) + K 3 PO 4 (aq) --->? For reaction which “goes to completion”: 2. Complete Product Formulas 3. Balance Equation 4. Do Total Ionic Equation 5. Do Net Ionic Equation 1. Decide which of above is NR (“no reaction”) by writing down all four ions involved

1. b) K 2 CO 3 (aq) + NH 4 NO 3 (aq) ---> ? K + (no ppt) CO 3 2- (ppt???) NH 4 + (no ppt) NO 3 - (no ppt) NO REACTION OCCURS 1. a) Na 2 CO 3 (aq) + Al(NO 3 ) 3 ( aq) ---> ? Na + (no ppt) CO 3 2- (ppt???) Al 3+ (ppt???) NO 3 - (no ppt) REACTION OCCURS! Group Work #1 Solution: 1a), b)

2. Complete and double check the product formulas: Al (CO 3 ) > Al 2 (CO 3 ) 3 Na + + (NO 3 ) - ---> NaNO 3 Unbalanced Equation: Na 2 CO 3 (aq) + Al(NO 3 ) 3 ( aq) ---> NaNO 3 ( aq) + Al 2 (CO 3 ) 3(s) NaNO 3 Al 2 (CO 3 ) Charge per ion Total charge Solution, 1b), continued

4. Total ionic equation: [6Na + (aq) + 3 CO 3 2- (aq) ] + [2 Al 3+ (aq) + 6 (NO 3 ) - ( aq) ] ---> [6Na + (aq) +6 (NO 3 ) - ( aq) ] + Al 2 (CO 3 ) 3(s) 3. Balance: 3Na 2 CO 3 (aq) + 2Al(NO 3 ) 3 ( aq) --->6 NaNO 3 ( aq) + Al 2 (CO 3 ) 3(s) 5. Net Ionic Equation: 2 Al 3+ (aq) + 3 CO 3 2- (aq) ---> Al 2 (CO 3 ) 3(s)

Group Work #1, Solution, 1c) 1c) CrCl 3 (aq) + K 3 PO 4 (aq) --->? 1. c) CrCl 3 (aq) + K 3 PO 4 (aq) ---> ? Cr 3+ (ppt???) Cl - (generally soluble) K + (no ppt) PO 4 3- (ppt???) REACTION OCCURS!

2. Complete and double check the product formulas: Cr PO > CrPO 4 K + + Cl - ---> KCl KCl CrPO Unbalanced Equation: CrCl 3 (aq) + K 3 PO 4 (aq) --->? KCl ( aq) + CrPO 4 (s)

3. Balance: CrCl 3 (aq) + K 3 PO 4 (aq) ---> 3 KCl ( aq) + CrPO 4 (s) 4. Total ionic equation: [ Cr 3+ (aq) + 3 Cl - (aq) ] + [3 K + (aq) + PO ( aq) ] ---> [3 K + (aq) + 3 Cl - ( aq) ] + CrPO 4 (s) 5. Net Ionic Equation: Cr 3+ (aq) + PO (aq) ---> CrPO 4 (s)

Reactions in Aqueous Solutions: #2 Acid/Base Reactions These reactions are double replacement, like the precipitation reactions we just studied. However, in this case, the reaction will go to completion because an un-ionized molecule, water, is formed. In both cases (precipitation; acid/base), the removal of ions from solution causes the reaction to go to completion... Acid + Base ---> Salt + H 2 O

ACIDS are defined in the most common (“Arrhenius”) system as substances which increase the H + ion concentration when dissolved in water. ACIDS may be recognized by the convention of writing H first in the formula of the compound; in general the formula contains H plus some cation except OH -. Strong acids are 100% ionized in aqueous solutions and therefore “strong electrolytes”; Weak Acids are generally <5% ionized in aqueous solutions and therefore “weak electrolytes”.

Let’s consider names and formulas of common acids and bases which we meet in these reactions: COMMON STRONG ACIDS: HCl Hydrochloric acid HBr Hydrobromic acid HI Hydroiodic Acid HNO 3 Nitric Acid HClO 4 Perchloric Acid H 2 SO 4 Sulfuric Acid Note names: learn! Add H to anions: go from ide, ite, ate to “ic acid”

COMMON WEAK ACIDS:* H 3 PO 4 Phosphoric Acid H 2 CO 3 Carbonic Acid H 2 SO 3 Sulfurous Acid CH 3 CO 2 H or HC 2 H 3 O 2 Acetic Acid * and many, many more.....

BASES are defined in the Arrhenius system as substances which increase the OH - ion concentration when dissolved in water. BASES may be recognized as the combination of some metal or ammonium cation plus the OH - or O 2- anion. The list of strong bases in water is severely limited by the lack of solubility of most metal hydroxides and oxides in water; all of these compounds do however react with acids to yield salts plus water.

COMMON STRONG BASES (LiOH Lithium Hydroxide) NaOH Sodium Hydroxide KOH Potassium Hydroxide These are the only bases classified as strong electrolytes because they are the only ones soluble in water. Other metal hydroxides and oxides are basic but insoluble.

AMMONIA AS A WEAK BASE As we have seen earlier: NH 3(g) + H 2 O ---> NH 4 + (aq) + OH - (aq) < % 1% This reaction with water produces a small number of hydroxide ions in water solution, so aqueous ammonia is considered a weak base.

SALTS IN AQUEOUS SOLUTION All common salts which are water soluble are strong electrolytes, 100% ionized in the aqueous solution. Salts can be recognized by their formulas, in which a metal or ammonium cation is written first and some anion second.

NET IONIC EQUATIONS FOR ACID BASE REACTIONS NOTE: If any acid and base are mixed together and at least one of them is in aqueous solution, a reaction will always occur and go quickly to completion: the formation of the water molecule from the H + of the acid and the OH - or O 2- of the base is very energy releasing and exothermic. Acid + Base -----> Salt + H 2 O Both strong and weak types react in this fashion!

Net Ionic Equations for Acid /Base Reactions 1. Recognize that all reactions between acids and bases go to completion if at least one of them is in aqueous solution. 2. Decide on formula of salt product to accompany the H 2 O formed. 3. Balance the Equation 4. Do a Total Ionic Equation 5. Do a Net Ionic Equation

2. Formulas of products: Combine the H and OH to form water, and the leftover ions to make the salt: Cl - + K > KCl 4. Total Ionic Equation: H + (aq) + Cl - (aq) + K + (aq) + OH - (aq) -----> H 2 O (l) + K + (aq) + Cl - (aq) 3. Balance the equation (ok as written): HCl (aq) + KOH (aq) ---> HOH (l) + KCl (aq) Reaction #1 HCl (aq) + KOH (aq) ---> ? ---> 1. H 2 O + salt

5. IDENTIFY THE SPECTATORS: H + (aq) + Cl - (aq) + K + (aq) + OH - (aq) -----> H 2 O(l) + K + (aq) + Cl - (aq) NET IONIC EQUATION: H + (aq) + OH - (aq) -----> H 2 O(l)

1. H 3 PO 4 (aq) + NaOH(aq) ---> ? -----> H 2 O + salt 2. Predict the salt formula and complete equation: ( PO 4 ) 3- + Na + ---> Na 3 PO 4 H 3 PO 4 (aq) + NaOH(aq) ---> H 2 O(l) + Na 3 PO 4 (aq) 3. balance the equation: H 3 PO 4 (aq) + 3 NaOH(aq) --->3H 2 O(l) + Na 3 PO 4 (aq) Now, another:

H 3 PO 4 (aq) + 3 NaOH(aq) --->3H 2 O(l) + Na 3 PO 4 (aq) 4. Convert into total ionic equation, noting that the phosphoric acid is a weak electrolyte, about 99% molecular, so we do not show its ions in water: H 3 PO 4 (aq) + 3 Na + (aq) + 3 OH - (aq) ---> 3H 2 O(l) + 3 Na + (aq) + PO 4 3- (aq) 5. NET IONIC EQUATION: H 3 PO 4 (aq) + 3 OH - (aq) ---> 3H 2 O(l) + PO 4 3- (aq)

Group Work #2 Do Net Ionic Equations for the following: 1. Fe 2 O 3 (s) + HCl(aq) ---> ? 2. H 2 CO 3 (aq) + KOH (aq) --->? 3. H 2 SO 4 (aq) + Ba(OH) 2 (s) -----> ?

Group Work #2 Solutions 1. Fe 2 O 3 (s) + HCl(aq) ---> ? -----> H 2 O + salt salt: Fe 3+ + Cl > FeCl 3 Unbalanced: Fe 2 O 3 (s) + HCl(aq) ---> FeCl 3 + H 2 O Balanced: Fe 2 O 3 (s) + 6 HCl(aq) ---> 2 FeCl H 2 O

Total Ionic: Fe 2 O 3 (s) + 6 H + (aq) + 6 Cl - (aq) -----> 2 Fe 3+ (aq) + 6 Cl - (aq) + 3 H 2 O (l) Net Ionic: Fe 2 O 3 (s) + 6 H + (aq) -----> 2 Fe 3+ (aq) + 3 H 2 O (l)

2. H 2 CO 3 (aq) + KOH (aq) --->? -----> H 2 O + salt salt: CO K > K 2 CO 3 Unbalanced: H 2 CO 3 (aq) + KOH (aq) ---> H 2 O(l) + K 2 CO 3 (aq) Balanced: H 2 CO 3 (aq) + 2 KOH (aq) --->2 H 2 O(l) + K 2 CO 3 (aq)

Total Ionic: H 2 CO 3 (aq) + 2 K + (aq) + 2 OH - (aq) -----> 2 H 2 O (l) + 2 K + (aq) + CO 3 2- (aq) Net Ionic: H 2 CO 3 (aq) + 2 OH - (aq) -----> 2 H 2 O (l) + CO 3 2- (aq)

3. H 2 SO 4 (aq) + Ba(OH) 2 (s) -----> ? -----> H 2 O + salt salt: SO Ba > BaSO 4 Unbalanced: H 2 SO 4 (aq) + Ba(OH) 2 (s) -----> H 2 O(l) + BaSO 4 (s) Balanced: H 2 SO 4 (aq) + Ba(OH) 2 (s) -----> 2 H 2 O(l) + BaSO 4 (s)

Total Ionic: 2 H + (aq) + SO 4 2- (aq) + Ba(OH) 2 (s) -----> 2 H 2 O(l) + BaSO 4 (s) No canceling spectators! Net Ionic: Same as Total Ionic!

Group Work #3 Limiting Reagent Review, Wed, Mon Quiz If an aqueous solution containing g of iron(III) chloride were reacted with a second solution containing g of silver nitrate, how many g of silver chloride precipitate would theoretically be formed? Complete your equation, balance, calculate.

Group Work #3 Solution 1. Equation, unbalanced: FeCl 3 (aq) + AgNO 3 (aq) -----> AgCl (s) + Fe(NO 3 ) 3 (aq) 2. Equation, Balanced: FeCl 3 (aq) + 3 AgNO 3 (aq) -----> 3 AgCl (s) + Fe(NO 3 ) 3 (aq) 3. Calculate molar mass of FeCl 3, AgNO 3, AgCl:  g/mol FeCl 3, g/mol AgNO 3, g/mol AgCl

FeCl 3 (aq) + 3 AgNO 3 (aq) -----> 3 AgCl (s) + Fe(NO 3 ) 3 (aq) g/mol g/mol g/mol g g ?g g FeCl 3 1 mol FeCl 3 3 mol AgCl g AgCl g FeCl 3 1 mol FeCl 3 1 mol AgCl = g AgCl (Too Much!) g AgNO 3 1 mol AgNO 3 3 mol AgCl g AgCl g AgNO 3 3 mol AgNO 3 1 mol AgCl = g AgCl (Correct Answer!)