CH 4 AP Reactions in Aqueous Solutions

Water Aqueous means dissolved in H 2 O Moderates the Earth’s temperature because of high specific heat H-bonds cause strong cohesive and adhesive properties Polar, therefore dissolves many substances both ionic and polar Solutions: with solvent (dissolving medium) and solute (being dissolved)

Electrolyte is a solution which will conduct electricity (solute makes ions in solution allow the flow of e - ) Ion Dissociation  Solvated is when an ionic compound dissolves in H 2 O and the ions become surrounded by the H 2 O molecules. Non-electrolyte is a solutions which will NOT conduct electricity (no ions present to permit flow of e - ) See page 126

Strong Electrolytes are made with substances which completely ionize in H 2 O, they have large equilibrium constants. 1. Soluble salts NaCl (s)  Na + 1 (aq) + Cl - 1 (aq) 2. Strong acids HCl  H + (aq) + Cl - 1 (aq) Arrhenius acids product H + 1 when dissolved in H 2 O 3. Strong bases NaOH (s)  Na + 1 (aq) + OH - 1 (aq)

Weak electrolytes exhibit small amounts of ionization when dissolved in H 2 O, have small equilibrium constants. 1. Weak acids HC 2 H 3 O 2(aq)  H + 1 (aq) + C 2 H 3 O 2 -1 (aq) 2. Weak bases NH 3 (aq) + H 2 O (L)  NH (aq ) + OH - 1 (aq) Non-electrolytes produce NO ions upon dissolving in H 2 O

Writing half reactions Ag  Ag + 1 Ba  Ba + 2 F  F - 1 S  S - 2 Fe + 2  Fe + 3

Do you know the subatomic particles? e-e- p+p+ e-e- p+p+ AgAg + 1 BaBa + 2 FF-1F-1 SS-2S-2 Fe + 2 Fe + 3

Do you remember how to write formulas? Oxide O - 2 Iodide I - 1 Sulfate SO Phosphate PO Cyanide CN - 1 Hydrogen H + 1 Barium Ba + 2 Chromium Cr + 3

Molarity (M): concentration of solution expressed as moles of solute per volume of solution in liters. Molarity = moles / liter Calculate the molarity of a solution prepared by dissolving 17.5 g of solid NaOH in enough H 2 O to make 1.75 L of solution.

Volumetric flask To use: Fill ½ full of distilled H 2 O Add solute Cap and invert to mix Add H 2 O until close to line on neck Cap and invert to mix Carefully, add H 2 O until meniscus is at the line on neck

Calculate the concentration of each type of ion in the following: 0.50 M AlCl 3 2 M (NH 4 ) 2 CO 3 Calculate the moles of Cl - 1 ions in 1.3 L of 0.05 M MgCl 2.

Typical blood serum is about 0.14 M NaCl. What volume of blood contains 1.0 mg NaCl? How many grams of AgCl must be used to make 75 mL of 0.1 M AgCl solution?

Dilutions:moles = moles M 1 V 1 = M 2 V 2 What volume of 0.25 M K 2 CrO 4 must be used to prepare 1.0 L of 0.05 M K 2 CrO 4 ? What volume of 12.0 M HCl must be used to prepare 2.5 L of 0.1 M HCl?

Types of Chemical Reactions 1. Precipitation Reaction is when 2 solutions are mixed resulting in an insoluble substance. (Insoluble & slightly soluble are used interchangeably here. Any substance that is less soluble than.01 mol /L is considered insoluble.) When writing the formula for the precipitate: a. Precipitate has a net charge of zero b. Contains a cation (+ ion) and an anion (- ion) c. Write cation first, anion second

Solubility Rules Soluble: the following are always soluble 1. Nitrates (NO ) and Acetates (C 2 H 3 O ) 2. Alkali Metals (Li + 1 Na + 1 K + 1 Cs + 1 Rb + 1 ) and Ammonium (NH ) 3. Halides (Cl - 1 Br - 1 I - 1 ) (except with Ag + 1 Pb + 2 Hg are insoluble ) 4. Sulfates (SO ) (except with Ba + 2 Pb + 2 Hg Sr + 2 Ca + 2 are insoluble)

Mostly Insoluble 1. Hydroxides (OH - 1 ) (except with Alkali metals or ammonium are soluble ) (except with Ca + 2 Sr + 2 Ba + 2 are slightly soluble ) 2. Sulfides (S - 2 ) (except with Alkali metals, ammonium are soluble ) (except with Alkali Earth metals Be + 2 Mg + 2 Ca + 2 Sr + 2 Ba + 2 are slightly soluble ) 3. Carbonates (CO ) and Phosphates (PO ) are slightly soluble ( except with alkali metals and ammonium are soluble ) Chromates are insoluble except with alkali metals or ammonium

Write the net ionic equation for all precipitates that can be made mixing the following chemicals: Ba(NO 3 ) 2 KCl Na 2 CO 3 Ag 2 SO 4 Mg(C 2 H 3 O 2 ) 2 NaOH

Writing chemical equations Formula equation: shows everything involved K 2 CrO 4 (aq) + Ba(NO 3 ) 2 (aq)  BaCrO 4 (s) + 2 KNO 3 (aq) Complete ionic equation: includes all spectator ions, those ions which are not involved in the chemical reaction 2K +1 ( aq) + CrO (aq) + Ba + 2 (aq) + 2NO (aq)  BaCrO 4 (s) + 2 K + 1 (aq) + 2 NO (aq) Net ionic equation: only shows chemicals involved in the reaction, no spectator ions Ba + 2 (aq) + CrO (aq)  BaCrO 4 (s) All of the above are the same reaction shown in different types of equations.

Try writing all 3 types of equations for: Aqueous potassium chloride is added to aqueous silver nitrate Aqueous potassium hydroxide is mixed with aqueous iron(III) nitrate

To calculate ion concentrations after reactions 1. Identify species present as reactants and products (the complete ionic equation) 2. Change to moles 3. Find the Limiting reactant 4. ICE chart I = initial concentration C = change in concentration E = ending concentration 5. Use stoichiometric relations 6. Calculate moles of ions for any soluble chemicals (or excess chemicals) 7. Change to Molarity using total volume

Calculate the concentrations of all ions and mass of precipitate after 100 mL of 1.0 M Mg(NO 3 ) 2 reacts with 200 mL of 1.0 M NaOH. Mg(NO 3 ) 2(aq) + 2 NaOH  I C E

1.0 L of 0.1 M Ba(NO 3 ) 2 is mixed with 500 mL of 0.1 M Na 2 SO 4. What are the final concentrations of ions in solution and mass of precipitate? Ba(NO 3 ) 2 (aq) + Na 2 SO 4 (aq)  I C E

Reaction Type 2 Acid – Base Reaction Theories:AcidBase ArrheniusH + OH - Bronsted – Lowry proton donor proton acceptor (proton = H +, a hydrogen without and e -) Lewis e - pair acceptor e - pair donor

Strong:ionize 100 % Need to memorize strong Acids & Bases both HCl & NaOH strong: H Cl Na OH - 1 spectator ions Net ionic equation: H OH - 1  H 2 O (L) Weak: ionize some weak acid (HC 2 H 3 O 2 stays mainly as molecule) & strong base(KOH ionizes 100 %) Net ionic equation: HC 2 H 3 O 2 + OH -   H 2 O + C 2 H 3 O 2 - Non – electrolyte: ionize 0% We will start with only strong acids and bases, therefore ionize 100%

strong acid+ strong base  salt + water HA + BOH  BA + H 2 O (net) H + + OH -  H 2 O Neutralization Reactions with gas formation: (Acid) 2 H S - 2  H 2 S (g) H HCO  H 2 CO 3 H 2 CO 3  H 2 O + CO 2 (g)

When calculating acid base equations, remember that moles A = moles B M A V A = M B V B What volume of M HCl solution is needed to neutralize 25.0 mL of 0.35 M NaOH?

moles A = moles B M A V A = M B V B This works using M x Liters = moles This also works using M x milliliters = millimoles

28.0 mL of 0.25 M HNO 3 reacts with 53.0 mL of 0.32 M KOH. Calculate the amount of H 2 O formed in the resulting reaction. What concentration of H + 1 or OH - 1 is in excess? KOH (aq) + HNO 3(aq)  I C E

Volumetric Analysis: Titration delivers a given amount of known concentration (titrant) to a given amount of unknown concentration (analyte). Equivalence point is the stoichiometric point when the moles = moles. An indicator chemical is added which changes color at the equivalence point (end point). Standardizing is using a solution of a very stable chemical to determine the M of an unknown.

A student standardizes an NaOH solution using Potassium Hydrogen Phthalate (KHC 8 H 4 O 4  KHP) (  =204.22). The student dissolves g KHP in H 2 O, adds phenolphthalein and titrates with mL NaOH of unknown concentration. Find the concentration. H OH - 1  H 2 O HC 8 H 4 O OH - 1  H 2 O + C 8 H 4 O 4 - 2

Reaction Types #3 Oxidation – Reduction (Redox) A reaction in which one or more e - are transferred. OxidationReduction Increase in os therefore Reducing Agent Loss of e - Gain of oxygen Loss of hydrogen Decrease in os therefore an Oxidizing Agent Gain of e - Loss of oxygen Gain of hydrogen 0 Oxidation Reduction

Oxidation State (o.s.) (oxidation number) *Is an imaginary charge. *It is a way to arbitrarily assign e - to atoms especially in covalent bonds *The e - are often not shared equally, one atom may have a stronger e - affinity or attraction.

Rules for assigning o.s. Whenever 2 rules appear to contradict one another, follow the rule that appears higher on the list! 1. The oxidation state (os) of an atom in the free (uncombined) element is zero (o). 2. The total of the os of all the atoms in a molecule is zero (o). For an ion, the total is equal to the charge on the ion. 3. In their compounds, Alkali metals (Li, Na, K, Rb, Cs, Fr) have os = +1 Alkali Earth metals (Be, Mg, Ca, Sr, Ba, Ra) have os = In their compounds, Hydrogen has os = +1 Fluorine has os = In it’s compounds, Oxygen has os = In 2 element compounds, Halogens (Cl, Br, I) have os = -1 (O, S, Se, Te, Po) have os = -2 (N, P, As, Sb, Bi) have os = -3

Determine the os for the following underlined elements. S 8 Cr 2 O Cl 2 O KO 2 S 2 O KMnO 4 H 2 CO 3

Write the formulas for the oxides with the following charges. Cr + 3 Cr + 4 Cr + 6 N+1N+2N+3N+4N+5N+1N+2N+3N+4N+5

Activity Series of Metals see page 144 A list of metals arranged in order of decreasing ease of oxidation. The higher on the chart, the more active the metal. Any metal on the list can be oxidized by the ions of elements below it. Will an aqueous solution of iron (II) chloride oxidize Mg (s)