AT Chapter 4 Notes 15.notebook...AT Chapter 4 Notes 15.notebook 3 October 16, 2015 Hydration When ionic substances (salts) dissolve in water, they break up into individual cations

AT Chapter 4 Notes 15.notebook

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October 16, 2015

Some key terms:

Solute:

Solvent:

Solution:

Solubility:

Precipitate:

Electrolyte:

Nonelectrolyte:

Molarity:

Dilution:

Titration:

Oxidationreduction:


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Water occurs in all three phases on earth

the universal solvent

polar molecule


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Hydration

When ionic substances (salts) dissolve in water, they break up into individual cations (positive ions) and anions (negative ions). The polar water molecules surround (hydrate) these ions such that the partial negative oxygen ends of water molecules surround the cations while the partial positive hydrogen ends surround the anions.


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Chemical equation representing hydration:

(NH4)3PO4(s) à 3 NH4+(aq) + PO4

3(aq)

hydrated ions

The phase in which a substance is dissolved in water is termed the aqueous phase and given the notation (aq)

Key idea: "like dissolves like"

Water will dissolve other polar substances as well as ionic compounds. Nonpolar solvents will dissolve nonpolar solutes.


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Electrolytes substances when dissolved in water “break apart” to yield aqueous ions

examples: acids, bases & salts

Ionic compounds dissociate into ions when they dissolve in water. Since these ions existed before the substance was dissolved, we call this process dissociation.

Some nonionic compounds also form ions when dissolved in water. These include acids and weak bases. In this case the ions did not exist before the substance was dissolved, and we this process ionization.

In order for a solution to conduct electricity, it must contain mobile ions.

Nonelectrolytes substances when dissolved in water do not dissociate or ionize to yield aqueous ions.

examples: sugars (CH2O); e.g. C6H12O6, C12H22O11

and alcohols (ROH); e.g. CH3OH, C2H5OH


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Strong Electrolytes:

certain salts dissociate completely in water (100% dissociation)

e.g. NaCl(s) à NaCl(aq)

note that NaCl(aq) actually does not exist in this solution; it exists as separate Na+(aq) and Cl(aq) ionsstrong acids ionize completely in water to yield H+ and anion (100% ionization)

e.g. HCl(g) + H2O(l) à H3O+(aq) + Cl(aq) strong bases dissociate completely in water to yield OH and a cation (100% dissociation)

e.g. NaOH(s) + H2O(l) à Na+(aq) + OH(aq)

Weak Electrolytes

weak acids ionize to a small extent (less than 5% ionization)

e.g. HC2H3O2(l) + H2O(l) ß H3O+(aq) + C2H3O2(aq)

weak bases ionize to a small extent (less than 5% ionization)

e.g. NH3(g) + H2O(l) ß NH4+(aq) + OH(aq)


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Solutions homogeneous mixtures

Solute substance dissolved in the solvent; present in smaller amount (by moles)

Solvent substance in which solute is dissolved; present in larger amount (bymoles)

moles mass

Problem: Calculate the molarity of a solution made by dissolving 23.4 g ofsodium sulfate in enough water to make a 125 mL solution.

Problem: Calculate the mass of sodium hydroxide present in 200. mL of a 1.50M solution.


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Standard Solution a solution whose concentration is accuratelu known.

Steps in preparing a standard solution:

a) mass solute needed on balanceb) add some water (solvent) to volumetric flaskc) transfer solute to volumetric and dissolve in waterd) add water to mark in volumetric

Problem: Describe how you would prepare 100.0 ml of a 0.50M solution of potassium nitrate. Specify amounts of materials to be measured.


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October 16, 2015

Dilution dilution with water doesn’t alter the number of moles of solute present:

moles solute before = moles solute after

MiVi = MfVf

Problem: Describe how you would prepare 250. mL of a 0.10M HCl solution by diluting a 16.0 M stock solution.


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Types of Solution Reactions: PrecipitationAcid/BaseOxidation/Reduction (Redox)

Precipitation Reactions involve double replacement reactions. In this reaction, aqueous ions from two different solutions are exchanged. (Recall that in order for a double replacement reaction to occur, one of the following must be formed: a precipitate, a gas, or water). In order to determine whether a precipitate forms, you must be familiar with the solubility rules. By the way, precipitation reactions are part of a broader class of reactions in which ions are exchanged, called metathesis reactions

Simple Rules for Solubility of Salts in Water

1. Most nitrates (NO3), chlorates (ClO3

), and acetates (C2H3O2) are soluble. Silver acetate is sparingly soluble.

2. Most salts of Na+, K+, and NH4+ are soluble.

3. Most chloride salts are soluble. Notable exceptions are AgCl, PbCl2, and Hg2Cl2.4. Most sulfates are soluble. Notable exceptions are BaSO4, PbSO4 ,and CaSO4,5. Most hydroxide salts are insoluble. The important soluble hydroxides are NaOH,

KOH, and Ca(OH)2.6. Most sulfide (S2), carbonate (CO3

2), and phosphate (PO43) salts are insoluble


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Describing Reactions in Solution: balanced molecular equation (BME)complete ionic equation (CIE)net ionic equation (NIE)

example:

Aqueous solutions of copper(II)chloride and sodium hydroxide are mixed.

1. Molecular Equation (BME):

2. Complete Ionic Equation (CIE):

3. Net Ionic Equation NIE):

Problem: Write the BME, CIE, anmd NIE for the reaction that occurs when aqueous solutions of plumbous nitrate and potassium sulfate are mixed:


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Selective Precipitation use the fact that salts have different solubilities to separate mixtures of ions used in qualitative analysis.

Problem: Given an aqueous solution containing the cations Ag+, Ba2+, Fe3+ and NO3anions. How could you separate the cations using the following “test” solutions:

NaCl(aq), Na2SO4(aq), NaOH(aq)

and in what order would you add the solutions?

Ag+

Ba2+

Fe3+


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Precipitation Reaction Stoichiometry

Problem: What volume of 0.100M sodium hydroxide is required to precipitate all the the nickel(II) ions from 150.0 mL of a 0.250M nickel(II)nitrate solution.


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AcidBase Reactions

Arrhenius acid – substance that ionizes in water to produce H+(aq) as the cationArrhenius base – substance that dissociates/ionizes in water to produce OH as the anion

BronstedLowry Acid = proton (H+) donorBronstedLowry Base = proton (H+) acceptor

Reactant Reacting Species Example

strong acid H+ HCl(g) à H+(aq) + Cl(aq)

or H3O+ or HCl(g) + H2O(l) à H3O+ + Cl(aq)

common strong acids:

hydrochloricnitricsulfuricperchloric

strong base OH KOH(s) + H2O(l) à K+(aq) + OH(aq)

common strong bases:

LiOHNaOHKOHCa(OH)2

weak acid HA HC2H3O2 = H+ + C2H3O2

reacting species = HC2H3O2

weak base NR3 NH3(g) + H2O(l) = NH4+(aq) + OH(aq) (R = H, CH3,etc)


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Acidbase reactions:

SA + SB: HNO3(aq) + NaOH(aq) à

CIE:

NIE:

WA + SB: HC2H3O2(aq) + NaOH(aq) à

CIE:

NIE:

SA + WB: HClq) + CH3NH2(aq) à

CIE:

NIE:


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AcidBase Titrations Volumetric Analysis

A very common laboratory procedure in which solution stoichiometry is used is titration. In a titration, a solution of accurately known concentration (the standard solution) is added gradually to another solution whose concentration is unknown until the reaction between the two solutions is complete.

define: equivalence point (stoichiometric point):

indicator:

end point:

Ideally, the end point and equivalence point should coincide.

Problem: What volume of 0.050 M HCl is needed to neutralize 50.0 mL of a 0.400 M Ca(OH)2 solution?


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OxidationReduction Reactions (Redox)

We have observed reactions involving the exchange of ions resulting in a precipitate. We also observed chemical compounds known as acids react with chemical compounds known as bases through the exchange of protons. Another class of chemical reactions involves the exchange of electrons. These reactions are called oxidationreduction or redox reactions.

Typical redox reactions involve the reaction between a metal and a nonmetal to form an ionic compound. The metal typically loses electrons while the nonmetal typically gains electrons. We can thus separate any redox reaction into two halfreactions; one halfreaction will show the loss of electrons by one species (the metal) while the other half reaction shows the gain of electrons by another species (the nonmetal). For example, the reaction

2Mg(s) + O2(g) à 2MgO(s) + energy

Can be written as the sum of two half reactions:

oxidation half:

reduction half:

net reaction:

to review,

oxidation = an increase in oxidation state (loss of e)

reduction = a decrease in oxidation state (gain of e)


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Rules for Assigning Oxidation States: review rules in textbook

Problem: assign oxidation states to each element in the following compounds:

K2Cr2O7 Fe3O4 SO42

oxidizing agent = substance which causes another substance to be oxidized by taking electrons from it; it itself is reduced.

reducing agent = substance which causes another substance to be reduced by giving it electrons; it itself is oxidized.

For the reaction below identify the substance oxidized, substance reduced, oxidizing agent and reducing agent:

CH4 + 2 O2 à CO2 + 2 H2O

substance oxidized:

substance reduced:

oxidizing agent:

reducing agent:


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Balancing OxidationReduction Reactions

I. The Oxidation States Method

In using the oxidation states method to balance a redox equation, we find the coefficients for the reactants that will make the total increase in oxidation state balance the total decrease.

* Assign the oxidation states of all atoms.

* Decide which element is oxidized and determine the increase in oxidation state.

* Decide which element is reduced and determine the decrease in oxidation state.

* Choose coefficients for the species containing the atom oxidized and the atom reduced such that the total increase in oxidation state equals the total decrease in oxidation state.

* Balance the remainder of the equation by inspection.

Example:

Cu(s) + HNO3(aq) à Cu(NO3)2(aq) + H2O(l) + NO(g)

Cu(s) + HNO3(aq) à Cu(NO3)2(aq) + H2O(l) + NO(g)


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The Half Reaction Method for Balancing Equations for Oxidation Reduction. Reactions Occurring in Acidic Solutions:

1. Write the equations for the oxidation and reduction half reactions. Show species as they actually occur in solution

2. For each half reaction:

3. a. Balance all of the elements except hydrogen and oxygen.4. b. Balance oxygen using H2O.5. c. Balance hydrogen using H+.6. d. Balance the charge using electrons.

7. If necessary, multiply one of both balanced half reactions by an integer to equalize the number of electrons transferred in the two half reactions.

8. Add the half reactions and cancel identical species.

9. Check that the elements and charges balance.


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Example:

As2O3(s) + NO3 (aq) à H3AsO4(aq) + NO(g)


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The Half Reaction Method for Balancing Equations for Oxidation Reduction Reactions Occurring In Basic Solution

* Use the half reaction method as specified for acidic solutions to obtain the final balanced equation as if H+ ions were present.

* To both sides of the equation obtained above, add a number of OH ions that is equal to the number of H+ ions. (We want to eliminate H+ ions by forming H2O).

* Form H2O on the side containing both H+ and OH ions, and eliminate the number of H2O molecules that appear on both sides of the equation.

* Check that elements and charges balance.


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Example:

MnO4(aq) + S2(aq) à MnS(s) + S(s)


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Redox Titration

Redox reactions are commonly used as the basis for volumetric analytical procedures. Three of the most frequently used oxidizing agents are aqueous solutions of:

potassium permanganate KMnO4

potassium dichromate K2Cr2O7

In acidic solution the permanganate ion undergoes the following reaction:

MnO4(aq) + 8 H+(aq) + 5 e > Mn2+(aq) + 4 H2O(l)

purple colorless

In acidic solution the dichromate ion undergoes the following reaction:

Cr2O72(aq) + 14H+(aq) + 6e à 2Cr3+(aq) + 7H2O(l)

orange green


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Problem: A solution of potassium permanganate is standardized by titration with oxalic acid. It required 29.97 mL of the permanganate solution to react completely with 0.1058 g of oxalic acid. The permanganate is reduced to manganese(II) ion and the oxalic acid is ozidized the carbon dioxide. Calculate the molarity of the permanganate solution.


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AT Chapter 4 Notes 15.notebook...AT Chapter 4 Notes 15.notebook 3 October 16, 2015 Hydration When ionic substances (salts) dissolve in water, they break up into individual cations