Classification of Acids and Bases PDF

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Classification of Acids and Base...

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Review of Chem 115 Key In order to succeed in Chem 115 students will need to be familiar with the following. NOTE: this is not an exhaustive list - you may find other skills and knowledge acquired in Chem 115 are needed as well. Conversions within the SI system Examples (with Answers): 1. Convert 1.29 x 10+5 mg into the following units: kilograms, grams, micrograms, and nanograms. In order from the smallest unit to the largest unit, the answers are: 1.29 x 10+11 ng = 1.29 x 10+8 microg = 1.29 x 10+5 mg = 1.29 x 10+2 g = 1.29 x 10-1 kg 2. Convert 7.44 x 10-6 m into the following units: angstroms, nanometers, millimeters, and centimeters. In order from the smallest unit to the largest unit, the answers are: 7.44 x 10+4 Angstroms = 7.44 x 10+3 nm = 7.44 x 10-3 mm = 7.44 x 10-4 cm = 7.44 x 10 -6 m Note: An Angstrom is 1 x 10 -10 meter. Angstrom is not a prefix for other units (like grams or liters). It refers only to length and is not an SI unit (However, it is a convenient and common unit of length for the atomic scale).

Solubility Rules and list of strong acids and strong bases Review from Chapter 4 and know the following: Solubility Rules

Na+, K+, Rb +, Cs +, NH 4+

NO3C2 H3O 2NO3-

ClBrI-

SO4 2- CO3 2- OH- S2-

s

s

s

s

s

s

2+

Mg

s

s

s

i

i

d

Ca 2+

s

s

sl s

i

i

d

Ba 2+

s

s

i

i

sl s

d

Al3+

s

s

s

*

i

d

Pb2+

s

i

i

i

i

i

Ag+

s

i

sl s

i

i

i

s

i

sl s

i

*

*

s

sl s

sl s

i

i

i

Hg2

2+

2+

Hg

s = soluble, sl s = slightly soluble, i = insoluble, d = decomposes in or reacts with water, * = compound does not exist

Classification of Acids and Bases Strong Acids and Strong Bases an acid or base that is virtually 100 percent dissociated into ions in aqueous solution (note that this means that strong acids and bases are also strong electrolytes) Weak Acids and Weak Bases an acid or base that partly dissociates into ions in aqueous solution (note that this means that weak acids and bases are also weak electrolytes) Common Strong Acids and Bases Acid

Name of Acid

Base

HCl

hydrochloric acid LiOH

lithium hydroxide

HBr

hydrobromic acid NaOH

sodium hydroxide

HI

hydroiodic acid

KOH

potassium hydroxide

HNO3 nitric acid

RbOH

rubidium hydroxide

H2SO 4 sulfuric acid

CsOH

cesium hydroxide

HClO4 perchloric acid

Ca(OH)2 calcium hydroxide Sr(OH)2

Name of Base

strontium hydroxide

Ba(OH)2 barium hydroxide Reactions: writing and balancing Examples (with Answers): 1. Write the balanced chemical equation for the combustion of C4 H6 O 4 (assume the only products are carbon dioxide and water). 2 C 4H6 O4 + 7 O 2 ----> 8 CO2 + 6 H 2O 2. Write balanced full molecular, full ionic, and net ionic equations for the following aqueous reactions: o phosphoric acid reacts with potassium hydroxide (Weak Acid - Strong Base) Molecular Representation H3PO 4 (aq) + 3 KOH(aq) ===> K 3 PO4 (aq) + 3 H2 O (l) Full Ionic Representation (only strong electrolytes are shown dissociated into ions)

H3PO 4 (aq) + 3 K+(aq) + 3 OH - (aq) ===> 3 K+(aq) + PO43-(aq) + 3 H2O (l) Net Ionic Representation H3PO 4 (aq) + 3 OH- (aq) ===> 3 H2O o

(l)

PO4 3-(aq)

hydrochloric acid reacts with ammonia (Strong Acid - Weak Base) Molecular Representation HCl (aq) + NH3 (aq) ===> NH4 Cl (aq) Full Ionic Representation (only strong electrolytes are shown dissociated into ions) H+(aq) + Cl-

(aq)

+ NH3 (aq) ===> NH4+ (aq) + Cl - (aq)

Net Ionic Representation H+(aq) + NH3 (aq) ===> NH4 + (aq) o

perchloric acid reacts with sodium hydroxide (Strong Acid - Strong Base) Molecular Representation HClO4 (aq) + NaOH (aq) ===> NaClO4 (aq) + H 2O (l) Full Ionic Representation (only strong electrolytes are shown dissociated into ions) H+(aq) + ClO4 -(aq) + Na +(aq) + OH-(aq) ===> Na+(aq) + ClO 4-(aq) + H2 O (l) Net Ionic Representation H+(aq) + OH-(aq) ===> H2 O (l)

o

copper (II) chloride reacts with sodium carbonate Molecular Representation CuCl2 (aq) + Na2 CO3 (aq) ===> CuCO3 (s) + 2 NaCl (aq) Full Ionic Representation Cu2+(aq) + 2 Cl-(aq) + 2 Na+(aq) + CO3 2-(aq) ===> CuCO3 (s) + 2 Na+(aq) + 2 Cl- (aq) Net Ionic Representation Cu2+(aq) + CO3 2- (aq) ===> CuCO 3 (s)

o

barium nitrate reacts with copper (II) sulfate Molecular Representation Ba(NO3) 2 (aq) + CuSO 4 (aq) ===> BaSO 4 (s) + Cu(NO 3) 2 (aq) Full Ionic Representation Ba2+ (aq) + 2 NO 3-(aq) + Cu2+ (aq) + SO42-(aq) ===> BaSO4 (s) + Cu2+(aq) + 2 NO3- (aq) Net Ionic Representation Ba2+ (aq) + SO 42- (aq) ===> BaSO 4 (s)

o

potassium hydroxide reacts with iron (III) chloride Molecular Representation 3 KOH (aq) + FeCl 3 (aq) ===> Fe(OH)3 (s) + 3 KCl (aq) Full Ionic Representation 3 K+(aq) + 3 OH - (aq) + Fe 3+(aq) + 3 Cl-(aq) ===> Fe(OH)3 (s) + 3 K+ (aq) + 3 Cl -(aq) Net Ionic Representation Fe3+(aq) + 3 OH- (aq) ===> Fe(OH)3 (s)

Mass, Moles, Molarity, and Stoichiometric Calculations Examples (with Answers): 1. Convert 25.49 grams of MgCl2 into moles of MgCl 2 . How many moles of Mg+2 and Cl- does this correspond to? 25.49 grams of MgCl2 = 0.2678 mole of MgCl2 = 0.2678 mole of Mg+2 = 0.5355 mole of Cl 2. What mass of MgCl2 is required to prepare 2.00 L of 0.150 M MgCl2? 28.6 grams of MgCl2 is required to prepare 2.00 L of 0.150 M MgCl2 3. What volume of 2.00 M MgCl2 (aq) would be required to prepare 2.00 L of 0.150 M MgCl2 (aq)? 0.150 L (or 150 mL) of 2.00 M MgCl2 (aq) is needed to prepare 2.00 L of 0.150 M MgCl2 (aq) 4. Refer to the balanced chemical equation you wrote for the reaction of perchloric acid with sodium hydroxide. If 25.00 mL of perchloric acid is titrated to neutralization by 20.51 mL of 0.09765 M NaOH, what is the molarity of the perchloric acid in the initial 25.00 mL sample?

0.08011 M (mole /L) HClO4 5. Refer to the balanced chemical equation you wrote for the reaction of potassium hydroxide with iron (III) chloride. When 50.0 mL of 0.300 M KOH reacts with 25.0 mL of 0.200 M iron (III) chloride, what is the maximum amount of precipitate that can form? (give your answer in grams) 0.532 g of Fe(OH)3 (s) is the maximum amount of ppt. Electronic structure and periodic trends Review sections of your text dealing with these topics and answer the following: 1. The arrangement of the elements in the periodic table are in order of increasing atomic number. 2. Elements in the same group of the periodic table have similar chemistry. Why? Same valence electron configuration leads to similar chemistry . . . For example: Li has a single 2s electron and Na has a single 3s electron and they both typically form +1 ions. 3. How does atomic size vary as you proceed down a group? Why? Atomic size increases down a group because the outermost electrons are stored in increasingly large orbitals . . For example: 3s orbital is larger than 2s orbital, therefore, Na is larger than Li 4. How does atomic size vary from left to right across a period? Why? [think about the effective nuclear charge that is experienced by the outermost electrons] The effective nuclear charge experienced by the valence electrons increases from left to right across a period because electrons in the same principle energy level (and particularly in the same subshell) do not screen each other well from the nucleus’ positive charge. 5. The equation for the first ionization step for Al can be written as: Al ---> Al+1 + ewrite equations for the second and third ionization steps. Explain why Al is not likely to undergo a fourth ionization process. Second step: Al+1 ---> Al+2 + eThird step: Al+2 ---> Al+3 + eFourth step: Al+3 ---> Al +4 + eAl has a valence shell electron configuration of 3s23p 1 . . . losing three electrons to form Al3+ only requires the removal of electrons from the valence shell. In order to form Al+4 an electron would have to be lost from the 2p energy level. This requires too much energy. 6. Describe and explain the variation in ionization energy down a group and across a period. Ionization Energy (IE) decreases as one proceeds down a group because the outermost valence electron is further from the nucleus and more loosely held (i.e. the valence electron is in a well shielded large orbital). IE increases from left to right across a period because the effective nuclear charge experienced by the valence electrons increases from left to right (see #4 above).

7. The equation for the first electron addition step for O can be written as: O + e- ---> O-1 write equations for the second electron addition step. Explain why O is not likely to undergo a third electron addition process. Second step: O-1 + e - ---> O-2

Third step: O -2 + e - ---> O-3

In order for oxygen to become O2- a neutral oxygen atom must gain two electrons. Since oxygen has a valence electron configuration of 2s22p 4 gaining two electrons completes the 2p energy level (where the electrons should feel a reasonably large effective nuclear charge). Therefore, the Electron Affinity (EA) for this process is favorable. In order to form O3- an additional electron would have to be gained and would occupy the 3s orbital (where it would feel a very weak effective nuclear charge). 8. Describe and explain the variation in electron affinity (the relative ability for a gas phase atom to gain an electron) down a group and across a period. Elements on the right-hand side of the periodic table have a greater ability to gain an electron. This is because the effective nuclear charge is higher on the right-hand side of the Periodic Table. Elements toward the bottom of the groups have lower ability to gain electrons. The electron is gained into a larger orbital that is better shielded and the energetic favorability of the process is not as great. 9. When metals react with nonmetals, ionic compounds are formed. Use the concepts of ionization energy and electron affinity to explain: (a) which gains and which loses electrons [metals or nonmetals] and (b) the relative charges acquired in these ionic compounds. Ex. Why is NaCl a correct formula but MgCl is incorrect? (a)Metals lose electrons because they have low IE and unfavorable EA (see # 4 - 7 above for the reasons). Nonmetals gain electrons because they have favorable EA and do not lose electrons easily due to their high IE. (see # 4 - 7 for the reasons). (b)Metals (representative) tend to lose all their valence electrons (but many transition metals do not and some representative metals do not). Nonmetals tend to gain enough electrons to complete their valence shell. For example: Cl (3s 2 3p5) + e - ---> Cl - (3s23p6 ) Mg (3s2 ) --> Mg2+ (3s 0 ) + 2 eThen, of course, when the ions combine to form compounds, they do so in ratios determined by their relative charges. For example: Mg2+ + 2 Cl- ---> MgCl2 Lewis structures and hybridization (examples done during in class reviews) 1. Review quizzes, exams, and the model building laboratory. Correct items on which you lost points. 2. Review organic functional groups, octet rule (and exceptions), formal charges, resonance structures, etc. Thermochemistry 1. Use the web simulation site to practice calorimetry. 2. Understand the application of Hess’ Law and the use of bond energies or ∆Hf to find ∆H for reactions....