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ISU / Chemistry / CHE 140 / what is Electrolytes?

# what is Electrolytes? Description

##### Description: Study guide for exam 2 on 10.1! Good luck :)
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CHE140 Exam 2 Study Guide

## what is Electrolytes? Chapter 4 (Sections 4.2-4.9)

Understand the concentration units (parts per million, parts per billion, and molarity) ∙ Parts per million (ppm) is grams of solute per million grams of solution ppm = g solute

1 x 106g solution

∙ Parts per billion (ppb) is micrograms of solute per kilograms of solution o Could also be used as milligrams of solute per L of solution

ppb = µg solute ppb = mg solute

kg solution L solution

∙ Parts per million is directly proportional to molarity

1. What is the concentration of Ca2+ in sea water (d = 1.025g/mol) if it is 411.9ppm Ca2+?

411.9g Ca2+

1mol Ca2+ We also discuss several other topics like social aggregate definition sociology

1.025g solution

1000mL

1 x 106solution

40.08g Ca2+

1mL solution

1L

= 0.01050M Ca2+

Dilutions of Solutions

M1V1 = M2V2

** V2 > V1

** M2 < M1

2. What volume of concentrated sulfuric acid, 18.4M H2SO4, is needed to prepare 2.00L of  0.250M H2SO4?

## what is Titrations? 18.4V1 = 0.250(2.00)

18.4V1 = 0.50 If you want to learn more check out costlessly

V1 = 0.0272L or 27.2mL

Electrolytes

∙ Electrolytes produce ions when dissolved in water

o Strong electrolytes: break down 100% into ions

▪ NaCl(s) ???? Na+ + Cl-

▪ Strong electrolytes are soluble ionic compounds and strong acids

∙ Strong acids: HCl, HBr, HI, HNO3, HClO4,H2SO4

o Weak electrolytes: partially break down into ions

▪ HC2H3O2(aq) ???? H+ + C2H3O2

Acid-Base Reactions

∙ Involves H+ (proton) transfer – hydrogen ion/proton transfer

∙ Acids produce H+ (hydronium) in solution

∙ Bases produce OH- (hydroxide) in solution

∙ Non-metal oxides react with water to produce acids

o SO3 + H2O ???? H2SO4

Bronsted-Lowry Acid/Base

∙ B-L acids are H+ donors

∙ B-L bases are H+ acceptors

HBr(aq) + KOH(aq) ???? KBr(aq) + H2O(l)

## what is Precipitation Analysis? ∙ Br loses, or donates H+; B-L Acid

∙ OH gains, or accepts H+(H2O = HOH); B-L Base

*** The B-L Acid/Base is always on the reactant side/left side

K2HPO4(aq) + KHSO4 ???? K2SO4 + KH2PO4

∙ PO4 gained, or accepted another H+; B-L Base

SO4 lost, or dontated H+; B-L Acid

Titrations

∙ Controlled acid/base neutralization reaction

H+ + OH- ???? H2O

∙ Add base to the endpoint when the OH added is equal to H+in sample 1. A 25.00mL sample of sulfuric acid is titrated with 0.1043M KOH, requiring 18.62mL to  reach the end point. What is the molarity of sulfuric acid? We also discuss several other topics like eumetazoa characteristics

H2SO4(aq) + 2KOH(aq) ???? K2SO4(aq) + 2H2O

18.62mL KOH

0.1043mol

KOH

1mol H2SO4

1

1000mL

1000mL KOH

2mol KOH

25.00mL

1L

= 0.03884M H2SO4

Solubility Rules

∙ Always Soluble:

o Cations: Group 1 (Alkali metals) and NH4

o Anions: NO3-and CH3OO-(acetate)

∙ Compounds containing the following anions are soluble except noted: o Group 17 ions (halides) except the halides of Ag+, Cu+, Hg22+, and Pb2+ o SO42-except the sulfates of Ba2+, Ca2+, Hg22+, and Sr2+

∙ Insoluble compounds include the following:

o All hydroxides except those of group 1 cations and Ca(OH)2, Sr(OH)2, and  Ba(OH)2

o All sulfides except those of group 1 cations and NH4+, CaS, SrS, and BaS o All carbonates except those of group 1 cations and NH4+

o All phosphates except those of group 1 cations and NH4+

Soluble – strong electrolytes ionize 100%

BaSO4(s) ???? doesn’t dissolve

K3PO4(s) ???? 3K+(aq) + PO43-

∙ Saturated – contains exactly the maximum solubility in solution

∙ Unsaturated – contains less than the maximum solubility

∙ Supersaturated – Contains more than the maximum solubility (solute dissolved in solvent  at a higher temperature, then carefully cooled)

Precipitation Reaction/ Ionic Equations

∙ AKA Double displacement reaction

∙ Mix two solutions and a solid precipitates out

2. What precipitate forms when solutions of silver nitrate and cesium iodide are mixed  together? If you want to learn more check out chbr3 bond angle

AgNO3(aq) + CsI(aq) ???? CsNO3(aq) + AgI(s)

Net Ionic Eq: Ag+(aq) + I-(aq) ???? AgI(s)

▪ Cesium and Nitrate are the spectator ions

o Silver Iodide (AgI) is the precipitate

Precipitation Analysis

∙ A neutralization reaction that precipitates

3. What is the concentration of bromide in a 25.0mL sample of water if excess silver ion  results in 361mg AgBr precipitate?

Volume Solution ???? molarity ???? Moles

Ag+(aq) + Br-(aq) ???? AgBr(s)

361mg AgBr

1000mL

1g AgBr

1mol AgBr

1mol Br

25.0mL

1L

1000mg AgBr

187.77g AgBr

Imol AgBr

= 0.00769M Br

Calculate Oxidation Numbers

Rules

1. The sum of the oxidation numbers is equal to the charge of the molecules or ion 2. The oxidation number for an element in its standard state is zero

a. Hg ???? O.N. = 0

b. N2 ???? O.N. = 0

3. The oxidation number for monoatomic ions is equal to the ion charge

a. Fe3+ ???? O.N. = +3

b. S2- ???? O.N. = -2

4. Flourine is always -1 in compounds

5. Hydrogen is +1 unless with a metal; oxygen is -2 unless with F or in a peroxide 6. If none of the previous apply, the most electronegative element has an oxidation number  equal to its anion charge

CF4 (F is always -1 in compounds, therefore, C is +4 in this case because -1(4)= -4 and to  get zero, C must be +4) We also discuss several other topics like dina zhabinskaya

Oxidation-Reduction Reactions

∙ Transfer of electrons

∙ Electrons lost equal electrons gained

∙ Reduction – gaining electrons

∙ Oxidation – loss of electrons If you want to learn more check out orh 1030 uf

∙ Reducing Agent – Reactant with oxidized element

∙ Oxidizing Agent – Reactant with reduced element

CCl4

+

5O2

????

CO2

+

4ClO2

C +4

Cl -1

[4(-1)]

O 0

C +4

O -2

[2(-2)]

Cl +4

O -2

+4 C +4

-1 Cl +4 [Losing 5 electrons]

0 O -2 [Gaining 2 electrons

10 O(+2 electrons) = +20 electrons

4 Cl(-5 electrons) = +20 electrons

Fe

+

Cl2

????

FeCl2

0

0

Fe +2

Cl -1

Fe loses 2 electrons (oxidized)

Cl gains 1 electron (reduced)

“Complex” Redox Reaction

∙ Uses the half reaction method

CrO42-(aq) + H2S(aq) ???? SO42-(aq) + Cr2+(aq) in acid solution

1. Break into half reactions

CrO42-

????

Cr2+

Cr +6

O -2

Cr +2

H2S

????

SO42-

H +1

S -2

S +6

O -2

2. Balance non-oxygen and non-hydrogen atoms

3. Balance oxygen with water

CrO4 ???? Cr + 4H2O 4H2O + H2S- ???? SO42- 4. Balance hydrogen with H+

8H+ + CrO4 ???? Cr + 4H2O 4H2O + H2S- ???? SO42- + 10H+ 5. Balance charge with electrons

2[4e- + 8H+ + CrO4 ???? Cr + 4H2O] 4H2O + H2S- ???? SO42- +10H+ + 8e

6. Combine the equations

6H+ + 2CrO42- + H2S ???? SO42- + 2Cr2+ + 4H2O

Chapter 5

Definitions

Kinetic energy – kinetic energy is the energy of motion

Potential energy – in exothermic chemical reactions, potential energy is the source of energy Electrostatic energy – a potential energy that results and is associated with a defined system and  its surroundings

Work – work is the energy required to move an object against its force

Enthalpy – enthalpy is the flow of energy into or out of a system at constant pressure State function – a state function is a property that depends only on the current state of the object,  not how the object got to that state

Law of Conservation of Energy – the total energy of an isolated system remains constant (it is  said to be conserved over time)

Calculations

∙ Potential Energy

o PE = mgh (mass x gravity x height)

∙ Kinetic Energy

o KE = ½ mu2(mass x speed)

∙ Electrostatic energy

o Eel α Q1 x Q2 (Q are charges, d is distance)

d

∙ Work

o w = f x d (force x distance)

Endothermic vs. Exothermic

∙ Endothermic processes gain heat from their surroundings (q > 0)

∙ Exothermic processes lose heat to their surroundings (q < 0)

Systems

∙ A system is the specific process you are looking at and the surroundings are everything  else

∙ Isolated system – neither energy nor matter is exchanged with the surroundings ∙ Closed system – heat can be exchanged with the surroundings, but not matter ∙ Open system – heat and matter can be exchanged with the surroundings

Enthalpy (H) and Enthalpy Change (∆H)

∆H = ∆E + P∆V

∙ enthalpy – the flow of energy into or out of a system at constant pressure ∙ ∆H>0 endothermic

∙ ∆H<0 exothermic

Heating and Cooling Curves

∙ Molar heat capacity (Cp) is the amount of energy required to heat one mole of a  substance 1oC

∙ For temperature changes

q = n x Cp x ∆T

o n = moles of a substance

o ∆T = temperature change

1. How much energy is needed to heat 24.1g water from 10.0oC to 75.0oC H2O(l) Cp = 75.3J/mol x C

∆T = 65.0oC

24.1g

1mol

75.3J

65.0oC

18.02g

mol x C

= 6.55 x 103J = q

Temperature Changes

q = n x Cp x ∆T

∆T = q

n x Cp

Phase Change

q = ∆Htransition x n

∆Hfus

S ???? ???? L

Fusion

For H2O 6.01kJ/mol

∆Hvap

L ???? ???? G

Vaporization

For H2O 40.67

kJ/mol

∆Hsub

S ???? ???? G

Sublimation

1. How much energy is required to melt 52.1g of water?

52.1g

1 mol H2O

6.01kJ

18.02g H2O

1 mol

= 17.1kJ

Cp H2O(s) = 37.1 J/mol x oC

Cp H2O(l) = 75.3 J/mol x oC

Cp H2O(g) = 33.6 J/mol x oC

Heating Curves

3.5

3

2.5

2

T

1.5

1

0.5

0

Heating Curve

1 1

0

3

2 2

Tb

Tf

From 0-1, the solid is heating From 1-1, the solid is melting From 1-2, the liquid is heating

From 2-2, the liquid is vaporizing From 2-3, the gas is heating

*** Endothermic reaction

Cooling Curve

7

6

6

5

4

T

3

2

1

0

Cooling Curve

5 5

Tb

4 4

Tf

3

q (lost) -

From 6-5, the gas is cooling

From 5-5, the gas is condensing

From 5-4, the liquid is cooling

From 4-4, the liquid is solidifying

From 4-3, the solid is cooling

*** Exothermic reaction

Solid heat/cool q = n x Cp(s) x ∆T

Melting/freezing q = n x ∆Hfus

Liquid heat/cool q = n x Cp(l) x ∆T

Vaporizing/condensing q = n x ∆Hvap

Gas q = n x Cp(g) x ∆T

∙ Calculate q for each step and add together for total energy

∙ All heating curves will be positive, all cooling curves will be negative

Heat Transfer

∙ Two systems in contact with each other will exchange heat until their temperatures are  the same

o This is thermal equilibrium

o qlost = -qgained

Calorimetry

qcalorimeter = Ccalorimeter x ∆T = ∆Hreaction

Standard Enthalpy formation, ∆Hof (of = standard condition)

∙ Standard enthalpy formation is the enthalpy of change at a constant pressure when one  mole of a substance is produced from its elements

*** ∆Hof for elements in their standard state is ZERO

∆Hof O2(g) = 0kJ/mol

∆Hof Hg(l) = 0kJ/mol

Hess’s Law

∙ Hess’s Law, or Hess’s Law of Constant Heat of Summation, states that the enthalpy of  reaction ∆Hrxn for a process that is the sum of two or more other reactions is equal to the  sum of ∆Hrxn values of the constituent reactions ***∆H1 = ∆H2 + ∆H3

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