GAMSAT 1 (High Value Chem Topics)

Question 1

*

Using k_w to find the pH of a Base

Answer 1

k_w=[H⁺][OH]

NaOH & KOH - Strong bases - donate 1 mole of OH^- ions per mole of base

• the value of kw @ 298K us 1.0 X 10^-14 mol²dm^-6
• find the pH of 0.1 moldm^-3 NaOH @298K
  
  • [OH^-] = 0.1moldm^-3 => [H⁺] = k_w/[OH^-] = 1.0 X 10^-14/.01
  • pH= -log10 1.0 X 10^-13 = 13

Question 2

*

Diprotic Acids

Answer 2

They release 2 protons when it dissociates

ex. H₂SO_4(l)+water -> 2H⁺_(aq) + SO₄^-2_(aq)

H⁺ = 0.2 moldm^-3 so….pH = -log₁₀[0.2] = 0.70

pH of sulfuric acid 0.25 moldm^-3

[H⁺] = 2 X 0.25 = 0.5 => pH = ^-log₁₀[.5] = 0.30

Question 3

*

pH Definition

Answer 3

• The measure (from 0-14) of Hydrogen Ion Concentration
  
  • 0 = Very Acidic
  • 7 = Neutral
  • 14 = Very Alkaline (base/basic)

*Expressed in -log₁₀ —> pH = -log₁₀[H⁺] ex. pH = -log₁₀[0.01] = 2

Or…

[H⁺] = 10^-pH ex. [H⁺] = 10^-1.52 = 0.03moldm^-3 = 3 X 10^-2moldm^-3

Question 4

*

Acids & Bases

Answer 4

• Protons are transferred (Acid -> Bases) when A&B react
• Acids can olny get rid of protons when there is a base to accept them
  
  • ex. HA_(aq)+B_(aq) ⇔BH⁺_(aq)+ A^-_(aq)
• If acid is addded to water the water acts as the base and accepts the proton
  
  • HA_(aq) + H₂O ⇔ H₃O⁺_(aq) + A^-_(aq)

Question 5

*

Bases: Strong & Weak

Answer 5

• Strong Bases - ionise almost completely in water too
  
  • ex. sodium hydroxide - NaOH_(s) + Water ⇒ Na⁺_(aq) + OH^-_(aq)

_​

• Weak Bases - only slightly dissociate in water
  
  • ex. Ammonia - NH_3(aq) + H₂0 ⇔ NH₄⁺_(aq) + OH^-_(aq)

_{​*Equilibrium lies well over to the left}

Question 6

*

Acids: Strong & Weak

Answer 6

• Acid releases a proton - A base accepts a proton

Strong Acids - dissociate (or ionise) almost completely in water - nearly all the H⁺ ions will be released

(Hydrochloric acid) ex. HCl_(g) + Water ⇒ H⁺_(aq) + Cl^-_(aq)

Weak Acids - dissociate only very slightly in water - so only small numbers of H⁺ ions are formed

(Ethanoic or citric) ex. CH₃COOH_(aq) ⇔CH₃COO^-_(aq) + H⁺_(aq)

Question 7

*

Kinetics: Rate Equation

Answer 7

Rate = k [A]ⁿ + [B]^m

^[A] - Concentration of A

n - rate order

k - Rate Constant

*Overall order of reaction is n + m

* Increase in temperature will increase k

Question 8

*

Orders of Reaction

Answer 8

* Order can be determined only by experiment using the Method of Initial Rates (the rate for a short time @ the beginning of the reaction is measured @ several different concentrations of reactants)

1. First Order - X¹ = Rate doubles when reactant doubles
   
   • X2 = X2; X3 = X3; …
2. Second Order - X² = Rate is (x4) when the reactant doubles
   
   • X2 = X4 (2²) ; X4 = X16; …..
3. Zero Order - X⁰ = rate stays the same regardless of reactant
   
   • X2 = 1; X4 = 1; …..

Question 9

*

Kinetics: Concentration of Catalysts

Answer 9

• Increase concentration of catalysts = Increase rate of reaction
• Increase concentration of reactants in a solution, on avg. the particles will be closer together
  
  • closer ⇒ collide more often ⇒ more collisions ⇒ more chances to react
• if gases are involved, and increase in pressure of the gas works the same way
• Catalysts increase rate of reactions too by providing an alternative reaction pathway w/ a lower activation energy​

*Catalyst is chemically unchanged @ the end

Question 10

*

Kinetics: Physical State of Reactants

Answer 10

• Particles must collide to react
  
  • in the right direction, facing the right way
  • must collide w/ the min. amt. of kinetic energy

(Collision Theory)

*Liquids & Gases best as particles move

*Increase in temp. = particles have more kinetic energy = faster reactions @ activation energy more particles have enough energy @ 35ºC > 25ºC

Question 11

*

Kinetics: Physical State of Reactants

Answer 11

• Solids - particles very close together.
  
  • High density & incompressible.
  • Particles vibrate about a fixed point & can’t move freely
• Liquids - Similar density to a solid & is virtually incompressible
  
  • Particles move freely & randomly w/in the liquid
• Gas - particles have lots more energy & are much further apart
  
  • Density is pretty low & it’s very compressible
  • Particles move freely, diffuse quickly, no alot of attraction between them

Question 12

*

Gibbs free energy (1)

Answer 12

• Free enthalpy, Gibbs energy, or Gibbs function
• ΔG = ΔH - TΔS
  
  • H - Heat energy in the system (kj)
  • S - Measure of Entropy (J/kmol)
  • T - temp (K)​​
    
    • Equation to determine how likely a reaction is to take place spontaneously

* If a reaction will take place it reduces Gibbs free energy (ΔG < 0)

*Gibbs energy is reduced if H is reduced

*Gibbs energy is reduced if S is increased

Question 13

*

Gibbs free energy (2)

Answer 13

• Reactions most likely to happen if: ΔH < 0 & ΔS > 0

Endo & Exothermic

• Endo - energy taken in so heat in system is increased ΔH > 0 = unfavourable
• Exo - energy given off so heat in system is reduced ΔH < 0 = favourable

* Endothermic reaction can still take place if it results in a large enough increase in entropy (ΔS>0)

Question 14

*

Gibbs free energy (3)

Answer 14

* at a phase change (gas → water )

• ΔG = 0
• ΔG = positive # = not spontaneous
• ΔG = negative # = spontaneous

ex. ΔG = ΔH -TΔS

• • • : ΔG= negative if temp high
• • • : ΔG= negative if temp low
• • • : ΔG= positive (non spontaneous) always
• • • : ΔG= negative always

Question 15

*

Feasibility of Reactions

Answer 15

• More Negative ( or less positive) E^{<span>Ø</span>} value moves left
• More Positive (or less negative) E^{<span>Ø</span>} value moves right

Ex. Fe(OH)_3(s)+ e^- ⇔ Fe(OH)_2(s) + OH^-_(aq) E^{<span>Ø</span>} = -0.56V

O_2(g) + 2H₂O + 4e^- ⇔ 4OH^-_(aq) E^Ø = ⁺0.40V

Question 16

*

Electrode Potential

(conditions affecting it)

Answer 16

• Half cell reactions are reversible
  
  • equilibrium position is affected by changes in:
    
    1. Temperature
    2. Pressure
    3. Concentration
• Standard Conditions are:
  
  1. Temp - 25°C (298K)
  2. Pressure - 100kPa
  3. Concentration - 1.00 moldm^-3

Question 17

*

Electrode Potential

(standard elctrochemical cell drawings)

Answer 17

• the potential difference between the electrode & its solution

E^øcell = (E^ø_{right side} - E^ø_{left side})

Zn/Cu cell short hand

Zn_(s) | Zn²⁺_(aq) || Cu²⁺_(aq) | Cu_(s) (Zn²⁺_(aq) + 2e^- ⇔ Zn_(s) )

—Charges go this way——► (Cu²⁺_(aq) + 2e^- ⇔ Cu_(s))

Reduced| Oxidised || Oxi. | Red.

Question 18

Substitution & Elimination (1)

(Rules & what influences then)

Answer 18

Most important factor: type of Halokane

• Primary - mostly substitution
• Secondary - both substitution & elimination
• Tertiary - mostly elimination
  
  • Can be influenced by changing conditions​

The Solvent = proportion of ethanol to water

• more water = more substitution
• more ethanol = more elimination

​Concentration - of sodium or potassium hydroxide solution

• higher concentration = higher elimination

Question 19

Substitution & Elimination

Reactions

Answer 19

Substitution - the halogen is replaced by an -OH group to give alcohol

• CH₃CHCH₃ ⇒ NaOH ⇒ CH₃CHCH₃ + NaBr
  
  • | |
  • Br OH

Elimination - also in the presence of Sodium &/or potassium

• hydrogen bond is removed from one of the end carbon atoms toghether w/ brownie from centre one
• ​CH₃CHCH₃ + NaOH ⇒ CH₂=ChCH₃ + NaBr + H₂O
  • ​|
  • Br

Question 20

S_N1 Reactions

Answer 20

S_N1 - Nucleophillic substitution

• S = Substitution ; N = Nucleophilic ; 1 = the initial stage involves 1 species
• Faster mechanism
• best with tertiary halokanes
  
  • ​ex. R₃C - X ⇒ R₃C + ⇒ R₃C - Nu
    
    • ​ ↑
    • Nu

Question 21

S_N2 Reactions

Answer 21

S_N2 - best with Primary Halokanes (initial stage 2 species)

• _X
• ex. R - X → R< → R - Nu
• ^Nu

Question 22

Fischer Projections

Answer 22

• 2-D representation of a 3-D organic molecule
• Used for carbohydrates but not non-carbohydrates ex. Fischer
• Rules H O
  
  • Carbon Chain - vertical, C1 at top \ //
  • Horizontal bonds project toward the viewer C
  • vertical bonds project away |
  • H — C — OH
  • |
  • OH — C — H
  • |
  • H — C — OH
  • |
  • H

Question 23

Optically Active

Answer 23

• Rotate plane - polarised light
  
  • normal light vibrates in all directions but plane - polarised light only vibrates in one direction
• One enantiomer rotates it in a clockwise direction & the other rotates it in an anti clockwise direction

Question 24

Racemates (Isomers)

Answer 24

• Racemic mixture
• Contains equal quantities of each enatiomer of an optically active compound
• don’t show any optical activity
  
  • ​they cancel each other’s light-rotating effect
• ​made typically by reactive 2 achrial things together to get a racemic mixture as the chances of each enantiomer is equal
  
  • ​ex. H CL H
  • | | |
  • C + Cl₂ → HCl + C or C
  • / | \ / | \ / | \
  • CH₃ | H CH₃ | H CH₃| Cl
  • C2H3 C2H3 C₂H₃

Question 25

Chirals

Answer 25

- Carbon w/ 4 different groups attached * ex. H H O * | | / * H—C—C\*—C \* = Chiral Centre * | | \ * H OH OH ​**_Enantiomers:_** * H | H * | | | * C | C * / | \ | / | \ * HOOC | CH₃ | H₃C | COOH * OH | OH

Question 26

Optical Isomers

Answer 26

![]()- type of stereoisomerism - same structural formula but atoms are arranged differently -_Chiral_ - (asymmetric) carbon atom is an optical isomer that has 4 different groups attached to it. The groups attached to it. The groups can be arranged in 2 different ways so that 2 different molecules are made \* called _enantiomers_ or optical isomers * ex. H | H * | | | * C | C * / | \ | / | \ * HOOC | CH₃ | H₃C | COOH * OH | OH

Question 27

Stereoisomers

Answer 27

- Same structural formula but a different arrangement in space - w/ a double bond molecules can't twist around so you get either E or Z isomers * _E (entgegen(opposite))_ - H CH₃ * \ / * C=C E-but-2-ene * / \ * H₃C H * _Z (zusammen (together))_ - H₃C CH₃ * \ / * C=C Z-but-2-ene * / \ * H H

Question 28

Isomer Formulas

Answer 28

* **_General Formula_** - C_nH_2n+1OH * **_Empirical (simplest ratio)_** - C₄H₁₀O * **_Molecular (actual #)_** - C₄H₁₀O * **_Structural_**- CH₃CH₂CH₂CH₂OH * H H H H * | | | | * **_Displayed_** H–C–C–C–C–O–H * | | | | * H H H H * **_Skeletal_** - shows the bonds of carbon only, w/ any functional group * /\/\_OH

Question 29

Isomers

Answer 29

* Molecules with the same molecular formula but different molecular structures **_3 Types_** * _Chain Isomers_ - straight & branched carbon skeletons ( * ex. butane | | | | * H– C–C–C–C–H * | | | | * _Positional Iso_ - same skeleton but w/ a functional group attached * ex. l-Chlorobutane | | | | * H–C–C–C–C–Cl * _Functional group Isomers_ - same atoms arranged into functional groups * ex. propanone H O H * | || | * H–C–C–C–H * | |

Question 30

Periodic Table: Atomic Radius & Ionisation

Answer 30

- Atomic radius decreases across a period - ↑ # of Protons = ↑ positive charge of nucleus = ↑ pull of centre = ↓ of radius - Ionisation ↑ across a period * ↑ attraction between outer shell & nucleus

Question 31

Periodic Table: Blocks

Answer 31

* blocks show sub-shells configuration * 1s |H| * 2sΠΠ 2pΠΠΠΠΠΠ * 3sΠΠ 3pΠΠΠΠΠΠ * 4sΠΠ 3dΠΠΠΠΠΠΠΠΠΠΠ 4pΠΠΠΠΠΠ * 5sΠΠ 4dΠΠΠΠΠΠΠΠΠΠΠ 5pΠΠΠΠΠΠ * 6sΠΠ 5dΠΠΠΠΠΠΠΠΠΠΠ 6pΠΠΠΠΠΠ * 7sΠΠ 6dΠ * ↖83-103 * _s-block_ - have an outershell config of s¹ or s² * ​Lithium (1s²2s¹) & Magnesium (1s²2s²2p⁶3s²) * _p-block_ - outhershell of s²p¹ → s²p⁶ * Chlorine (1s²2s²2p⁶3s²3p⁵) * _d-block_ - d sub-shells are filled * Cobalt (1s²2s²2p⁶3s²3p⁶3d⁷4s²)

Question 32

Electron Shells: Subshells & Orbitals

Answer 32

- Subshells are divided into orbitals - Orbitals can hold up to 2 electrons * **_Subshell | # of Orbitals | Max # of electrons_** * s | 1 | 1 X 2 = _2_ * p | 3 | 3 X 2 = _6_ * d | 5 | 10 * f | 7 | 14 * **_Shell | Subshells | Total # of electrons_** * 1st | 1s | _2_ * 2nd | 2s 2p | 2+(3x2) = 8 * 3rd | 3s 3p 3d | 18 * 4th | 4s 4p 4d 4f | 32

Question 33

Periodic Table Trends: periods, groups & atomic #

Answer 33

* all elements in a period (row) have the same # of electron shells * (ignore s&p subshells) * |H| 0 * G1 2 3 4 5 6 7Π * 2ΠΠ ΠΠΠΠΠΠ * 3ΠΠ ΠΠΠΠΠΠ * 4ΠΠΠΠΠΠΠΠΠΠΠΠΠΠ * 5ΠΠΠΠΠΠΠΠΠΠΠΠΠΠ * 6ΠΠΠΠΠΠΠΠΠΠΠΠΠΠ * 7ΠΠΠ * All groups have same # of electrons in outer shells * G3 = 3 electrons in outer shells (G1→7,G0) * Similar outer shells = similar properties

Question 34

Buffer (pH)

Answer 34

* A solution that resists changes in pH when small amounts of acid or alkalai are added * doesn't stop the pH from changing but it does make it very slight * only works with small amounts of acids or bases * salt is a common buffer as it takes a few of the H⁺ ions

Question 35

Indicators (pH)

Answer 35

* Change colour at range that lies entirely on the vertical part of the pH curve * Methyl orange & Phenolphthalein are common indicators * Methyl orange - | Start | 3.1 - 4.4pH | High pH | * | Red | Orange | Yellow | * * Phenolphthalein | Start | 8.3 - 1.0pH | High pH | * | White | White/Pink | Pink | * * Weak base/ weak acid - no sharp curve → indicators won't work * Strong base/ strong acid - Either is fine

Question 36

8 Strong Bases

Answer 36

1. LiOH - Lithium hydroxide 2. NaOH - Sodium hydroxide 3. KOH - Potassium hydroxide 4. RbOH - Rubidium hydroxide 5. CsOH - Cesium hydroxide 6. Ca(OH)₂ - Calcium hydroxide 7. Sr(OH)₂ - Strontium hydroxide 8. Ba(OH)₂ - Barium hydroxide

Question 37

7 Strong Acids

Answer 37

1. HCl - Hydrochloric acid 2. HNO₃ - Nitric acid 3. H₂SO₄ - Sulfuric acid 4. HBr - Hydrobromic acid 5. HI - Hydroiodic acid 6. HClO₃ - Chloric acid 7. HClO₄ - Perchloric acid

Question 38

Naming acids

Answer 38

* **_Binary acids_** - 2 elements * (prefix) hydro + compound + -ic * ex. HCl = hydrochloric acid * **_Tertiary acids_** - typically hydrogen + a non-metal + oxygen * named by Oxygen amount * 2 common form = (prefix) hypo + compound + (suffix) -ous * ex. Hypochlorous acid | HClO * 1 common form = suffix = -ous * ex. Chlorous acid | HClO₂ * most common form = suffix -ic * ex. Chloric acid | HClO₃ * +1 common form = (prefix) -per + compound + (suffix) -ic * ex. Perchloric acid | HClO₄

Question 39

pH Curves

Answer 39

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Question 40

Calculate pK_a & also pH from pK_a

Answer 40

* pK_a = -log_10Ka & k_a = 10^-pK_a * pH = -log₁₀[H⁺] & [H⁺] = 10^-pH * * ex. Calculate the pH of 0.050 moldm^-3 HCOOH * pK_a of 3.75 @ 298k * K_a = 10^-pK_a = 10^-3.75 = 1.78 X 10^-4 moldm^-3 * * K_a = [H⁺]²/[HCOOH] ⇒ [H⁺]² = K_a[HCOOH] = 1.78 X 10^-4 X 0.050 =8.9 X 10^-6 * ^​⇒[H⁺] = √8.9 X 10^-6 = 2.98 X 10^-3 moldm^-3 * pH = -log₁₀ 2.98 X 10^-3 = **_2.53_**

Question 41

Calculate [H⁺] from pH

Answer 41

* [H⁺] = 10^-pH = ? X 10 ^-? moldm^-3 * * ex. pH = 3.02 * [H⁺] = 10^-3.02 = 9.55 X 10^-4 moldm^-3 *

Question 42

Dissociation Constant

Answer 42

* used to find the pH of a weak acid * K_a = _[H⁺][A^-] ex. HA_(aq) ⇔ H⁺_(aq) + A^-_(aq) * ^{---------------} \*can assume that all acids come from the * ^[HA] acid so [H⁺] = [A^-] ⇒ [H⁺]² * * ​K_a = _[H⁺]² * ^--------- pH = -log₁₀[H⁺] * ^[HA]

Question 43

Electrochemical Cells

Answer 43

* \* make electricity * Can be made from 2 different metals dipped in sale solutions of their own ions & connected by a wire (external circuit) ​ ![]() ![]() ![]() * **_How it works_** 1. Zinc loses electrons more easily than copper. The zinc is oxidised to form Zu²⁺_(aq) ions thus releaseing electrons into the external circuit 2. In the other half, the same # of electrons are taken from the external circuit, reducing the Cu²⁺ ions to copper atoms

Question 44

Good Reducing Agents (Common)

Answer 44

* Active metals (sodium, magnesium, aluminium, & zinc) * relatively small ionization energies & low electro- negativities * Metal hydrides (NaH, CaH₂, & LiAlH₄) which formally contain the H^- ion are good too * Hydrogen gas = reducing & oxidation agent * reduces non-metals * oxidises metals

Question 45

Common Oxidizing Agents (Good)

Answer 45

* **_Flourine (F₂)_** * the strongest oxidizing agent * even water will burst into flame in its presence * **_O₂, O₃, & Cl₂_** * good oxidising agents * 2^nd & 3^rd most electro negative elements * High oxidizing states = good oxidizing agents * Permanganate (MnO₄^-) * Chromate (CrO₄^2-) * Nitric acid (HNO₃) * Perchloric acid (HClO₄) * Sulfuric acid (H₂SO₄)

Question 46

Ionic half-equations

Answer 46

* Ionic half-equations show oxidation or reduction * Can combine them for different oxidizing or reducing agents together to make full equations for redox reactions * * ex. Magnesium burns in oxygen to form magnesium oxide * Oxygen is reduced to O^2-: O₂ + 4e^- ⇒ 2O^2- * Magnesium is oxidised to Mg²⁺: Mg ⇒ Mg²⁺ + 2e^- * (Both equations need to contain the same number of electrons so double everything) * 2Mg ⇒ 2Mg²⁺ + 4e^- * The electrons aren't included in the full equation. There are 4 on each side so they cancel * 2Mg + O₂ + 4e^-- ⇒2MgO + 4e^- * Final = 2Mg + O₂ ⇒ 2MgO

Question 47

Redox Reactions: Oxidation States (Rules)

Answer 47

* **_Oxidation states = Oxidation number_** 1. All atoms are treated as ions for this, even if they're covalently bonded 2. Uncombined elements have an oxidation state of 0 (zero) 3. Elements just bonded to identical atoms (O₂ & H₂) also have an oxidation state of 0 (zero) 4. Ox. state of a simple monoatomic ion (eg Na⁺) is the same as its charge (eg 1) 5. In compounds or compoind ions, the overall ox. state is just the ion charge * ex. SO₄^2- - overall ox. state = -2 * Ox state for O = -2 ( -2 x 4 = -8) * Ox state for S = +6 ( -2 = X - 8) 6. The sum of the oxidation states for a neutral compound is 0 (zero) * Fe₂O₃ - overall ox. state = 0 1. ox st. for O = -2 (-2 X 3 = -6) 2. Ox st. for Fe = -3 (-3 X 2 = -6 7. Combined Oxygen is nearly always -2 * except * peroxides where it's -1 * Flourides - OF₂ = +2 * O₂F₂ = +1 * O₂ = 0 * (In H₂O ox state of O = -2, H₂O₂ ox state = -1 b/c Hydrogen can't give 2 electrons, only 1) 8. Combined Hydrogen is +1 except in metal hydrides where it is -1 ( and H₂ where it is 0 * In HF ox st of H = +1 * In NaH ox st of H = -1 9. Roman Numerals give oxidation state (ox #) * ex. Copper (II) sulfate ox state Cu = 2+ * Manganate (VII) ion (MnO₄^-) ox state Mn = 7+

Question 48

Redox Reactions

Answer 48

* if electrons are transferred its a **_Redox reaction_** * If electrons are lost its called **_Oxidation_** * If electrons are gained its called a **_Reduction_** * An _Oxidising agent_ accepts electrons & gets reduced * A _Reducing agent_ donates electrons & gets oxidised * Na + ¹/₂Cl₂ ⇒ Na⁺Cl^- * Na is oxidised (loses an electron thus + in result) * Cl is reduced (gains an electron thus - in result)