Flashcards
Aqueous Flashcards
Flashcards do conjunto (47)
- 1Define EquilibriumThe rate of the forward reaction
Equals
The rate of backwards reaction - 2Define KsSolubility constant
No unit - 3Ks formulaKs = [C]c [D]d
- 4s formula (gL-1) given gmol-1 and mol L-1s = mol L-1 x gmol-1
- 5s formula (mol L-1) given gmol -1 and gL-1s = gL-1/gmol-1
- 61:1 ratio KsKs = (s) x (s)
= s2
s = square root Ks (mol L-1) - 71:2 ratio Ks 2:1 ratio KsKs = (s) x (2s)2
= 4s3
s = 3square root Ks/4 (mol L-1) - 8Define solubilityAmount of substance
Dissolve in a given volume
form saturated solution - 9Describe Common Ion EffectIonic solution
Added to a solution containing one of common ion
Solubility is reduced - 10Explain Common Ion EffectState common ion e.g Br-
Br- added to solution
Increase in [common ion]
Reverse reaction favoured
To restore equilibrium
Decrease in [other ion] e.g Ag+
Salt becomes less soluble e.g AgBr - 11Complex Ions - NH3[Ag(NH3)2]+
[Cu(NH3)4]+2
[Zn(NH3)4]+2 - 12Complex Ions - OH-[Al(OH)4]-
[Pb(OH)4]-2
[Zn(OH)4]-2 - 13Complex Ion - cation (e.g Ag+) reacting with base NH3 addedProduce Complex Ion:
Ag+ + 2NH3 >< [Ag(NH3)2]+
Base added to the system (NH3)
Reduces concentration of solution
Product (Ag+) remove from equilibrium
Ag2CO3 --> 2Ag+ + CO3^2-
Increase in the forward reaction
Equilibrium shifts to favour products
More salt is dissolved e.g Ag2SO4 (s)
Solubility is increased - 14Complex Ion - cation (e.g Zn(OH)2) reacting with base OH- addedZnOH2 --> Zn2+ + 2OH-
Small amount OH- added
- decrease solubility
- reaction moves to left
- removes added product
- produce more reactant
Excess OH- added
- react with cation
- produce complex ion
[Zn(OH)4]-2
- reduces concentration of solution
- product removed
- reaction increase in forward direction
- more salt dissolve
- solubility increase - 15Complex Ion - salt (e.g Ag2CO3) contains carbonate (CO3) when acid addedAcid added to system
Reduces concentration of solution
Dissociate into ions - cation and carbonate
The acid reacts with carbonate (neutralises)
Product (CO3) removed from equilibrium
Ag2CO3 >< Ag+ + CO3^2-
Increase in the forward reaction
Equilibrium shifts to favour products
More salt is dissolved
Solubility is increased - 16Describe Strong Acids/BasesDissociates completely
Base - readily accepts H+
Acid - readily donates H+
A lot of particles per unit volume - 17Describe Weak Acids/BasesPartially dissociates
Base - partially accepts H+
Acid - partially donates H+
Few particles per unit volume - 18Calculate conc from pH[C] = (10-pH)2/Ka
- 19Kb formulaKb = Kw/Ka
- 20Strong Acid Water Dissocation[C] = [D] > [H3O+] = [OH-]
No additon of H3O+ or OH - pH remains neutral
Acid fully dissociates into ions - 21Strong Acid Water Dissocation when H+ added in ionisation[H3O+] = [C] > [OH-]
Addition of H3O+ - pH decreases
Ions added in dissociation - 22Strong Base Water Dissocation when OH- added in ionisation E.g NaOH --> Na+ + OH-[OH-] = [Na+] > [H3O+]
Addition of OH in ionisation - pH increases
Solution basic
Ions added in dissociation - 23Explaining pH- depends on [H3O+] and [OH-] ratio
- dissociation
- pH is high [OH-] high
- pH is low [H3O+] high - 24Describe conductivityDepends on the concentration of ions
Complete dissolve of ions
- good conduct
Partially dissolves (if weak substance)
- bad conduct - 25Calculate Ionic Product (precipitate)Calculate concentration of each ion
- mol-1 of ion x inital ml/total ml
IP = [calculated conc] x [calculated conc]
Calculate IP no units
Compare IP with given Ks
- a precipitate forms or not? - 26Comparing IP to KsIP < Ks < IP
- 27Define AcidProton (H+) donor
- 28Define BaseProton (H+) acceptor
- 29Calculate pH (given [H30+])-log [H30+]
- 30Calculate [H30+] (given pH)inv log - pH
- 31Calculate pH (given [OH-])Kw (1x10^-14)/[OH-] = [H30+]
pH = -log [H30+] - 32Calculate pH (weak acid)Ka = [C] [H30+]/[A] - H20 not included
assume [C] = [H30+]
[H30+]^2 = [A] x Ka
[H30+] = square root [H30+]^2
pH = -log [H30+] - 33Calculate pH (weak base) (given Ka)Kb = Kw (1x10^-14)/Ka
Kb = [C] [OH-]/[A] - H20 not included
assume [C] = [OH-]
[OH-]^2 = [A] x Kb
[OH-] = square root [OH-]^2
pOH = -log[OH-]
pH = 14 - pOH - 34Calculate pKa- log Ka
- 35Calculate Ka (given pKa)inv log - pKa
- 36Describe Buffer SolutionsAbsorbs the addition
Small amounts
Acids or base
Without changing pH significantly - 37Explain the role of the BufferNeutralise solution of acid by adding strong base
Dissolve salt of conjugate base into solution of acid - 38When is a Buffer Solution most effective?[Base] = [Acid]
pH = pKa + log 10
pH = pKa + log 10 [conjugate base]/[acid]
pH = pKa - 39Calculate pH of a Buffer Solution given KaKa = [H30+] [D] / [A]
[H30+] = Ka x [A] / [D]
pH = -log [H30+] - 40Calculate pH of a Buffer Solution given pKa + conc of [A] and [D]pH = pKa - log [A] / [D]
or
pH = pKa + log [D] / [A] - 41Define Equivalence PointAll of the weak acid has reacted with base being added
Most vertical point on the graph - 42Describe the features of a Titration Curve1. Start of pH of acid - initial pH due to starting acid or base
2. E.P - weak acid has reacted with base
3. Mid-point - half way in volume to E.P, pKa is pH of midpoint
4. Buffer Zone - area 1 pH either side of mid-point
5. After E.P - pH depends on conc of solution being added from burette - 43Calculate pH of a solution at Start Point (before base added)Equation
Ka = 10 - pKa
Ka = [H30+]^2 / [A]
[H30+]^2 = Ka x [A]
sqaure root [H30+]^2 = [H30+]
pH = -log [H30+]
pH = 3 sf - 44Calculate volume of base at Equivalence PointEquation
Calculate moles of substance where conc is given
n = c x v (given)
n(conc given) = n(base we want to know)
v = n/c
v(base) = ml (3 sf) - 45Calculate Mid-Point of BufferpH = pKa
v(base) /2 = ml 3sf
pH = pKa when __ml of (base) has been added - 46Calculate pH of Equivalence Point (after Mid-Point)v(base) + v(given) = ml 3 sf
c(conc given initially) / new ml 3sf
Kb = 1x10^14 / Ka
Kb = [C] [OH-]/[A]
Kb = [OH-]^2 /[c(conj base)]
[OH-]^2 = Kb x [A] or [c(conj base)]
sqaure root [OH-]^2 = [OH-]
[H30+] = 1x10^-14 / [OH-]
pH = - log [H30+]
pH = 3 sf - 47Calculate pH of a solution at Final Point (after base added)Amount added - E.P volume = excess volume
Total volume caluclate (before v + added v)
c(base) = excess v/total v X original conc of base
[OH-] = c(base)
[H30+] = 1x10^14 / [OH-]
pH = -log [H30+]
pH = 3 sf
Informações
- Descrição
- NCEA Level 3 Chemistry (Equilibrium) FlashCards sobre Aqueous Flashcards, criado por Shivani Kershaw em 31-07-2018.
- Criado por
- Shivani Kershaw
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