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Chemistry IA > Week 6 Module 3 > Flashcards

Week 6 Module 3 Flashcards

Refer to lecture for examples.

1
Q

What is the definition of a Bronsted acid?

A

They donate a proton (H+) to water to produce hydronium ions (H3O+)

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2
Q

What is the definition of a Bronsted base?

A

Accepts a proton from water, forming hydroxide ions (OH-)

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3
Q

What are the acid, base forms of water?

A

H3O+ is an acid, OH- is a base. Water is amphiprotic.

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4
Q

What is often used instead of H3O+?

A

H+.

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5
Q

What is a conjugate base/acid?

A

Related by the loss or gain of a proton. Their reversal roles come about because both forward and backward reactions are occuring. When an acid loses a proton it becomes its conjugate base and vice versa.

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6
Q

What is the reaction of water with itself?

A

H2O(l) + H2O(l) = H3O+(aq) + -OH(aq)

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7
Q

What is the autoionisation/autoprotolysis (equilibrium) constant of H2O?
Then work out the pKw

A

Kw = [H3O+][-OH] = 10^-14 at 25 degrees celsius. (conc = 1 as it is a pure liquid)
Take the -log of Kw
-log10Kw = -log10(10^-14) at 25 degrees celsius. p is used to denote -log10 of quantity.
pKw = 14 at 25 degrees celsius.

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8
Q

What values can the pH scale range from in M?

A

10^0 to 10^-14 M (mol/dm^3).

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9
Q

What scale is used for pH?

A

The logarithmic scale: pH = -log10[H+] and [H+] = 10^-pH (works the same for pOH)
As a result, ranges from 1 to 14 with a difference of 1 pH being a 10fold difference.

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10
Q

Calculate the value of the pH of water.

A

Kw = [H3O+/H+][-OH] = 10^-14 at 25 degrees celsius.
As [H+] = [-OH] as 1 molecule of water reactions with 1 to make 1 mole of H+ and -OH Kw = [H+]^2 = 10^-14
Therefore [H+] = root of 10^-14 = 10^-7 M
pH = -log10[H+] = 7 = Neutral

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11
Q

What does it mean if a pH is greater than 7?

A

It is basic.

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12
Q

What does it mean if a pH is less than 7?

A

It is acidic.

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13
Q

What are the calculations for the pOH scale?

A

pOH = -log[OH-]
[OH-] = 10^-pOH

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14
Q

What does pHw = pH + pOH =?

A

14.

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15
Q

Work out the pH of HCl reacting with water.

A

HCl [H3O+] = 10^-12M
H2O [H3O+] = 10^-7M
10^-7 + 10^-12 = around 10^-7
pH of 7.

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16
Q

Calculate the pH of Ba(OH)2(aq) in water.

A

Ba(OH)2(aq) - > Ba2+(aq) + 2OH-(aq)
[Ba(OH)2] = 0.5M
[OH-] = 2x 0.05 = 0.1
pOH = 1
pH = 14 - pOH = 13

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17
Q

Besides water, what is another species that auto-ionises?

A

Ammonia.
NH3(l) + NH3(l) = NH2- + NH4+
(NH2- is a strong base)

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18
Q

What is the measure of the extent of a reaction?

A

The equilibrium constant K.
Kc is the funcion of concentrations at equilibrium.

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19
Q

Show an example of how Kc works for gaseous systems.

A

Kc = [N2O4]/[NO2]^2
All concentrations are relative to standard conc, c^θ = 1mol /dm^3 (M, molL-1)
The real equation is ([N2O4]/C^θ)/([NO2]/C^θ)^2
This makes Kc unitless/dimentionless.

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20
Q

What do the values of Kc mean?

A

Kc<1 -> reactant favoured
Kc>1 -> product favoured

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21
Q

What are heterogenous systems?

A

Solids reacting with liquids
eg. CaO(s) + H2O(l) = Ca(OH)2(s)
Solids reacting with gases
eg. CaCO3(s) = CaO(s) + CO2(g)
Liquids reacting with gases
eg. CO2(g) + H2O(l) = H2CO3(s)

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22
Q

What must you consider with the concentrations of solids and liquids?

A

Amounts may change but conc/density stays constant.
Nearly pure liquid solvents have a constant conc.

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23
Q

Give an example of constant concs.

A

CaCO3(s) = CaO(s) + CO2(g)
Kc = [CaO][CO2]/[CaCO3]
The [CaO] and [CaCO3] remain constant during the reaction as they are solids, set to be 1.
-> Kc = [CO2]
Thats why conc of reactants is left out in self-ionisation of water.

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24
Q

What is the pH of natural water (example)

A

Water in nature absorbs CO2, makes carbonic acid, becoming acidic.
CO2(g) + H2O(l) = H2CO3(aq) K1 = 1.7 x 10^-3
H2CO3(aq) + H2O(l) = HCO3-(aq) + H3O+(aq) K2 = 4.2 x 10^-7
Atmospheric CO2 holds pH of water at 5.6.
For 426ppm CO2 (average atmospheric CO2 conc Feb 2024)
[CO2] = 9.68 x 10^-3 mol dm^-3

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25
How are sig. figs. dealt with?
If we have log10 of a number with N sig figs, the result will have N decimal places. eg. If [H+] = 0.00564 mol/dm^3 = 5.64 x 10^-3 mol/dm^3 Therefore, pH = -log10[H+] = 2.249 Also true for natural logs. Inverse is the same
26
What is Ka?
The acid ionisation constant. Ka dictates the strength of an acid.
27
Provide an example of working out Ka from an acid HA reacting with water.
Acid base reactions are essentially equilibrium reactions, thus have an equilibrium constant. Ha(aq) + H2O(l) = A-(aq) + H3O+(aq) Thus, the equilibrium constant is Ka = [A-][H3O+]/[HA] <- pure water = 1
28
What would Ka be like for a strong and weak acid?
For a strong acid, the equation tends to lie on the product side, producing a lot o conjugate base and hydronium ions. Therefore, Ka is large. For a weak acid, there would be more reactnat so Ka would be small.
29
How can you work out pKa?
-logKa. Gives a value betwen 1-14.
30
How do you discuss the pKa for strong acids?
It often isn't quoted. Just stated that it is strong and has a lot of dissociation.
31
How can you tell when a base is strong?
When there is more ionisaiton.
32
What is the equilibrium constant for base strength?
Kb is the base ionisation constnat. Kb = [BH+][OH-]/[B]
33
What would Kb be like for a strong and weak base?
Large product, large Kb. Weak: small.
34
How can you work out pKb?
-logKb.
35
What would be the strength of the conjugate base of a strong acid?
Weak.
36
Consider the weak acid acetic acid.
CH3COOH(aq) + H2O(l) = CH3COO-(aq) + H3O+(aq) The conjugate base can accept a proton to form the acid: CH3COO-(aq) + H2O = CH3COOH(aq) + OH-(aq) Ka and Kb for the conjudate base are related by Ka x Kb or [H3O+][HO] as A- and HA are cancelled out. = [H+][OH-] = Kw
37
What happens when you take the negative log of Ka and Kb?
pKa + pKb = pH + pOH = pKw = 14. (check page 44 of notes)
38
What does it mean if Ka is large?
There is high ionisation and a low pKa. Conjugate base must be a poor proton acceptor (Kb is low, pKb is high)
39
What are the relationships between strength of something and its conjugate??
Conj base of a strong acid is (very) weak. Conj base of a (very) weak acid is strong. Conj base of a (moderately) weak acid is (moderately) weak.
40
What do mono, di, and polyprotic mean?
Monoprotic: can donate 1 proton. Diprotic: can donate 2 protons Polyprotic: can donate more than 1 proton.
41
How does Ka work for polyprotic acids?
Protons are donated 1 at a time with a Ka for each step. Ka gets smaller for each proton lost - getting weaker. More difficult to remove 2nd H+ and so on. K for all 3 protons (in a triprotic acid) is the multiplication of the 3 K values.
42
Example of the question: What is the pH of a 0.010M aqueous solution of HCl?
HCl(aq) + H2O(l) = Cl-(aq) + H3O+(aq) As HCl is strong, it fully dissociates. [HCL] = [H3O+] and pH = -log(0.01) = 2
43
What is the limitation for calculating the pH of weak acids?
They do not fully dissociate, thus [HA] does NOT = [H3O+]
44
Refer to page 45 for another worksed examples of working the pH of solution of weak acid.
Study up how to use the ICE table again.
45
What is a titration used for?
It is used to find the concentration of an unknown solution by reaching the end/equivalence point or using the pH.
46
How is the equivalence/end point reached? What is an alternative name?
It is reached when the number of moles of H+ in the solution is equal to the moles of OH- originally (or vice versa). Also called stoichiometric point.
47
What is a titrant?
The known solution in a titration.
48
What is an analyte?
The solution in a titration with a known volume but unknown titration.
49
What happens at the end point?
Sufficient titration has been added to react with all of the analyte. For strong acid base titrations, the equivalence point is pH=7, neutral. Strong acids and bases undergo a complete H+ transfer. Solution contains salt and H2O <- neither contribute to pH.
50
Consider a solution of the weak base ammonia and the strong acid HCl.
NH3(aq) + HCl(aq) -> NH4+ + Cl- At aquivalence point, all ammonia and HCl has been reacted so the solution consissts of a strong acid and a very weak base. Excess of H3O+ exists in solution as NH4+ + H2O = H3O+ + NH3 End point occurs on the acidic side of pH 7.
51
What would the titration curve look like for Ammonia and HCl?
pH drops slowly at first until there is a sudden change in pH. pH slowly converges to value of strong acid. Stoichiometric point S occurs on acidic side of 7. pH around = pKa of conjugate acid of base at half equivalence point.
52
Consider a solution of weak acid formic acid and strong base NaOH.
NaOH(aq) + CH3COOH(aq) ->CH3COO-(aq) + Na+(aq) + H2O(l) At equivalence point, solution consists of Na+, CH3COO-, and H2O. CH3COO- is conjugate base of a weak acid = basic. Excess of OH- in solution: CH3COO- + H2O = CH3COOH + OH- Thus, toichiometric point is on the basic side of pH=7.
53
What would the titration curve look like for Formic Acid and NaOH?
pH rises slowly at first, then there is a sudden change in pH. pH slowly converges to value of strong base. pH around = pKa at half-equivalence point (point C0).
54
Consider a weak acid and weak base.
CH3COOH(aq) + NH3(aq) = CH3COO-(aq) + NH4+(aq) but also CH3COO-(aq) + H2O(l) = CH3COOH(aq) + OH-(aq) and NH4+(aq) + H2O(l) = NH3(aq) + H3O(aq) At equivalence, pH depends on the relative strength of acid and base.
55
Refer to page 47 for ecervise of a weak acid and base
Sorry but i aint doing all that - rememebr ICE table again
56
What are equivalence points like in polyprotic acid titrations?
There are multiple equivalence points. Eg. Histidine has 4 protons
57
What is the purpose of buffer solutions?
To resist a change in pH. Vital in places like homeostasis.
58
What is the titration curve like in a buffer solution?
pH changes slowly halfway to stoichiometric point. Addition of small amounts of acid or base to water reduces pH gradient. Buffer solutions: resist changes in pH.
59
What is an acid buffer (pH<7) made of?
It is a mixture of weak acid and its conjugate base.
60
What is a base bugger (pH>7) made of?
It is a mixture of weak base and its conjugate acid.
61
Why do buffers work?
Because: eg. if you have the weak acid in solution with its conjugate base, neutralises any added acid or base. Add base: weak acid HA donates H+ to OH- Add acid: Conjugate base A- accepts H+.
62
How to work out pH of buffer solution?
In terms of equilibrium: HA(aq) = A-(aq) + H+(aq). Ka = [H+][A-]/[HA] [H+] = Ka x ([HA]/[A-]) Take the - log of both sides of the equation: -log10[H+] = -log10 (Ka x ([HA]/[A-])) = -log10 (Ka) - log10([HA]/[A-]) Thus, pH = pKa + log10([HA]/[A-]) (Henderson-Hassenbach equation)
63
What are the best buffering conditions?
As long as added acid/base amount is relatively small, solution acts to neutralise addition so pH change is small. - [Conjugate base (A-)] = [Acid (HA)] -Concentration of added base or acid << [Conjugate base] and [Acid]
64
What is required to use H-H to calculate the exact pH?
The exact pH requires considering the approach to equilibrium uin the reaction.
65
What is the buffer capacity?
The measure of amount of acid (or base) that canbe added without substantial pH change. High conc buffers have higher capacity. Dilution is opposite.
66
What is buffer exhaustion?
It occurs when most of the weak acid/base is converted to base/acid.
67
What is the general range buffers are effective in?
pH = pKa +- 1 - Remember: change in 1 pH unit = 10 x change in concentration.
68
What shouldn't the ideal ratio of a buffer exceed?
10:1 or 1:10.

Chemistry IA (6 decks)

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