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Physical chemistry > FPC May 23 exam > Flashcards

FPC May 23 exam Flashcards

1
Q

Give an example of a strong acid and a strong base

A

HCl (aq) -> H+ (Aq) + Cl- (aq)

100% dissociated, equilibrium lies to the right.

NaOH (Aq) -> Na+ (aq) + OH- (aq)

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2
Q

Define the term ‘acid oxide’ and give an example

A

An acid oxide is a type of oxide that reacts with water to produce an acidic solution.

An example is when carbon dioxide is reacted with water it produces carbonic acid (H2CO3) which is a weak acid

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3
Q

Define the term ‘base oxide’ and give an example

A

A base oxide is a type of oxide that reacts with water to produce an basic solution.

An example is sodium oxide which forms sodium hydroxide when reacted with water.

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4
Q

What is the term for an oxide that show both acidic and basic properties ? Give example.

A

Amphoteric oxides

ZnO + 2H+ (aq) => Zn2+ (aq) + H2O
ZNO + 2OH-(aq) + H20 => (Zn(OH)4)2-(aq)

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5
Q

What is a Bronsted-Lowry acid ?

A

A substance that donates a proton

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6
Q

What is a Bronsted-Lowry base ?

A

A substance that accepts a proton

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7
Q

Write the basic acid dissociation constant (Ka)

A

Ka = (H+)(A-) / (HA)

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8
Q

How do you identify conjugate acid-base pairs ?

A
  1. identify acid and base
  2. identify products
  3. compare products to original acid and base, conjugate acid that has accepted a proton and the conjugate base that when formed has donated a proton.
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9
Q

What is Kw ?

A

Kw is the ionic product of water, at 25degrees Kw = 1.02x10*-14mol2md6

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10
Q

Calculate pH from H+

A

pH = -log(H+)

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11
Q

Calculate H+ from pH

A

(H+) = 10*-pH

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12
Q

Calculate OH from pOH

A

(OH-) = 10*-pOH

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13
Q

Calculate pOH from OH-

A

pOH = -log(OH)

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14
Q

What is the pH of a 0.01 mol dm-3 HCl

A

-log 0.01 (1x10*-2) = 2.00

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15
Q

What is the pH of a 0.01 mol dm-3 NaOH solution

A

14 - (-log 0.01) = 12.00

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16
Q

What is the pH of a 0.01 mol dm-3 ethanoic acid given that Ka for ethanoic acid is 1.8x10*-5 at 298K ?

A

Ka = (H+)(CH3COO-) / CH3COOH)

1.8x10*-5 = (x)(x) / 0.01 - x

because x is so small assume 0.01 - x = 0.01

1.8x10-5 = x2 / 0.01

1.8x10-7 = x2

sqrt 1.8x10*-7 = x (H+)

pH = 3.4

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17
Q

What is the relationship between pKa and Acid strength ?

A

The larger the pKa value, the weaker the acid.

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18
Q

How do you calculate pKa from Ka ?

A

pKa = -log Ka

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19
Q

How do you calculate Ka from Pka ?

A

Ka = 10 *-pKa

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20
Q

Write the Kb expression for

NH4OH(aq) <=> NH4 + (aq) + OH-

A

Kb =(NH4+)(OH-) / NH4OH

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21
Q

What is the relationship between Kb value and the strength of of the base ?

A

The larger the value of Kb, the stronger the base.

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22
Q

Calculate the pKb of N2H4 (Hydrazine)

Kb = 9.8 x 10*-7

A

pKb = -log 9.8x10*-7 = 6.01

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23
Q

How do you calculate pH of a weak base using concentration of weak base and, pKb ?

A

pH = 14 - 1/2pKb + 1/2logc

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24
Q

What is the affect of adding H+ and OH- to a buffer solution

A

If H+ added = no change in pH

If OH- added = equilibrium shifts to the right

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25
How do you calculate the pH of a buffer solution using pKa
pH = pKa + log (salt / acid)
26
Calculate the pH of a weak acid at a concentration of 0.1 mol dm-3 ? Ka = 10*-6
Use both (H+) = square root KaC & pH = 1/2 pKa - 1/2 log c pKa = -log Ka = 6.0 pH = 1/2 pKa - 1.2 log c pH = 0.5 x 6.0 - (0.5 x log 0.1) = 3.0 - (-0.50)
27
How do you convert between pKa and pKb ?
pKa + pKb = 14
28
The pH of a solution undergoing titration and containing the indicator Hln was found to be at one point 3.52. Spectrophotometric data at this point revealed that 19% of the indicator was in the ln- form. Calculate the pKind of the indicator.
pH = pKln + log (ln-) / (Hln) if 19% in the ln- form then 81% is in the form Hhln pKin = 3.52 - log 19/81 = 3.52 - (-0.63) = 4.15
29
What are the names for substances that can and cannot form ions in solution, how does this relate to conductivity ?
Substance does form ions in solution - electrolyte. Solution conducts electricity as current is carried by ions. Substance does not form ions in solution - non electrolyte. Solution does not conduct electricity, e.g sugar.
30
What is the Faraday constant ?
Faraday (F) = quantity of charge equivalent to 1 mole of electrons, is equal to 96,500 coulombs (C)
31
What is Ohm's law and what substances obey it ?
I=E/R I = current strength, E = emf, R = resistance
32
Define conductance (G)
Conductance G is defined as 1/R (S) where S is the Siemen unit and is equivalent to Ω –¹
33
What are the rearrangements for I = E/R ?
E = I*R R = E/R
34
Define the 'cell constant' and identify the constituent parts ?
d/A is termed the cell constant. A is the area of the electrode and d is the distance between the electrodes.
35
What is the equation for specific conductance or conductivity ?
K = G x (d/A) K = G x cell constant
36
The conductance of a solution of 0.1 mol dm-3 KCl was measured as 0.0286 S. Given that the conductivity of 0.1 mol dm-3 KCl is 1.29 S m-1, calculate the cell constant of the cell used for the measurement.
cell constant = K / G = 1.29 S m-1 / 0.0286S = 45.1 m-1
37
What is the calculation for molar conductivity ?
Molar conductivity = K (conductivity) / c (concentration in mol)
38
What is the difference between a microscopic and a macroscopic system ?
Macroscopic - Large measurable amounts of matter Microscopic - Molecular study
39
What is a state function and give two examples ?
A state function is a system property that only depends on the current state of the system and not how this property was achieved. Energy, mass and pressure are examples
40
What is a path function and give two examples ?
A path function is a systems property that depends on the state of the system and how the state was reached. Heat and work are examples.
41
What is the difference between intensive and extensive properties ?
An extensive property is a property that depends on the amount of matter in a sample. Mass and volume are examples of extensive properties. An intensive property is a property of matter that depends only on the type of matter in a sample and not on the amount.
41
At 298K the internal energy for elements at one atmosphere is .....
0 joules
42
What is a 'system' ?
A system is what you are studying or the part of interest everything else is known as the 'surroundings'
43
Describe the difference between an isolated system a closed system and an open system
Isolated system - No exchange of matter or energy Closed system - Fixed material (Atoms, molecules etc) but allows the transfer of E Open system - Matter and energy can be exchanged
44
Explain what happens to a volume of a mixture in an isolated system
The volume in an isolated system does not change.
45
Outline Boyles law (Volume and Pressure)
At a constant temperature, the volume of a fixed amount of gas is inversely proportional to its pressure. eg if pressure = x and volume = y then 2x = 1/2y
46
Outline Charle's Law (Volume and temperature)
The volume of a fixed amount of gas , at a constant pressure, is proportional to the absolute temperature. Ie if the temperature is doubled at fixed pressure the volume will double.
47
If the volume of a fixed mass of argon gas at 1.00 x 10*5 Pa is 0.0312M3 at 273K calculate its volume at a pressure of 10 x 10*5 Pa and a temperature of 373K
use initial and final state so P1V1/T1 = P2V2/T2 = T2P1V1/T1P2 = V2 V2 = 4.26 x 10*-3
48
What is the ideal gas equation and define each constituent part
pV = nRT p = gas pressure v = volume occupied n = amount of gas R = Gas constant (8.314 J K) T = Absolute temperature
49
Calculate the volume of one mole of an ideal gas under the standard condition of temperature and pressure. T = 298k , p = 1 x 10*5 pa, N = 1.00 , R = 8.314
T = 298k , p = 1 x 10*5 pa, N = 1.00 , R = 8.314 1.00x8.314x298 / 1x10*5 = 227 x 10*-3
50
Outline Daltons law of partial pressures
Daltons law of partial pressures The total pressure exerted by a mixture of gases is the sum of the partial pressures of each individual gas PT = P1+P2+P3
51
In a sample of dry air weighing 1.00g there was found to be 0.75g nitrogen and 0.25 oxygen. Calculate the partial pressures of nitrogen and oxygen in the sample if the overall pressure of the air is 1atm.
Total = 1atm O2 = .25g atm = 0.75atm m/mr = n N = 0.027 mol O2 = 0.0078 mol Total = 0.0348 mol N = 0.027 / 0.0348 = 0.78 atm O2 = 0.0078 / 0.0348 = 0.22 atm
52
In a closed system what is the calculation for internal energy (Delta U) ?
ΔU = q + w q = transferred heat w = work
53
A reaction is carried out in which 1.23 kJ of heat is released and 235 J of work is done on the surroundings. What is the delta U for the process ?
ΔU = 1.23 kJ + 235 J = 1.465 KJ -ve because lost from the system
54
What is the calculation for work (w) involving external pressure (Pext) and change in volume ?
w = -pext (Vfinal-Vinitial)
55
Calculate w for a reaction that produces 1.00dm3 of CO2 at one bar
1 bar = 1 x 10*5 external pascals 1.00dm3 = 1 x 10*-3 m w = -1 x 10*5 (1 x 10*-3) = 100nm
56
What is ΔH ?
ΔH = Enthalpy change Enthalpy change is the heat transferred between a reaction and the surroundings at constant pressure. This applies to many reactions carried out in the laboratory in open systems.
57
What is the equation for enthalpy change involving ΔH, ΔU and PΔV ?
Enthalpy change = the sum of the change in internal energy, ΔU, plus the product of the constant pressure, P, and the change in volume, ΔV ΔH = ΔU + PΔV.
58
What is the equation for internal energy from enthalpy change and work ?
Internal energy = ΔH + w
59
At constant pressure what is ΔH equal too ?
At constant pressure ΔH = q q = amount of energy transferred in a system
60
What happens to the internal energy change for a closed system if a reaction is carried out at constant volume ?
ΔU = q Because the internal energy of the system is constant so ΔU = 0
61
What is the calculation for Gibbs free energy change ?
ΔG = ΔH - TΔS
62
The resistance of a conductance cell containing 0.10 mol dm-3 lithium nitrate solution is 192ohm. The resistance. of the same cell containing 0.02 mol potassium chloride solution is 620ohms. The standard conductivity is 0.02mol dm-3 potassium chloride is 0.242 S m-1. Calculate i) the cell constant of the conductance cell ii) the conductivity of the lithium nitrate
Conductance = G=1/R = 1.1613 X 10*-3 Cell constant = 0.242 / 1.613 x 10 * -3 = 150m Conductivity = G / cell constant = 5.208x10*-3 x 150.310559 = 0.79

Physical chemistry (3 decks)

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