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Chemistry > Equilibria > Flashcards

Equilibria Flashcards

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1
Q

conjugate acid

A

transfers proton to its conjugate base

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2
Q

conjugate base

A

accepts a proton from its conjugate acid

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3
Q

what is Kw

A 
Kw = [H+][OH-]
Kw = [H+]^2
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4
Q

Calculate the pH of a solution H+ is 5.32 x 10^-4 moldm-3

A

pH = -log[H+]
= -log(5.32 x 10^-4)
= 3.27

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5
Q

what is Ka of CH3COOH –> H+ + CH3COO-

A

[H+][CH3COO-]/[CH3COOH]

[H+][A-]/[HA]

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6
Q

what is pKa

A

pKa = -logKa

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7
Q

Calculate the [H+] from pH 5.40

A

10^-5.4

= 3.98 x 10^-6 mol dm-3

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8
Q

Calculate the value of Ka for methanoic acid. A solution of 0.010 mol dm-3 HCOOH, has a pH of 2.90

A

pH = 10^-2.90
= 1.26 x 10^-3
(1.26 x 10^-3)^2/0.010
= 1.59 x 10^-4 moldm-3

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9
Q

Calculate the pH of 0.1 mol dm-3 CH3COOH (Ka = 1.74 x 10^-5 mol dm-3)

A 
Ka = [H+]^2/[CH3COOH]
1.74 x 10^-5 = [H+]^2/ 0.1
[H+]^2 = 1.74 x 10^-6
[H+] = 1.32 x 10^-3
pH = -log[1.32 x 10^-3]
pH = 2.88
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10
Q

Calculate the hydrogen ion concentration of a solution whose pH is 10.5

A 
[H+] = 10^-10.5
[H+] = 3.16 x 10^-11
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11
Q

Calculate the pH of a solution of sodium hydroxide 0.05 mol dm-3 / Kw = 1.00 x 10^-14 mol2dm-6

A 
Kw = [H+][OH-]
1.00 X 10^-14 = [H+][0.0500]
[H+] = 2.00 x 10^-13 mol dm-3
pH = -log[2.00 x 10^-13]
pH = 12.7
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12
Q

buffer solutiona

A

a solution in which resists the changes in pH when small amounts of H+ or OH- are added / weak conjugate pair

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13
Q

how to make a buffer solution

A

2 ways

  1. weak acid and conjugate base
  2. weak base and conjugate acid
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14
Q

weak ethanoic acid and sodium ethanoate

A

weak ethanoic acid CH3COOH only partially dissociates. CH3COONa is fully ionised and forms CH3COO-.
Conjugate pair to form buffer solution
CH3COOH H+ + CH3COO-
when H+ is added, equilibrium moves to the left but the high concentrations of the conjugate pair won’t be changed significantly so pH doesn’t change too much
when OH- is added, H+ is reduced so equilibrium move to the right

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15
Q

calculate pH of buffer solution

0.05 mol dm-3 methanoic acid and 0.1 moldm-3 sodium methanoate / Ka = 1.60 x 10^-4

A 
[H+] = Ka x [acid]/[salt]
= 1.60 x 10^-4 x 0.05/0.1
= 8 x 10^-5
pH = -log(H+)
pH = 4.10
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16
Q

uses of buffer solution

A

the blood contains HCO3- , haemoglobin, plasma proteins, H2PO4- and HPO42-
CO2 + H20 –> H+ + HCO3- with carbonic anhydrase.
H+ + HCO3- forms CO2 and released at the lungs.
Too much H+ in the blood causes acidiosis where there is too much acid in the body.

When there is increase in [H+], shifts to left, forms CO2 + H2O, equilibrium restored, reduces [H+]

When there is decrease in [H+], shifts to right, forms H+ + HCO3-, equilibrium restored, increases [H+]

17
Q

What is Ksp for AgCl dissolving

A

Ksp = [Ag+][Cl-]

18
Q

what is the Ksp for Fe(OH)2

A

[Fe2+][OH-]^2

19
Q

calculate Ksp where MgF2 has solubility of 1.22 x 10^-3 moldm^-3

A

MgF2 –> [Mg2+][F-]^2

[Mg2+] = 1.22 x 10^-3
[F-]^2 = 1.22 x 10^-3 x 2
= 2.44 x 10^-3

Ksp = [1.22 x 10^-3][2.44 x 10^-3]^2
= 7.26 x 10^-9 mol3dm-9

20
Q

Calculate solubility from Ksp

Ksp for CuS = 6.3 x 10^-36

A

6.3 x 10^-3 = [Cu2+][S2-]
6.3 x 10^-3 = [Cu2+]^2
[Cu2+] = 2.5 x 10^-18

21
Q

predicting solubility

A

Ksp = 5.5 x 10^-10
when greater than Ksp, ppt will form
when less than Ksp, no ppt will form

22
Q

common ion with AgCl and NaCl

A

AgCl (s) <=> Ag+ + Cl-
there is more Cl- since we have more chloride ions.
Shifts to the left, more solid, so ppt will form, reduces solubility

23
Q

what is Kpc

A

the conc of a solute partitioned between two immiscible solvents at a particular temperature

24
Q

calucaltions of Kpc

A 
first calculate moles of substnace (NH3) 
then calculate moles of layer1 (aqueous)
calculate the difference for layer 2 (organic)
caluclate the conc for each layer.
Kpc = [organic]/[aqueous]
the bigger value means more soluble.
Kpc > 0 is soluble in top
Kpc < 0 is soluble in bottom
25
Q

what affects Kpc

A

depends on the bonds between solute and solvent
e.g NH3 is a polar molecule so it can hydrogen bonds and is very soluble in water. However, NH3 is less soluble in organic because the form pd-pd which is weaker than H bonds.

e.g I2 is molecule so it doesn’t form many hydrogen bonding and so doesn’t dissolve. dissolves in organic as they are both non polar and can form id-id bonds

26
Q

Kw meaning

A

ionic product of water, equilibrium constant for the ionistion of water

27
Q

Ka meaning

A

acid dissociation, equilibrium constant for the dissociation of weak acid

Chemistry (11 decks)

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