Equilibria Compiled + Transition Metals

Question 1

Is solubility and Ksp the same? Define the terms and explain the difference, if any

Answer 1

Solubility refers to the amount of solute (ie the salt) that can dissolve in 1dm3 of the solution to form a saturated solution at a given temperature.

In other words, solubility is the concentration of the salt in a saturated solution at a given temp

Ksp, the solubility product of a salt, is the product of the molar concentrations of its constituent ions in a saturated solution, raised to its respective stoichio coefficients at a given temperature.

Question 2

How to find Ksp from solubility? (And vice versa)

Answer 2

Recall that solubility is essentially the concentration of the SALT (ie, AgCl, not Ag+ and CL-) in a saturated solution. And Ksp = [Ag+][Cl-], the product of the concentrations of the IONS in a saturated solution.

1. Write equation (with reversible arrow)
   AgCl (s) —> / —< Ag+ (aq) + Cl- (aq)
2. At equilibrium, if (x) amount of salt per dm3 of solution is present in the saturated solution, then product of concentration of ions should be (x)(x) = x^2 !! (Here we define solubility as x, which will be given in the question)
3. Calculate Ksp

Question 3

Define dynamic equilibrium

Answer 3

Dynamic equilibrium is a state of a reversible reaction whereby rates of forward and backward reaction are equal but non zero

Question 4

Common ion effect: Explain what happens to the solubility of Ag2SO4 when it is added to AgNO3 aqueous

Answer 4

1m:
Ag+ is the common ion. Concentration of Ag+ increases. Ionic product of silver sulfate increases such that IP> Ksp. (For 1m, stop here: Hence solubility decreases)

1m:
Position of equilibrium shifts to the side of the Ag2SO4 to reduce concentration of Ag+. Solubility of Ag2SO4 decreases/ the precipitation of Ag2SO4 is favoured

Question 5

Describe (with words) how to find solubility of AgCl when dissolved in 0.1moldm3 of NaCl

Answer 5

Write equation
Draw equilibrium table: for concentration of Cl-, write “x+0.1” (add the extra concentration directly to x, the concentration of Cl- in a saturated solution.

Equate Ksp = [Ag+] [Cl-]
Ksp = [x] [0.1+ x] (assume x«<0.1 ie negligible)
= 0.1x
Find x

Question 6

Common ion effect on solubility vs complex ion formation on solubility

Answer 6

Common ion decreases solubility while complex ion formation increases it

Question 7

(TM) List common transition metal complex ions with coordination number of 4 (ie, tetrahedral or square planar arrangement)

(Solubility) List common complex ions which are not transition metal complexes

Answer 7

1. [NiCN4]2-
2. [CuCl4]2-

Non transition metal complexes
1. Al(OH)4-
2. (PbCl4)2-

Other commonly tested complex ion formations:
1.(Ag(NH3)2) +
2. (Cr(OH)4)-
3. (Cr(OH)6)3-

Question 8

[TM] Explain why cations like Al3+ and Cr3+ can form complexes

Answer 8

Metal cations like these have
(1) high charge density and thus polarising power. They can strongly attract the lone pairs of electrons on ligands towards themselves
(2) vacant low lying 3D orbitals to accept the electrons donated by ligands, resulting in dative bond formation

Question 9

What complex ion does Al3+ form in excess NaOH?

Answer 9

[Al (OH)4]-

Question 10

When do you bring in complex ion formation? Explain the effect of complex ion formation on solubility.

Answer 10

When metal cation has high charge density / and or is a transition element

Complex ion formation decreases the concentration of the metal cation in the solution. POE shifts to the side of the metal cation, favouring the dissolution of the salt. Hence solubility increases

To show:
1. Write out your salt dissociation equation
2. Write how the metal cation interacts with the added solution

Question 11

How to find pH of salt solution? If you’re only given the equation of the formation of the salt, eg A+B —> C

Answer 11

1.Identify your cations and anions (ie, the species present in the solution at equilibrium. Determine which one is the CA/CB of the weak species out of the two reactants
2. Remember that the conjugate bases and conjugate acids of weak species can hydrolyse in water to give H+ or OH-. This is the only source of OH- or H+ ions that are present at equilibrium
3. Write out the Kb/Ka equation for the conjugate base/acid. Make sure reactant is the species that is actually present in the solution
4. Kb/Ka will be given. Use it to calculate [OH-] [H+] and thus pH or pOH.

Question 12

What is the Kc expression for this equation?
A(g) + B (g) —>/< C(s)

What is the effect on [Ag] when [C] is increased?

Answer 12

Kc = 1/([A][B])

No effect. C is not included in the Kc equation and thus POE does not change.

Question 13

What is a buffer solution?

Answer 13

A buffer solution is a solution that can maintain a fairly constant pH when small amounts of acid/base are added to it

Question 14

What is a transition element?

Answer 14

A transition element is a d block element that is able to form one or more stable ions with partially filled d subshells

Question 15

Explain why transition metals have relatively invariant atomic radius across the period

Answer 15

Nuclear charge increase due to inc in no of protons
Electrons added to penultimate 3d sub shell, increase in shielding effect
Increase in shielding effect nullifies inc in nuclear charge, leaving Zeff and atomic radius relatively invariant

Question 16

Explain why transition metals have higher densities than s block elements (group 1 and 2 metals)

Answer 16

Transition metals have smaller atomic radiuses and larger atomic masses than s block elements.

Additionally, TM more closely packed due to stronger metallic bonding as compared to s block elements

Atoms per unit volume is higher —> density is higher

Question 17

Explain why TM have higher boiling points than period 4 s block elements

Answer 17

Both have giant metallic lattice and exhibit metallic bonding
TM able to delocalise electrons from both 4s and 3d subshells, because the energy level difference between 4s and 3d is relatively similar. forming stronger metallic bonds which require more energy to break
For s block elements, only electrons from 4s subshells can delocalise from the metallic lattice structure.

Question 18

Why can TM exhibit variable oxidation states (hint, 4s and 3d electrons)? And why can’t s block elements do so?

Answer 18

TM can exhibit variable oxidation states due to the close similarity in energy levels of the 3d and 4s electrons

Once 3d electrons have been removed, 4s electrons can also be removed without requiring much more energy.

For s block elements, subsequent removal of e would be from inner shell p orbitals —> require a lot more energy

Question 19

Why can Transition metals act as heterogeneous catalysts? Why can they act as homogeneous catalysts?

Answer 19

Transition metals can act as het catalysts due to availability of partially filled d orbitals which can accept electron pairs from, or donate e pairs to reactant molecules, forming temporary weak bonds with reactants at the catalyst surface

Good homogeneous catalysts as they can form variable oxidation states, and can easily convert from one oxidation state to another, enabling them to remain in the same phase as the reactants throughout the reaction.

Question 20

Define complex ions

Answer 20

Complex ions are central metal ions which are surrounded by ligands which are bonded to it through dative bonds

Question 21

Define ligands

Answer 21

Ligands are neutral molecules or anions which have at least one atom with a lone pair of electrons that can be donated to form dative bonds with a metal atom/ion