CE- Equilibrium in aqueous systems notes

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Solubility Equilibrium

Equilibrium expression, aA+bB ⇌ cC+dD

The reaction occurs in both directions

The equilibrium constant (Kc) measures how much product compared to how much reactant is present when the reaction is complete for an equilibrium system.

Kc= [C]^c.[D]^d/[A]^a.[B]^b

where [C]^c.[D]^d equals the total concentration of the products and [A]^a.[B]^b equals the total concentration of the reactants.

When Kc is small (Kc<0) the reverse reaction (where there are more reactants produced) is favored

When K is large (Kc>0) the forward reaction (where there are more products produced) is favored

Note: Never include solids or H2O in Kc as they have a constant concentration.

Le Chatelier's principle- Equilibrium will always in the direction the opposes change.

A saturated solution- no more solid will dissolve in the given amount of solvent at a fixed temperature.

The solubility product (Ks) measures how much product compared to how much reactant is present in a saturated solution.

Ks= [C]^c.[D]^d/[A]^a.[B]^b

The common ion effect

A solution can only take a certain amount of a given pair of ions before it becomes saturated, and if one of those pairs are already present because of another previous reaction less can be dissolved before Ks is reached.

When the ionic solid dissolving in a solution shares a common ion, the solubility will be reduced.

Solubility will be measured by the ion that is not shared.

solubility

when Ks=[A].[B] has an equal ratio of moles in the solubility constant the solubility, S=√ks or Ks=s^2

When the ks=[A].[B]^2 has a 1:2 ratio of moles the solubility, S=∛(ks /4) or Ks=4s^3

Sparingly soluble materials are those, which have lowered solubility.

The presence of H30+ ions can greatly increase the solubility of the sparingly soluble compounds if they contain an ion at reacts with an acid>basic ion.

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complex ions

If a complex ion can form, removing some of one of the ions in equilibrium, the equilibrium will move to the right, increasing the solubility of the salt.

Cu(OH)2(s)⇌Cu^2+(aq)+20OH-(aq)

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Predicting precipitation

Concentration=number of moles/Volume, C=n/v

number of moles=mass/Molar mass, n=m/M

To predict if a precipitate will form we use the general ionic product (Qs), this measures how much product compared to how much reactant is present in a solution.

Qs= [C]^c.[D]^d/[A]^a.[B]^b

when Qs is less than Ks the solution is unsaturated.

When Qs=Ks the solution is saturated, at equilibrium and no more solid can dissolve.

When Qs>Ks the solution is supersaturated and a precipitate will form.

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Acid Base Equilibrium

An Acid is a proton (H)+ donor

A Base is a proton (H)+ acceptor

A conjugate base is a substance formed by the donation of a proton.

A conjugate acid is a substance formed by the acceptance of a proton.

Amphiprotic substances can donate and accept protons.

Strong acids/bases fully dissociate when in aqueous solutions

Acid-base reactions

strong Acid + H2O → conjugate base + H3O+

strong base + H2O → conjugate acid +OH-

Calculating PH of strong acids/bases

PH= -log[H3O+], [H3O+]=inverse log(-PH)

for a strong acid, the hydrogen ion concentration is the same as the concentration of acid as it fully dissociates.

for a strong base, the process is similar as it also fully dissociates hower Kw=10^-14=[H3O+].[OH-] is used to find the hydrogen ion concentration when the concentration of acid is the same as [OH-].

PH of water

dissociation of water- H2O(l)⇌OH-(aq)+H3O+(aq)

the ionic product for water (Kw) = [OH-].[H3O+] =10^-14

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Weak acids and bases

Weak acids/bases partially dissociate when in aqueous solutions and so are in equilibrium

Acid-base reactions (weak)

weak Acid + H2O ⇌ conjugate base + H3O+

weak Base + H2O ⇌ conjugate acid + OH-

Calculating PH of weak acids/bases

[H3O+]= √Ka.[ acid]

for a weak acid, the hydrogen ion concentration is related to both the acid concentration and the equilibrium constant, because it partially dissociates.

[OH-]=√kb.[base]

For a weak base, the process is similar but the [OH-] is used to find the [H3O+] using (Kw) = [OH-].[H3O+] =10^-14

Acid dissociation constant- Ka=[conjugate base].[H3O+]/[weak acid]

PKa measures the strength of the acid pka= -log[Ka]

Base dissociation constant- Kb=[conjuate acid].[OH-]/[weak base]

pkb measures the strength of the base pkb= -log[kb]

Ka.Kb=Kw=10^-14

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Buffer solutions

A buffer solution is a solution which resists a change in its PH when a small amount of acid or base is added to it

This is because a buffer solution contains a weak acid mixed with its conjugate base ( or a weak base mixed with its conjugate acid)

Calculating the PH of a buffer-

[H30+]=ka.[conjugate acid or base]/[acid or base]

Buffer solutions work best for resisting the change in added acid and added base when [acid]=[base] and so when the PH=pka. This is the most efficient buffer as it can equally resist added acids and bases.

when the buffer solution contains a higher concentration of acid than base the PH is less than the pka the solution will more effective at resisting added base.

When the buffer solution contains a higher concentration of base than acid the PH>pka and the solution will be more effective at resisting added acid.

buffers are only effective in a range of plus-minus 1 of the pka value

the [conjgate acid/base]/[base/acid] part of this equation is a ratio of relative concentrations and so the moles present can be used to form this ratio for some questions without changing the answer.

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Acid base indicators

are weak acids which have a different coloured conjugate base.

this allows a colour change to be observed when the acidic conditions are changed to basic conditions

so for acid-base titrations an indicator should change colour (and so the conditions) at the same point in the titration as the equivalence point of the acid-base titration.

to choose an indicator

indicators have a PH range for colour change, so you want to choose an indicator which has a pka similar to the PH at the equivalence point.

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Titration curves

Steps for titration calculations

write out the chemical equation to determine the mole ratio

calculate the amount of the known reactant

calculate the amount of the unknown reactant

calculate the concentration of the second reactant

Graphs have PH on the vertical x-axis and the volume of the base or acid of the horizontal y-axis

the equivalence point is when just enough acid/base has been added to react with all the available base/acid

for strong acid-strong base titrations at the equivalence point the PH=7

for weak acid-strong base titrations at the equivalence point the PH>7

for weak base-strong acid titrations at the equivalence point the PH<7

Initial PH- at the start the PH equals the pH of the solution which that titration is starting with.

Buffer zone PH- only occurs with weak acids or weak bases. At the halfway equivalence point the PH=pka (POH=pkb) and so the [weak acid]=[conjugate base]

PH at the equivalence point- acid and base have completely reacted with one another.

Final PH- at the end equals the PH of the leftover acid or base where the volume is twice that at the equivalence point

Initial PH-

strong acid-strong base

[H3O+]=[acid]

PH=-log[H3O+]

Strong base-strong acid

[OH-]=[base]

Kw=[H3O+].[OH-]

PH=-log[H3O+]

weak acid-strong base

[H30+]=√(Ka[acid])

PH=-log[H3O+]

weak base-strong acid

[OH-]=√(kb[base])

Kw=[H3O+].[OH-]

PH=-log[H3O+]

Buffer zone PH

weak acid-strong base

[H3O+]=Ka.[conjugate base]/[acid]

PH=-log[H3O+]

weak base-strong acid

[OH-]=kb.[conjugate acid]/[base]

Kw=[H3O+].[OH-]

PH=-log[H3O+]

PH at the equivalence point

strong acid-strong base

PH=7

strong base-strong acid

PH=7

weak acid-strong base

[H30+]=√(Ka[acid])

PH=-log[H3O+]

weak base-strong acid

[OH-]=√(kb[base])

Kw=[H3O+].[OH-]

PH=-log[H3O+]

Final PH-

strong acid-strong base

[OH-]=[base]

Kw=[H3O+].[OH-]

PH=-log[H3O+]

strong base-strong acid

[H3O+]=[acid]

PH=-log[H3O+]

weak acid-strong base

[OH-]=[base]

Kw=[H3O+].[OH-]

PH=-log[H3O+]

weak base-strong acid

[H3O+]=[acid]

PH=-log[H3O+]

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