Acid-Base Equilibria (1) Definitions of acids and bases

Edexcel A level Chemistry (2017)

Topic 12: Acid–-Base Equilibria:

Here are the first learning objectives:

12/1. To know that a Brønsted–Lowry acid is a proton donor and a Brønsted–Lowry base is a proton acceptor

12/2. To know that acid-base reactions involve the transfer of protons

12/3. To be able to identify Brønsted–Lowry conjugate acid-base pairs

Definitions of acid and base:

Bronsted and Lowry defined acids as proton donors.

The definition applies to all solvents, not just aqueous systems.

So         HCl(g)      +      H₂O(l)      ⟶    H₃O⁺(aq)     +      Cl^—(aq)

When hydrogen chloride gas dissolves completely in water, all its protons (H+ ions) bond to water molecules to form a hydrated proton or oxonium ion. 

Bronsted defined acids broadly so that in the above equation the HCl is an acid in the forward direction but H₃O+ is an acid in the reverse direction. 

Bronsted in other words saw acids and bases as pairs, strictly speaking conjugate pairs. 

So     HCl(g)       +       H₂O(l)        ⟶    H₃O⁺(aq)   +        Cl^—(aq)

         Acid                    base           conjugate acid   conjugate base

The chloride ion is the conjugate base of the acid HCl.

The water molecule is the base of the conjugate acid H₃O+ which introduces an immediate conceptual leap in our understanding for who thinks of water as a base!!

Then again consider this aqueous system:

NH₃(aq)    +       H₂O(aq)     ⇌      NH₄⁺(aq)   +       OH^—  (aq)

Base                   acid                    conjugate acid   conjugate base

Here ammonia accepts a proton from a water molecule so ammonia is a base and that means that water acts here as an acid!!! (whereas in the previous example it acted as a base). 

What’s more the ammonium ion is also an acid but in the reverse direction. 

And as this is an equilibrium reaction not all the ammonia protons are donated to water so ammonia is a weak base.

Within this topic, students can consider how the historical development of theories explaining acid and base behaviour show that scientific ideas change as a result of new evidence and fresh thinking.

Historical background:

Back in the day it was Lavoisier (1777) (of discovering oxygen fame)who first attempted a theory about acids.

Lavoisier—I guess as you might expect since it might be argued he was in love with the element—suggested that all acids contain oxygen. 

When he discovered oxygen he named the gas oxygen because the name means “acid producer”.

It fell to Sir Humphrey Davy (of miners safety lamp fame) to put a spanner in Lavoisier’s theory and tell us all that “oxymuriatic acid” is nothing but hydrochloric acid and contains no oxygen.

Davy (1816) instead was the first to suggest that all acids contained hydrogen not oxygen.

The first true attempt at a theory of acidic action came with Arrhenius (yes he of the rate equation fame!!).

Arrhenius (1887) defined an acid as a compound that could produce hydrogen ions in aqueous solution. 

Conversely an alkali was a compound that could produce hydroxyl ions in aqueous solution.  

This is about where you are as a student when you study acids at GCSE or High School level. 

And you may extend Arrhenius’ definition to say that strong acids, like hydrochloric acid, produce many hydrogen ions in aqueous solution—they ionise or dissociate completely.

At GCSE or in High School you might have shown how this means that a solution of hydrochloric acid exhibits a very high conductivity.  

Bronsted (1923) as we discussed above, extended the Arrhenius definition to apply to other solvents besides water such as liquid ammonia NH_3.

In liquid ammonia there are NH₄⁺  ammonium ions and NH₂^— amide ions in an equilibrium:

NH₃   +    NH₃    ⇌     NH₄+    +   NH₂^—

This self–ionisation is analogous to that of water:

H₂O    +    H₂O      ⇌     H₃O⁺     +    OH^—

And in both cases the water molecule and the ammonia molecule is amphiprotic in that it can act as both an acid and a base accepting and/or donating protons. 

Bronsted said that acid–base behaviour involves the transfer of protons but in any solvent not just water.

Finally 1938 G.N. Lewis (of dot and cross diagram fame!!!) extended the definitions of acid and base even further.

Lewis defined an acid as an acceptor of a pair of electrons and a base as a donor of a pair of electrons. 

This broadens the definitions even further.

Not only is a hydrogen ion an acid but in this reaction below BF₃ is an acid:

H₃N:    +       BF₃     ⇌   H₃N⟶BF₃

base           acid

Since BF₃ is deficient of an electron pair in its outer shell is accepts the lone pair from the nitrogen atom. 

A dative covalent bond results between the two species.

Can you pick out then the acids and their conjugate bases in these equations?

HSO₄^—     +   H₂O   ⇌  H₃O+    +    SO₄ ^2—

CH₃COOH     +    H₂O     ⇌  H₃O+      +      CH₃COO^—

Equilibrium (2) The significance of the value of Kc and Kp

 Edexcel A level Chemistry (2017)

Topic 11: Equilibrium II:

Here is the third learning objective:

11/II/5. To be able to understand that the value of the equilibrium constant is not affected by changes in concentration or pressure or by the addition of a catalyst

Within this topic, students can consider how chemists can use the concept of equilibria to predict quantitatively the direction and extent of chemical change.

The equilibrium constant is related to both temperature and the enthalpy change of a chemical reaction:

There is linear relationship between K and 1/T provided ΔH^⦵ remains constant.  This is in fact the case to all intents and purposes, since ΔH^⦵ changes very slowly with temperature.

But concentration and pressure changes have no effect on K for one reason is that changes in pressure on a gaseous equilibrium are only effective if there is a difference in the number of moles of reactants and product as here:

N₂(g)    +     3H₂(g)   ⇌      2NH₃(g)

But if the reaction is this:

                             H₂(g)   +    I₂(g)   ⇌    2HI(g)

Then changes in pressure on this system have no effect on the position of equilibrium or therefore on K_p.

K_p remains constant if the pressure (or the concentration) changes. 

Addition of a catalyst is a different matter.

Catalysts lower the activation energy of a reaction and bring more molecules into a state where they might collide with others and break and make new bonds and form new products.

Catalysts are very inclusive.

But how do they work in equilibria?

Here is the typical energy profile of catalysed reaction:

The catalyst lowers the activation energy of the forward AND the backward reaction.

Therefore it cannot affect the position of equilibrium.

All it can do is reduce the time it takes for the reaction to reach equilibrium since the catalyst speeds up both forward and backward reactions.

Therefore a catalyst or pressure or concentration have no effect on the value of the equilibrium constant.

What the equilibrium constant does tell us is how far a reaction has gone.

From the value of an equilibrium constant we can tell if a reaction has gone to completion or not.

If

N₂O₄(g)    ⇌   2NO₂(g)  ΔH= +57.2kJmol^–1

At 600K, K_p = 13800 atm

This means that the reaction has virtually gone to completion.

The partial pressure of N₂O₄ must be very low and that of NO₂ very high for this value of K_p.

Similarly at 100K, K_p  =  3.61×  10^–21 atm which is incredibly small and means that in the equilibrium mixture there is mainly N₂O₄ and very little NO₂, in fact the decomposition of N₂O₄ has hardly begun.

So K the equilibrium constant is a measure of the extent of chemical change and therefore will be related to ΔG, the free energy change, ΔS_total the total entropy change and E^o the redox potential because all these thermodynamic quantities are also measures of the extent of reaction. 

Equilibrium (2) Effect of temperature change on Kp and Kc

Edexcel A level Chemistry (2017)

Topic 11: Equilibrium II:

Here is the third learning objective:

11/II/3. To know the effect of changing temperature on the equilibrium constant (Kc and Kp), for both exothermic and endothermic reactions

If    N₂O₄(g)    ⇌   2NO₂(g)  ΔH= +57.2kJmol^–1

Then the position of equilibrium moves left if the system is cooled.

Also if the system is warmed the position of equilibrium moves to the right forming more NO₂ and the colour of the gas mixture darkens.

The reaction left to right is endothermic so Le Chatelier’s principle would suggest that the reaction is moving in such away to evolve heat when the temperature is lowered and to absorb heat when the temperature is raised.

But what is the theoretical basis for this conclusion?

Experimental values show there is an approximate relationship between temperature and K_p.

The data in the table below reflects this relationship.

A plot of log₁₀K_p vs 1/T  gives a straight line as the graph shows.

If it is assumed (and this is where the relationship becomes approximate) that ΔH is constant over the range of temperatures used then the graph shows that:

The slope of the graph is – ΔH^⦵/2.303 R.

This equation confirms quantitatively Le Chatelier’s qualitative principle.

If ΔH^⦵   is positive the forward reaction is endothermic, then as temperature increases, 1/T decreases so Kp increases and so the position of equilibrium must be moving to the right to favour products.  As was said above when the temperature is raised the system absorbs heat and moves to favour products.

A summary is given in the table below:

ΔH⦵ of forward reaction	Temperature change applied	Effect on value of K	Change in equilibrium position
Endothermic	Increase	Increase	Favours products
Endothermic	Decrease	Decrease	Favours reactants
Exothermic	Increase	Decrease	Favours reactants
Exothermic	Decrease	Increase	Favours products