14.3 Relative Strength of Acids & Bases

Acids and Bases strengths are related to the dissociation or ionization of these acids and bases in water (aqueous solution). A strong acid and strong base will dissociate or ionize completely in water while a weak acid and weak base will dissociate or ionize partially.

The table below shows some strong acids and bases:

Reference: https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/15%3A_Acid%E2%80%93Base_Equilibria/15.3%3A_Acid_and_Base_Strength

Strong Acids and Bases Dissociation:

HCl(aq)      +      H₂O( (l)       H₃O ⁺ (aq)     +    Cl ^– (aq)                               complete dissociation

NaOH(aq)       +     H₂O(l)     Na ⁺(aq)    +   OH ^– (aq)   +   H₂O(l)                  complete dissociation

Weak Acids and Bases Dissociation:

H₃PO₄(aq)   +  3 H₂O(l)        3 H₃O ⁺ (aq)    +   PO₄ ^3- (aq)                  partial dissociation  

NH₃(aq)      +     H₂O(l)        NH₄ ⁺ (aq)    +     OH ^–(aq)                      partial dissociation

Note that strong acids and bases dissociation in water is recognized with one single arrow and the weak acids and bases dissociation in water is recognized with double arrows.

The strength of the strong and weak acids and bases are determined by measuring their equilibrium constant in aqueous solutions.

Example:

HA(aq)     +   H₂O( (l)       H₃O ⁺ (aq)     +    A ^– (aq)    

Ka =  { [ H₃O ⁺ (aq) ]  x [ A ^– (aq) ] }  /   [ HA(aq) ] 

Examples:

CH₃COOH(aq)   +  H₂O( (l)       H₃O ⁺ (aq)  +  CH₃COO ^– (aq)        Ka  = 1.8 x 10 ^{– 5}

HNO₂(aq)    +  H₂O( (l)       H₃O ⁺ (aq)  +  NO₂ ^– (aq)        Ka  = 4.6 x 10 ^{– 4}

HSO₄ ^–(aq)   +  H₂O( (l)        H₃O ⁺ (aq)  +  SO₄  ^2– (aq)        Ka  = 1.2 x 10 ^{– 2}

Ka of HSO₄ ^–(aq) > Ka of HNO₂(aq) > Ka of CH₃COOH(aq)

The Percent Ionization is another tool is used to express the strengths of the acids and bases:

Let us consider the acid dissociation:

HA(aq)     +   H₂O( (l)        H₃O ⁺ (aq)     +    A ^– (aq)    

Percent Ionization %  =  { [ H₃O ⁺ (aq) ] / [ HA(aq) ]_o } x 100%

[ HA(aq) ]_o  = Initial Acid Concentration

Example:

Calculate the % ionization of an acetic acid CH₃COOH That has the molarity of 5.50 M and a pH of 4.5

Percent Acid Ionization %  =  { [ H₃O ⁺ (aq) ] / [ HA(aq) ]_o } x 100%

First one has to calculate the [ H₃O ⁺ (aq) ]:

[ H₃O ⁺ (aq) ] = 10 ^{– pH} = 10 ^{– 4.5} = 3.2 x 10  ^-5 M

[ HA(aq) ]_o = 5.50 M

Percent Ionization %   = { [ 3.2 x 10  ^-5 ]  /  [ 5.50 ] } x 100% = 5.75 x 10  ^{– 4} %

The strengths of the bases are measured by measuring their equilibrium constant Kb. Kb is also called Base Dissociation or Ionization constant. Strong bases dissociate completely 100% or near 100%. Weak bases dissociate partially (less than 100%).

Examples:

NO₂  ^–  (aq)   +     H₂O  (l)  HNO₂ (aq)    +   OH ⁻ (aq)                Kb  = 2.17 x 10 ⁻¹¹

CH₃COO ^–  (aq)  +  H₂O (l)  CH₃CO_OH (aq)    +    OH⁻ (aq)        Kb = 5 .6 x 10 ⁻¹⁰

NH₃ (aq)  +  H₂O (l) NH₄ ⁺  (aq)   +   OH−(aq)                           Kb = 1.8 x 10 ^{– 5}

Kb of NH₃ (aq) > Kb of CH₃COO ^–  (aq) > Kb of NO₂  ^–  (aq)

The Percent Base Dissociation or Ionization is a tool to express the relative strengths of the bases as well.

Let us consider the base dissociation:

B (aq)     +   H₂O( (l)        BH ⁺ (aq)     +    OH ^– (aq)    

Kb = { [ BH ⁺ (aq) ] x [ OH ^– (aq) ] }  /  [ B (aq) ]

Percent Base Ionization %  =  { [ OH ^– (aq) ] / [ B (aq) ]_o } x 100%

It can be seen from both Percent Ionization %’s that the acid percent ionization is based on the concentration of H₃O ⁺ (aq) and the base ionization is based on the concentration of OH  ^– (aq).

Relative Strengths of Conjugate Acid  –  Base Pairs:

Let us consider the two reactions:

HA(aq)     +   H₂O( (l)       H₃O ⁺ (aq)     +    A ^– (aq)     Ka =  { [ H₃O ⁺ (aq) ] x [ A ^– (aq) ] } / [ HA (aq) ]

A ^– (aq)    +   H₂O( (l)     HA (aq)  +  OH ^– (aq)              Kb =  { [ HA (aq) ] x [ OH ^– (aq) ] } / [ A ^– (aq) ]

Adding both reactions:

HA (aq)    +   H₂O( (l) + A ^– (aq)    +   H₂O( (l)  H₃O ⁺ (aq)     +    A ^– (aq)   +  HA (aq)  +  OH ^– (aq)

The Net Ionic Chemical Equation is:

2 H₂O( (l)     H₃O ⁺ (aq)  +  OH ^– (aq)

Kw = { [ H₃O ⁺ (aq) ] x [ OH ^– (aq) ]  }. Note that the H₂O( (l) concentration is not included.

The above Kw expression can be obtained by multiplying Ka with Kb

Ka =  { [ H₃O ⁺ (aq) ] x [ A ^– (aq) ] } / [ HA (aq) ]

Kb =  { [ HA (aq) ] x [ OH ^– (aq) ] } / [ A ^– (aq) ]

Ka x Kb =  {{ [ H₃O ⁺ (aq) ] x [ A ^– (aq) ] } / [ HA (aq) ] } x {{ [HA (aq) ] x [ OH ^– (aq) ] } / [ A ^– (aq) ] }

Ka x Kb = { [ H₃O ⁺ (aq) ] x [ OH ^– (aq) ]  } = Kw

Considering the log of both sides:

(-Log Ka) + (-Log Kb)  = (-Log Kw)

 p Ka + p  kb = ( – log 10 ^-14) = 14

From the above expression, one can deduce other sub formulas:

Ka  =  Kw / Kb

Kb = Kw  /  Ka

The diagram illustrates the relative acids strength and  the strengths of their corresponding conjugate pairs.

Note that the inverse relationship between Ka and Kb expresses the relative strengths of the acids and the bases:

The stronger the acid or base, the weaker its conjugate pair.

Reference: https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/15%3A_Acid%E2%80%93Base_Equilibria/15.3%3A_Acid_and_Base_Strength

In the above table, the conjugate acid – base pairs shown is arranged to express the relative strength of each conjugate acid – base species as compared with water,

The conjugate acid – base species shown below water are weaker acids than water and hence do not undergo acid ionization in water. The conjugate acid – base species above water are stronger acids than water

Below hydronium ion are weak acids which are undergoing partial acid ionization. Above hydronium ion are strong acids which are undergoing complete ionization in aqueous solution. This can be seen at the acid column.

This is quite different when looking at the base column.

The conjugate acid – base species shown below water are strong bases than water and hence do undergo partial base ionization in water. The conjugate acid – base species above water are weaker bases than water and do not undergo any ionization.

Below hydroxide ion are strong bases which are undergoing complete base ionization. Above hydroxide ion are weaker bases which are undergoing partial ionization in aqueous solution.

The videos below illustrate the concepts of the conjugate acid – base pairs:

A simulated video explains the concept of conjugate acid base pair in a very simple way:

Calculation of Ionization Constants of Conjugate Acid – Base pairs:

Example:

A 0.500 M solution of formic acid is prepared and its pH is measured to be 2.04. Determine the K_a for formic acid.

Given:

initial [HCOOH] = 0.500 M

pH = 2.04

Asked:

           Ka = ?

First: we have to set up the equilibrium equation for the dissociation/ionization of formic acid:

HCOOH (aq)    +    H₂O(l)   HCOO ^– (aq)   +   H₃O ⁺ (aq)

Second: we have to write the expression of Ka and calculate the H₃O ⁺ (aq) concentration:

Ka = { [HCOO ^– (aq)]   *  [H₃O ⁺ (aq)] } /  { [HCOOH] }

[H₃O ⁺ (aq)] =  10  ^{– pH}  = 10 ^– 2.04 = 9.12 × 10 ^{– 3}

Third: we have to use the Table of ICE (Initial, Change and Equilibrium):

HCOOH (aq)    +    H₂O(l)   HCOO ^– (aq)     +     H₃O ⁺ (aq)

Name	Formic Acid	H30+	Formate A–
Initial	.5 M	0	0
Change	– 9.12 * 10-3	9.12 * 10-3	9.12 * 10-3
Equilibrium	.491	9.12 * 10-3	9.12 * 10-3

Ka = { [HCOO ^– (aq)]   *  [H₃O ⁺ (aq)] } /  { [HCOOH] }

Ka = { [9.12 × 10 ^{– 3}] * [9.12 × 10 ^{– 3}] } / { [0.491] } = 1.70 x 10  ^{– 4}

Calculation of Acids Ionization Constants from the Equilibrium Concentrations:

The HSO4 ^– ion, weak acid used in some household cleansers:

HSO₄ − (aq) + H₂O (l) ⇌ H₃O + (aq) + SO₄ ^2- (aq)

What is the acid ionization constant for this weak acid if an equilibrium mixture has the following data?

[H₃O +] = 0.027 M

[HSO₄ ^–] = 0.29 M

[SO₄ ²⁻] = 0.13 M

Ka = { [H₃O + (aq)] * [SO₄ ^2- (aq)] } /  { [HSO₄ − (aq)} }

No need to use the ICE table since we have the equilibrium concentrations already:

Ka =  { [0.13 M] * [0.027 M] } /  [0.29 M] = 1.2 x 10 ^-2

Calculation of Bases Ionization Constants from the Equilibrium Concentrations:

Example:

The pH of a 0.1000 M basic solution of aqueous ammonia is be 11.63.  Determine the value of Kb, the ionization constant for this solution.  The equilibrium concentration of NH₄ ⁺ (aq) is 0.0500 M.

Given:

initial [NH₃] = 0.1000 M

pH = 11.63

Asked:

           Kb = ?

First: we have to set up the equilibrium equation for the dissociation/ionization of formic acid:

NH₃ (aq)    +    H₂O(l)   NH₄ ⁺ (aq)   +   OH ^– (aq)

Second: we have to write the expression of Kb and calculate the H₃O ⁺ (aq) concentration:

Kb = { [NH₄ ⁺ (aq)]   *  [OH ^– (aq)]  /  { [NH₃ (aq)] }

pH = 11.63,  

pOH = 14.00 – 11.63 = 2.37

[OH ^– (aq)]  =  10  ^{–  2.37}  = 0.00427

Kb = { [NH₄ ⁺ (aq)]   *  [OH ^– (aq)]  /  { [NH₃ (aq)] }

Kb =  [{ 0.0500 M] * [0.00427 M] }  /  [0.1000 M] =  0.00505995 = 5.05 x 10 ^{– 3}

Calculation of Acids Ionization Constants Ka or Kb from pH

Example:

Calculate the Ka value of a 0.200 M aqueous solution of propionic acid (CH₃CH₂COOH) with a pH of 4.88.

CH₃CH₂COOH (aq)    +    H₂O (l)    H₃O ⁺ (aq)   +   CH₃CH₂COO ^– (aq)

Ka =  { [H₃O ⁺ (aq)]  *   [CH₃CH₂COO ^– (aq)] }  /  { [CH₃CH₂COOH (aq)] }

pH = 4.88,

[H₃O ⁺ (aq)]  = 10  ^{– pH} = 10  ^{– 4.88}   = 1.32×10 ⁻⁵

Using ICE table:

CH3CH2COOH (aq)   +   H2O (l) H3O + (aq)     +       CH3CH2COO – (aq)
CH3CH2COOH (aq)     +   H2O(l)                H3O + (aq)             +       CH3CH2COO – (aq)
Initial:	0.200 M                      0	0	0
Change:	– 1.32×10 −5	+1.32×10 −5	+1.32×10 −5
Equilibrium:	0.1999868 = (0.200)	+1.32×10 −5	+1.32×10 −5

Ka =  { [H₃O ⁺ (aq)]  *   [CH₃CH₂COO ^– (aq)] }  /  { [CH₃CH₂COOH (aq)] }

Ka = { [1.32×10 ⁻⁵] * [1.32×10 ⁻⁵] } / { [0.1999868] } = 8.69 x 10 ^{– 10}

Calculating Equilibrium Concentrations in a Weak Acid Solution Using Ka

Example

Calculate the equilibrium concentration and the pH of a weak acid solution of 0.2 M HOBr, given:

HOBr (aq)  + H₂O (l)   H₃O ⁺ (aq)  +   OBr ^– (aq)

Ka = 2 × 10 ^{– 9}

The ICE Table

HOBr (aq)  + H2O (l)   H3O + (aq)       +     OBr – (aq)
Initial:	0.2 M	0	0
Change:	– X	+X	+X
Equilibrium:	0.2 – X	+X	+X

Ka =  { [H₃O ⁺ (aq)] * [OBr ^– (aq)] } /  { [HOBr (aq)} } = 2 × 10 ^{– 9}

Ka = [X] * [X]    /   [0.2 – M]   = 2 × 10 ^{– 9}

X ²  /  [0.2 – X]   = = 2 × 10 ^{– 9}

Solving for X:

X  ²   +     (2 X 10 ⁻⁹) X  −   (4 × 10  ⁻¹⁰)   =   0

Using the Quadratic Equation:

a = 1

b = 2 × 10  ^{– 9}

c = −4 ×1 0 ⁻¹⁰

X =

X = 2.0 x 10 ^{– 5}   =  [H₃O + (aq)]

pH = – log [H3O + (aq)]  =  – log [ 2.0 x 10 ^{– 5}  ]   =  4.69

Using the Simplifying Assumption Method:

Assume that 2.0 M is much greater than X

1. M >>>> X

Then Ka = [X] * [X]    /   [0.2 – M]   = 2 × 10 ^{– 9}

Ka  =  X  ²  /  [0.2]   = 2 × 10 ^{– 9}

X ² =  0.4 × 10 ^{– 9} = 4 x 10 ^{– 10}

X = √ 4 x 10 ^{– 10}  = 2 x 10  ^{–  5}

Bond Strengths of Binary Acids

The bond strengths of binary acids depend on the structure of these acids. In general, the bond strengths of binary acids HA’s or BH ⁺ decrease within the periodic table for the halides from top F (fluoridone) to the bottom I (iodine).  HF is considered as the weakest acid while HI is considered as strongest acid.

The table illustrates this trend:

Reference: https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/16%3A_Acids_and_Bases/16.8%3A_Molecular_Structure_and_Acid-Base_Behavior

The figure below and the video explain this trend:

https://www.khanacademy.org/science/ap-chemistry/acids-and-bases-ap/acids-bases-and-ph-ap/v/acid-strength-anion-size-and-bond-energy

Ternary Acids Strengths

A ternary acid is an acid made of 3 elements: hydrogen, oxygen and a third element which is a none metal that is considered as a central atom.  The central atom is the least electronegative atom.

Example of a ternary acid is a carboxylic acid below:

https://courses.lumenlearning.com/boundless-chemistry/chapter/acid-strength-and-molecular-structure/

Another of example of a ternary acid is nitric acid:

The general molecular formula of ternary compound is O_mE(OH)_n. This general formula covers acidic, basic and amphoteric compounds depending on the electronegativity and the property of the central atom E.

Examples are give above in the carboxylic acid and the nitric acid. Also basic compounds such barium hydroxide and sodium hydroxide can be considered as ternary compounds.

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In general, the lower the electronegativity of the central atom E, then the least the attraction to the valence electrons of the terminal atoms can observed. The bond between the oxygen and the least central atom E leads to a weaker covalent bond and attraction and the bond broken and a hydroxide ion is produced. This is an ionic bonding and the ternary compound is a base. The breakage of the bonding on Bond a as can be seen from the figure above

On the other hand, the higher the electronegativity of the central atom E, then the stronger the attraction to the valence electrons of the terminal atoms can observed. The bond between the oxygen and the central atom E will be stronger covalent on the expense of the weak attraction between the oxygen atom and the terminal hydrogen atom leading to the release of hydrogen and the solution becomes covalent. Bond b becomes more polar. This is a ternary bonding and the ternary compound is an acid. The breakage of the bonding on Bond b as can be seen from the figure above as well.

Similar trend can be seen in the Oxyacids.  The higher the oxidation number on the central atom the stronger the acid (the higher the acidity).

The higher the oxidation number leads to higher attraction of the central atom E to electrons it shares with oxygen and hence the oxygen bonding is weekend with the hydrogen a=which facilitates the dissociation and the release of the hydrogen of the Oxyacids into the solution.

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