Chemistry IONIZATION OF ACIDS AND BASES, The Ionization Constant of Water and its Ionic Product and pH Scale.

### Topic to be covered

star Ionization of acids and bases.
star The Ionization Constant of Water and its Ionic Product.
star The pH Scale.

### IONIZATION OF ACIDS AND BASES

=> Arrhenius concept of acids and bases becomes useful in case of ionization of acids and bases as mostly ionizations in chemical and biological systems occur in aqueous medium.

=> Strong acids like perchloric acid color{red}((HClO_4)), hydrochloric acid color{red}((HCl)), hydrobromic acid color{red}((HBr)), hyrdoiodic acid color{red}((HI)), nitric acid color{red}((HNO_3)) and sulphuric acid color{red}((H_2SO_4)) are termed strong because :

•They are almost completely dissociated into their constituent ions in an aqueous medium, thereby acting as proton (H+) donors.

=> Similarly, bases like lithium hydroxide color{red}((LiOH)), sodium hydroxide color{red}((NaOH)), potassium hydroxide color{red}((KOH)), caesium hydroxide color{red}((CsOH)) and barium hydroxide color{red}(Ba(OH)_2) are strong because:

•They almost get completely dissociated into ions in an aqueous medium giving hydroxyl ions, color{red}(OH^–).

=> According to Arrhenius concept they are strong acids and bases as they are able to completely dissociate and produce color{red}(H_3O^+) and color{red}(OH^–) ions respectively in the medium. Alternatively, the strength of an acid or base may also be gauged in terms of Brönsted- Lowry concept of acids and bases, wherein a strong acid means a good proton donor and a strong base implies a good proton acceptor. Consider, the acid-base dissociation equilibrium of a weak acid color{red}(HA),

color{red}(undersettext(acid)(HA (aq)) + undersettext(base)(H_2O(l)) ⇌ undersettext(conjugate acid)(H_3 O^+ (aq)) +undersettext(conjugate base)(A^- (aq)))

•If color{red}(HA) is a stronger acid than color{red}(H_3O^+), then color{red}(HA) will donate protons and not color{red}(H_3O^+), and the solution will mainly contain color{red}(A^–) and color{red}(H_3O^+) ions. The equilibrium moves in the direction of formation of weaker acid and weaker base because the stronger acid donates a proton to the stronger base.

=> It follows that as a strong acid dissociates completely in water, the resulting base formed would be very weak i.e., strong acids have very weak conjugate bases.

=> Strong acids like perchloric acid color{red}((HClO_4)), hydrochloric acid color{red}((HCl)), hydrobromic acid color{red}((HBr)), hydroiodic acid color{red}((HI)), nitric acid color{red}((HNO_3)) and sulphuric acid color{red}((H_2SO_4)) will give conjugate base ions color{red}(ClO_4^–, Cl, Br^(–), I^(–), NO_3^(–)) and color{red}(HSO_4^(–)) , which are much weaker bases than color{red}(H_2O).

=> Similarly a very strong base would give a very weak conjugate acid.

=> On the other hand, a weak acid say color{red}(HA) is only partially dissociated in aqueous medium and thus, the solution mainly contains undissociated color{red}(HA) molecules.

=> Typical weak acids are nitrous acid color{red}((HNO_2)), hydrofluoric acid color{red}((HF)) and acetic acid color{red}((CH_3COOH)). It should be noted that the weak acids have very strong conjugate bases. For example, color{red}(NH_2^(–), O^(2–)) and color{red}(H^–) are very good proton acceptors and thus, much stronger bases than color{red}(H_2O).

Certain water soluble organic compounds like phenolphthalein and bromothymol blue behave as weak acids and exhibit different colours in their acid color{red}((HI n)) and conjugate base color{red}((In– )) forms.

color{red}(undersettext(acid indicator colour A)(HIn(aq)) +H_2O (l) ⇌ undersettext(conjugate acid)(H_3O^+(aq))+undersettext(conjugate base colourB)( In^(-)(aq)))

Such compounds are useful as indicators in acid-base titrations, and finding out color{red}(H^+) ion concentration.

### The Ionization Constant of Water and its Ionic Product

=> Some substances like water are unique in their ability of acting both as an acid and a base.
•In presence of an acid, color{red}(HA) it accepts a proton and acts as the base.
• While in the presence of a base, color{red}(B^–) it acts as an acid by donating a proton. In pure water, one color{red}(H_2O) molecule donates proton and acts as an acid and another water molecules accepts a proton and acts as a base at the same time. The following equilibrium exists:

color{red}(undersettext(acid)(H_2O(l)) +undersettext(base)(H_2O(l)) ⇌ undersettext(conjugate acid)(H_3O^+(aq))+ undersettext(conjugate base)(OH^(-)(aq)))

The dissociation constant is represented by,

color{red}(K = [H_3O^+ ] [OH^- ] // [H_2O]) .............(7.26)

The concentration of water is omitted from the denominator as water is a pure liquid and its concentration remains constant. color{red}([H_2O]) is incorporated within the equilibrium constant to give a new constant, color{red}(K_w), which is called the ionic product of water.

color{red}(K_w = [H^+ ] [OH^-]) ...............(7.27)

The concentration of color{red}(H^+) has been found out experimentally as color{red}(1.0 × 10^(–7) M) at color{red}(298 K). And, as dissociation of water produces equal number of color{red}(H^+) and color{red}(OH^–) ions, the concentration of hydroxyl ions, color{red}([OH^–] = [H^+] = 1.0 × 10^(–7) M). Thus, the value of color{red}(K_w) at color{red}(298K),

color{red}(K_w = [ H_3 O^+] [OH^-] = (1xx10^(-7))^2 = 1xx10^(-14) M^2) ..............(7.28)

The value of color{red}(K_w) is temperature dependent as it is an equilibrium constant.

The density of pure water is color{red}("1000 g / L") and its molar mass is color{red}("18.0 g /mol"). From this the molarity of pure water can be given as,

color{red}([H_2O] = (1000 g // L)(1 mol//18.0 g) = 55.55 M.)

Therefore, the ratio of dissociated water to that of undissociated water can be given as:

color{red}(10^(-7) // (55.55) = 1.8xx10^(-9))  or color{red}(2) in color{red}(10^(-9)) (thus, equilibrium lies mainly towards undissociated water)

We can distinguish acidic, neutral and basic aqueous solutions by the relative values of the color{red}(H_3O^+) and color{red}(OH^–) concentrations:

color{green}("Acidic"): color{red}([H_3O^+] > [OH^–])
color{green}("Neutral"): color{red}([H_3O^+] = [OH^– ])
color{green}("Basic") : color{red}([H_3O^+] < [OH^–])

### The pH Scale

Hydronium ion concentration in molarity is more conveniently expressed on a logarithmic scale known as the color{red}(pH) scale.

color{purple}(✓✓)color{purple} " DEFINITION ALERT"
The color{red}(pH) of a solution is defined as the negative logarithm to base 10 of the activity color{red}(a_(H^(+)) of hydrogen ion.

In dilute solutions color{red}(< 0.01 M), activity of hydrogen ion color{red}((H^+)) is equal in magnitude to molarity represented by color{red}([H^+]). It should be noted that activity has no units and is defined as:

color{red}(a_(H^(+)) = [H^+]// mol L^(-1))

From the definition of color{red}(pH), the following can be written,

color{red}(pH = -loga_(H^+) = - log { [H^+] // mol L^(-1) })

Thus, an acidic solution of color{red}(HCl (10^(–2) M)) will have a color{red}(pH = 2). Similarly, a basic solution of color{red}(NaOH) having color{red}[OH^– ] =10^(–4) M) and color{red}([H_3O^+] = 10^(–10) M) will have a color{red}(pH = 10).
At 25 °C, pure water has a concentration of hydrogen ions, color{red}([H^+] = 10^(–7) M). Hence, the color{red}(pH) of pure water is given as:

color{red}(pH = - log (10^(-7) ) = 7)

=> Acidic solutions possess a concentration of hydrogen ions, color{red}([H^+] > 10^(–7) M), while basic solutions possess a concentration of hydrogen ions, color{red}([H^+] < 10^(–7) M). thus, we can summarise that Acidic solution has color{red}(pH < 7) Basic solution has color{red}(pH > 7) Neutral solution has color{red}(pH = 7)

Now again, consider the equation (7.28) at 298 K

color{red}(K_w = [H_3 O^+] [OH^-] = 10^(-14))

Taking negative logarithm on both sides of equation, we obtain

color{red}(-log K_w = - log { [H_3O^+] [OH^-]})

 color{red}(= - log [ H_3O^+] - log [OH^- ])

 color{red}(= - log 10^(-14))

color{red}(pK_w = pH +pOH = 14) .......(7.29)

Note that although color{red}(K_w) may change with temperature the variations in color{red}(pH) with temperature are so small that we often
ignore it.

color{red}(pK_w)  is a very important quantity for aqueous solutions and controls the relative concentrations of hydrogen and hydroxyl ions as their product is a constant. It should be noted that as the color{red}(pH) scale is logarithmic, a change in color{red}(pH) by just one unit also means change in color{red}([H^+]) by a factor of 10. Similarly, when the hydrogen ion concentration, color{red}([H^+]) changes by a factor of 100, the value of color{red}(pH) changes by 2 units. Now you can realise why the change in color{red}(pH) with temperature is often ignored.

Now-a-days color{red}(pH) paper is available with four strips on it. The different strips have different colours (Fig. 7.11) at the same color{red}(pH). The color{red}(pH) in the range of 1-14 can be determined with an accuracy of ~0.5 using color{red}(pH) paper.

For greater accuracy color{red}(pH) meters are used. color{red}(pH) meter is a device that measures the color{red}(pH)-dependent electrical potential of the test solution within 0.001 precision. color{red}(pH) meters of the size of a writing pen are now available in the market. The color{red}(pH) of some very common substances are given in Table 7.5 .
Q 3100101018

The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10^(–3)M. what is its pH ?

Solution:

pH = – log[3.8 × 10^(–3) ]
= – {log[3.8] + log[10^(–3)]}
= – {(0.58) + (– 3.0)} = – { – 2.42} = 2.42
Therefore, the pH of the soft drink is 2.42 and it can be inferred that it is acidic.
Q 3170201116

Calculate pH of a 1.0 × 10^(–8)M solution of HCl.

Solution:

2H_2O (l) ⇌ H_3O^+ (aq) + OH^(–)(aq)

K_w = [OH^–][H_3O^+] = 10^(–14)
Let, x = [OH^–] = [H_3O^+] from H_2O. The H_3O^+

concentration is generated (i) from the ionization of HCl dissolved i.e.,
HCl(aq) + H_2O(l) ⇌ H_3O+ (aq) + Cl^–(aq), and (ii) from ionization of H_2O. In these very dilute solutions, both sources of
H_3O^+ must be considered:
[H_3O^+] = 10^(–8) + x
K_w = (10^(–8) + x)(x) = 10^(–14)
or x^2 + 10^(–8) x – 10^(–14) = 0
[OH^– ] = x = 9.5 × 10^(–8)
So, pOH = 7.02 and pH = 6.98