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`

 
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