Chemistry pH SCALE
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pH SCALE

pH of Acids and Base :

Many properties of aqeous solution depend on the concentration of `H^(+)` ions of the solutions and therefore there is a need to express these concentrations in simple terms. For this purpose we introduce the concept of `pH`.

`pH = - log a_H^(+)` (where `a_H^(+)` is the activity of `H^(+)` ions).

Activity of `H^(+)` ions is the concentration of free `H^(+)` ions in a solution. By free we mean those that are at a large distance from the other ion so as not to experience its pull. We can infer from this that in dilute solution, the activity of an ion is same as its molar concentration since more number of solvent molecules would separate the two ions. For concentrated solution the activity would be much less than the concentration itself.

Therefore, the earlier given expression of `pH` can be modified for dilute solution as `pH = - log [H^(+)]`. This assumption can only be made when the solution is very much dilute, i.e., `[H^(+)] < 1M`. For higher concentration of `H^(+)` ions, one needs to calculate the activity experimentally and then calculate the `pH`.

`pH` Calculation for Strong Acids :

Let us now see how to calculate the `pH` of a solution of a strong acid in water (It should be noted that `pH` calculations are only rnade for aqueous solutions). Let the strong acid be `HCl`. If we take `10^(-1) M HCl,` the `[H^(+)]` would be `10^(-1) M` as `HCl` is a strong acid and would dissociate completely. Therefore the `pH` would be,

`pH = - log 10^(-1) = 1 `

See Table.

We can see that for `10^(-7) M` of `HCl` we have some hesitation in talking about the `pH`. This is because if we use our expression of `pH`, it works out to be `7` which is somehow associated with neutrality. We shall now explain how to calculate the `pH` of `10^(-7) M HCl.` Before we do this we shall discuss the dissociation of water.

When we add `10^(-7) M HCl` in water, the `[H^(+)]` from `HCl` would be `10^(-7) M.` But `pH` is the negative logarithm of the total `H^(+)` ion concentration of a solution and not that part which comes from only `HCl` (you might wonder why we have changed our stand. This will become clear in a short while). Therefore, we need to add the concentration of `H^+` ions coming from water also. This can be done as

In presence of `HCl`, the dissociation of water will be suppressed & hence the concentration of `H^(+)` would be less than `10^(-7) M`.

Let the conc. of `H^(+)` due to the dissociation of water be `x`.

`H_2O ⇋ H^(+) + OH^(-)`

At equilibrium: `x` `+` `10^(- 7)` `x`

`[H^(+)] [OH^(-)] = 10^(-14)`

`(x+ 10^(- 7)) (x) = 10^(-14)`

Calculating for `x` we get `x = 0.618 xx 10^(- 7)M`.

Therefore, `[H^(+)]_(text(Total)) = 10^(-7) + 0.618 xx 10^(-7)`

`= 1.618 xx 10^(-7)M`

`pH = - log 1.618 xx 10^(-7) = 6.7910`

Now, lets answer the question as to why we take the `H^(+)` ions of water into account for calculating the pH of `10^(-7) M` `HCl` while we never considered it for calculating the pH of `10^(- 6)` `M`, `10^(-5)` `M`, `10^(- 4)` `M`, `10^(-3)` `M`, `10^(-2)` `M` and `10^(-1)` `M` `HCl`. It can be seen that the `H^+` ions from water has decreased due to the common ion effect. Greater the concentration of the common ion added, greater will be the effect. Therefore for concentrations higher than `10^(- 6)M` (and inclusive of `10^(-6)M`, the `H^(+)` from water will be even less than `0.618 xx 10^(- 7) M` and would be so small in comparison to the [`H^+`] from `HCl`, that we can ignore the contribution from it. So, finally we conclude that `H^(+)` ions of water needs to be considered only if [`H^(+)`] acid `< 10^(- 6)M`.

pH calculation for mixture of two Weak acid

Let us have two weak monoprotic acids as HA and HB and their concentrations are `C_1` and `C_2` M respectively. Let their ionization constants are `K_(alpha_1)` and `k_(alpha_2)` , respectively. The degree of dissociation of each would be affected in presence of other due to common ion effect. Let `alpha_1` be the degree of dissociation of HA in presence of HB and `alpha_2` be the degree of dissociation of HB in presence of HA. In an aqueous solution of a mixture of HA and HB, following equilibria exist.

`tt((HA, ⇋ ,H^(+) , + A^(-)),(c_1(1-alpha_1) , quad , (c_1alpha_1 + c_2alpha_2), c_1alpha_1))`

`tt((HB, ⇋ ,H^(+) , + A^(-)),(c_2(1-alpha_2) , quad , (c_2alpha_2 + c_1alpha_1), c_2alpha_2))`

`k_(alpha_1) = ([H^(+)][A^(-)])/[[HA]] = ((c_1alpha_1+c_2alpha_2)c_1alpha_1)/(c_1(1-alpha_1)) = ((c_1alpha_1+c_2alpha_2)alpha_1)/(1-alpha_1)..................(i)`

and

`k_(alpha_2) = ([H^(+)][B^(-)])/[[HB]] = ((c_1alpha_1+c_2alpha_2)c_2alpha_2)/(c_2(1-alpha_2)) = ((c_1alpha_1+c_2alpha_2)alpha_2)/(1-alpha_2)..................(ii)`

Knowing the values of `K_(alpha_1)`, `K_(alpha_2)` , `c_1` and `c_2` the values of `alpha_1` and `alpha_2` can be calculated from the above equations.

The `[H^(+)]` can then be calculated as

`[H^(+)]_T = c_1alpha_1 + c_2alpha_2`

Finally, if the total `[H^(+)]` from acid is less than `10^(-6) M,` the contribution of `H^(+)` from water should be taken into account for calculating pH while if it is `10^(-6) M,` then `[H^(+)]` contribution from water can be ignored.

Using this `[H^(+)]`, `pH` of the solution can be calculated.

`text(Approximation:)`

If `alpha_1` and `alpha_2` are very small as compared to unity then

`1- alpha_1 > > 1 & 1 - alpha_2 > > 1`

Therefore, `[ H^(+) ]` can be calculated from the expression

`[H^(+)] = sqrt(k_(alpha_1)c_1 +k_(alpha_2)c_2)`
 
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