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Concept of PH

In order to cover a wide variation of H3O+ in aqueous solution, Sorenson, in 1909, introduced a logarithmic scale for the sake of convenience and gave it a symbol pH. It is expressed as
pH = - log {[H3O+]/mol dm-3} ....
that is, pH of a solution is the negative of the logarithm to base 10 of [H3O+]/mol dm-3 or it is equal to the logarithm of reciprocal of [H3O+]/mol dm-3. Note that pH is simply defined in terms of a mathematical expression. Given the value of [H3O+], one can calculate pH using Eq. (9.15). Its value may be negative, zero or positive depending upon the provided value of [H3O+]. For example, pH of a solution containing 2mol dm-3 of H3O+ will have a pH of - 0.301 (= - log 2). However, the concentrations of H3O+ commonly met within solutions vary over the range 10-14 to 1 mol dm-3. The pH of such solutions will lie in range of 0 to 14.

Pure water at 25 ° C has hydronium ion concentration of 10-7 mol dm-3 and hence its pH will be

pH = -log (10-7) = 7
For an acidic solution, [H3O+] > 10-7 mol dm-3 and thus its pH will be less than 7 (e.g. pH of 10-5 M HCl solution is 5). On the other hand, an alkaline solution has [H3O+] < 10-7 mol dm-3 and hence its pH will be more than 7 (e.g. pH of 10-5 M NaOH is 9 at 25 ° C as it contains 10-9 M hydronium ion concentration*).

In a similar manner, we can define a pOH scale as the negative of the logarithm of the [OH-]/mol dm-3. However, the acidity or alkalinity of an aqueous solution is often expressed in terms of pH of a solution. Both pH and pOH are related to each other through an expression derived as follows.

[H3O+] [OH-] = Kw
or {[H3O+]/mol dm-3} {[OH-]/mol dm-3} = Kw/(mol dm-3)2
taking the logarithm of the above expression and multiplying the resultant expression by -1, 
we get
-log {[H3O+]/mol dm-3} - log {[OH-]/mol dm-3}
= - log {Kw/mol dm-3)2}
or pH + pOH = pKw
where pKw, like pH and pOH, is equal to - log {Kw/(mol dm-3)2}.

In general for an acidic solution (say, HCl), there are two sources of H3O+, namely, (1) from acid, and (2) from water. For a solution of molarity 10-5 M or more (say, 10-4 M or 10-3 M, etc.), the contribution of H3O+ from water may be taken to be negligible in comparison to that coming from acid. For very dilute solutions (molarity £ 10-6 M), such a contribution is not negligible. In order to compute pH of such solutions, the concentration of H3O+ coming from both the sources should be taken into account. Regardless of how much an acidic solution is diluted, its pH will always be less than seven. For example, pH of 10-7 M HCl at 25 ° C is 6.79. Similarly, pH of alkaline solution will always be greater than seven however dilute the alkaline solution is made.

Note that pH = 7 for a pure water holds good only at 25 ° C, since at this temperature water ionizes to produce 10-7 M of H3O+. The extent of ionization of water varies with temperature; it increases with increase in temperature. Thus, the concentration of H3O+ present in pure water increases with increase in temperature and hence its pH decreases with increase in temperature. For example, we have


0 ° C

10 ° C

25 ° C

40 ° C

50 ° C







In order to predict the acidic or alkaline nature of an aqueous solution at a given temperature, we must compare its pH value with that of pure water at that particular temperature. For example, an aqueous solution of pH 7 will have the following characteristics at different temperatures.


PH of Water

PH of solution

Nature of solution

10 ºC



Acidic as pH of solution is less than the pH of pure water at 10 ºC.

25 ºC



Neutral as pH of solution is equal to the pH of pure water at 25 ºC.

40 ºC



Alkaline as pH of solution is more than the pH of pure water at 40ºC.

At 298 K, the pH of a solution of lemon juice is 2.32. What are the [H3O+] and [OH-] in this solution?
Since pH = -log {[H3O+]/mol dm-3}, we will have
log {[H3O+]/mol dm-3} = -2.32
Taking the antilogarithm, we get
[H3O+]/mol dm-3 = 4.786 × 10-3
or [H3O+] = 4.786 × 10-3 mol dm-3
At 298 K, [H3O+] [OH-] = 10-14 (mol dm-3)2. Hence

= 0.209 × 10-11 mol dm3

What is the pH of a solution which has [H3O+] = 1.84 × 10-5 mol dm-3?
We have
pH = -log {[H3O+]/mol dm-3} = - log (1.84 × 10-5)
= - log (1.84) - (-5) = -0.26 + 5 = 4.74.

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