The Power of Protons

An Ongoing Attempt to Simplify the Understanding of Calculations involving pH

(Andrew Pearson and Robert Lancashire, University of the West Indies, Mona Campus, Jamaica.)

Note: The use of the term proton refers to H+ or what is more accurately described as the hydronium ion (H3O+ in its many extended forms).


  1. Introduction to protons in biochemistry
  2. Introduction to logarithms
  3. Proton concentrations and the pH scale
  4. The Henderson_Hasselbalch Equation

Introduction to protons in biochemistry

Protons are important in biochemistry for many reasons including:

The binding of a proton to an acceptor changes the electrical charge on the acceptor, and this often changes the biochemical properties of the molecule. For example:


The willingness of a particular acceptor to bind a proton is an intrinsic property of the acceptor, governed largely by the behaviour of the electrons in that part of the acceptor molecule close to the proton binding site. This is called the pK of the group.


It is possible to know how many potential acceptor groups are protonated once the proton_binding affinity (pK) of the acceptor and the number of available protons is known. This latter is described in terms of pH.

Small numbers

By international convention the standard number of molecules that is used by biochemists and chemists is Avogadro's number (6.022 x 1023); this is called a mole of the molecules.

When a solution is made up containing 1 mole of solute in 1 litre (1 dm-3) of solvent the concentration of the solute is said to be 1 molar or 1 mol dm-3 (in out_dated notation: 1 M), and this is the standard unit of concentration.

The concentrations of most biochemicals in vivo range between about 10-2 mol dm-3 and 10_12 mol dm-3.

The concentration of protons in a neutral (neither acidic nor basic) aqueous solution is 10_7 mol dm-3.

Manipulating such small numbers is facilitated by using logarithms.

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Introduction to logarithms

The principle of logarithms is to express a number as another number with an index or exponent:

actual number = another numberindex, or exponent

e.g. 100 = 102

16 = 24

The "another number" mentioned above is called the "base" of the logarithm; the most common base numbers used are 10, 2 and a value 'e' ( the base for the so_called natural logarithms).
For most biochemistry calculations, logarithms to the base 10 are used.

In the days before electronic calculators were readily available to school children (not so long ago!) logarithm and antilogarithm tables were easily available, and we had to have a set handy for maths, physics and chemistry classes and homework.

The following table, in conjunction with the logarithm table, will require little explanation:

Number Scientific Notation Log10 No.
1 1 x 100 0.0000
6 6 x 100 0.7782
10 1 x 101 1.0000
20 2 x 101 1.3010
300 3 x 102 2.4771
4,000 4 x 103 3.6021
5,000,000 5 x 106 6.6990

The log(arithm)10 of the number has two distinct parts separated by a full stop called "point", which is commonly mistaken for, but is not a simple decimal point: the integer to the left of the point is called the characteristic and corresponds to the power to which the multiplier 10 must be raised; the number to the right of the point is called the mantissa and is determined either by calculation or by looking up in the tables . The mantissa is always positive, whereas the characteristic is negative for numbers that are less than 1.0.

The log and antilog tables shown are four figure tables, and these are accurate enough for most applications. For more exacting applications, tables are available with which greater accuracy is possible.

To find the log10 of 45.67 we first inspect the number and determine the value of the characteristic by converting the number to scientific notation : 4.567 x 101. The characteristic is thus 1.

We now look up 4567 in the log tables as follows:
the first two digits are to be found in the first column of the table. If the number that we were looking for were 4500 the mantissa would have been: .6532 (always include the point in the mantissa).
The mantissa for 4560 is .6590, and the one we really want is .6597 _ having added the number from the appropriate column in the right hand set of columns to the mantissa of the first three digits.

Log10 45.67 = 1.6597

Practice looking up the log10 for a number and type your answer in the box alongside.

Log10 of a number between 1.0 and zero.

A complication arises for numbers that are less than 1.0, which when expressed in scientific notation have 10 raised to a negative power such as 0.005 = 5 x 10-3.

The characteristic of the log of this number is called bar three, and the whole log is called:

"bar three point six nine nine zero"

'Bar three point ...........' is not exactly the same as 'minus three point ...........', and calculations involving pH become very difficult if this is not clearly understood: much confusion in the correct use of logarthims arises over the difference.

It is only the characteristic that can be negative; the mantissa is always positive.

This confusion is compounded by the facts that:

  1. electronic calculators and standard keyboards cannot show bar numbers
  2. because of the ubiquity of electronic calculators, any formal treatment of logarithms has dropped out of many high school mathematics syllabuses _ perhaps somebody should think again!
Since is it not currently possible to use a standard keyboard to enter a bar number, we will follow the current convention for displaying the characteristic for negative powers of ten by using the minus sign - .

Thus log10 0.004545 = -3.6585.

Practice looking up the log10 for a number less than 1 and type your answer in the box alongside.

Now you should be able to reverse the process and use the table to find the number for which the log10 is 2.8388, however it is slightly easier to look up the mantissa in the antilog tables, the characteristic merely tells you where to put the decimal point.

Where a bar characteristic is involved, as with some pH calculations, some understanding is required. For assistance with examples of the latter we recommend that you continue with this document up to the end of the next section on the pH scale; or you can jump immediately to some pH exercises.

These days scientists and their students expecting to have to manipulate logarithms are already equipped with smart phones/tablets with built-in calculators, but here we find another difficulty: not all such devices are easy to use or interpret. We strongly urge that you take time here to practice 'logging' and 'antilogging' numbers with your own calculator application, checking the outputs against the logarithm and tables.

If you are unable to reconcile results, you are advised to:

  1. either find (if possible!), read and understand (if possible!) the instructional 'manual' that came with your calculator app,
  2. and/or seek direct, expert, human advice,
  3. or get a better calculator app! There are many that are freely available
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Proton concentrations and the pH scale

Many biochemicals are Brønsted_Lowry acids or bases: they can donate protons if acidic and accept protons if basic such as the carboxylate and amino groups on the amino acid alanine respectively.

The aqueous environment of most biochemical reactions can exchange protons with such groups. The proton exchange reaction is governed, as are others, by the concentrations of the reactants, their willingness to participate and the temperature of the system (which is really an indication of the speed of the reactants).

Of the willingness to participate (pK in reactions involving protons), strong acids are more willing to donate their protons than are weak acids: by definition the protons of a strong acid are always fully dissociated in aqueous solutions, whereas the protons of a weak acid show a greater affinity for their conjugate base.

We encourage you to look at the accompanying (free) Windows Titrations program.

Water is both an acid and a base:

H2O + H2O = H3O+ + OH_
H3O+ is the hydronium ion, and is a simple form in which protons can be thought to occur in aqueous solutions. For the sake of simplicity we shall assume the presence of the H2O molecule associated with the proton and only write H+. The reaction can thus be written: H2O = H+ + OH_

To what extent does this dissociation of water molecules (protolysis) occur? If we could fill a bucket with pure H2O and allowed it to dissociate (for an infinite length of time) we could then measure the concentrations of the protons or hydroxyl ions present and determine the acid dissociation constant Ka:

Ka = ([H+] + [OH_]) /[H2O]

(Note that the equilibrium constant is actually a ratio of the activities of the reactants, and activities are equal to concentrations only in solutions that are ideal , i.e. which obey Raoult's Law. In practice, provided that we are dealing with dilute aqueous solutions, we assume that the concentrations of the solutes are the same as their activities because the former are easier to determine than the latter.)

At 25oC the Ka = 1.8 x 10_16 : there is so little dissociation that it is reasonable to say that [H2O] is overwhelming and constant.

We can re_arrange the above equation and introduce a new term Kw the ion product of water:

Kw = Ka[H2O] = [H+] x [OH_] = 10_14 mol dm-3 at 25oC

Now in pure water since [H+] must equal [OH_] and multiplied together they = 10_14mol dm-3, therefore [H+] = [OH_] = 10_7mol dm-3.

10_7 x 10_7 = 10_7+(_7) = 10_14

To demonstrate the manipulation of indices and logs10:

10_7 x 10_7 = 10_14 _7 + (_7) = _14
therefore 10_7 = 10_14 / 10_7 _7 = _14 _(_7)
therefore 10_7 = 10_14 _(_7) _7 = _14 _(_7)
therefore 10_7 = 10_14 +7 _7 = _14 +7

Sørensen had the idea of eliminating the minus signs from the log10 of [H+] and invented the pH scale.

The view that Sørensen's idea was misguided has been expressed.

Now in pure water

[H+] = [OH_] = 10_7 mol dm-3
therefore the log10 [H+] of pure water = _7
therefore the pH of pure water = 7

If we increase the [H+] of an aqueous solution 100_fold, from 10_7 mol dm-3 to 10_5 mol dm-3, the pH value changes from 7 to 5.


the more acid the solution, the lower the pH value;


the pH value rises as the solution becomes more alkaline.

Try some calculations

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The Henderson_Hasselbalch Equation

As stated earlier, it is possible to determine the degree of protonation of an acidic or basic group in an aqueous solution once we know the pK of the group and the pH of the solution. It is often important to know this because the degree of protonation of a group will determine its electrical charge and may influence its reactivity.

The pK of such a group is defined as the pH at which the group is exactly 50% protonated, and is named by analogy to pH: it is -log10K (K = equilibrium constant).

Dissociation of Acids in Water

HA = H+ + A_
Ka = [H+][A_] /[HA]
In aqueous solution:HA + H2O= H3O+ + A_
Keq = [H3O+][A_] /[HA][H2O] = Ka /[H2O]

If we now repeat the earlier assumptions about the constancy and overwhelming concentration

of H2O, and the equivalence of H3O+ and H+, we can write:

Ka = [H+][A_]/[HA]
which is what we had four lines higher. This shows that the dissociation constant of an acid has the same value as its equilibrium constant in water.

This can be written:

[H+] = Ka x [HA] / [A_]
Taking logs log10[H+] = log10Ka + log10([HA] / [A_])
so:_log10[H+] = _log10K a _ log10([HA] / [A_])
andpH = pKa _ log10([HA] / [A_])

Note that since _ log10a / b = + log10b / a, it could be written:

pH = pKa + log10([A_] / [HA])

Both are called the Henderson_Hasselbalch Equation: beware of the last term in the equation since many mistakes are made with the sign and decision of which concentration is to be divided by the other.

Consider the effect of putting pH = pKa


pKa = pKa _ log10 ([HA] / [A_])


0 = _ log10([HA] / [A_])
Since antilog 0.0000 is 1.0000, then (remembering the negative sign!)1 = [A_] / [HA]
Therefore:[HA] = [A_]

This means that the acid is exactly half dissociated.

Consider the effect of putting pH = pKa + 1. This involves a ten_fold change in [H+]:

[H+] is now ten times less than the [H+] at 50% dissociation.

pKa + 1 = pKa _ log10([HA] / [A_])


1 = _ log10([HA] / [A_])

Since antilog 1 is 10, then (remembering the negative sign!)

10 = [A_] / [HA]


10[HA] = [A_]

This means that if we had eleven molecules of acid, only one would be protonated at that [H+], and ten would be unprotonated.

Let us consider the amino acid alanine which has an acidic carboxyl group and a basic amino group.

When the [H+] is high, say pH = 1, both groups will be protonated and alanine will have a net charge of +1.

When pH = pK of the carboxyl group at 2.34, protons will have dissociated from half of the alanine carboxyls present and the net charge will be + 0.5, since the amino group will still be fully protonated and thus carry a positive charge but the unprotonated half of the carboxyl groups present will carry a negative charge.

Further reducing the [H+] of the solution, by adding NaOH for example, will drag the rest of the protons from the alanine carboxyls present and the net charge will be zero. The pK of the amino group of alanine is 9.69 so that when the pH of the solution reaches this value, the net charge will be _0.5.

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Free Titrations Windows program.
Created and maintained by Andrew Pearson and Robert Lancashire,
University of the West Indies, Mona Campus, Jamaica.
Comments to authors at Andrew Pearson and/or Robert Lancashire.
Created Monday Apr 15th 1996. Last modified 22nd February 2015.