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Atoms and Molecules

Acids, bases and the Henderson-Hasselbalch Equation


An acid is a substance which produces hydrogen ions (H+) by dissociation.
    For example         HCl [goes to, arrow] H+ + Cl-

Bases are substances which can combine with H+ like ammonia (NH3)
    For example         H+ + NH3 [goes to, arrow] NH4+


They may produce hydroxide ions (OH- ) either by direct dissociation or subsequent to reaction with water,
    e.g.                         KOH [goes to, arrow] K+ + OH-
    and                     NH3 + H2O [equilibrium arrows] NH4OH [equilibrium arrows] NH4+ + OH- 

Hydrogen ions (H+) associate with water to form H3O+ (and H11O5+ etc.)

Water dissociates to a tiny extent: H2O [equilibrium arrows] H+ + OH-
    The equilibrium constant = Keq
                                               = 1.8 x 10-16 = [H+]x[OH-] = [H+]x[OH+]     (see footnote 1)
                                                                           [H2O]          (1000/18)

Therefore                 [H+]x[OH- ] = 1.8 x 10-16 x 55.5 = 10-14

In pure water [H+] must equal [OH-] (the solution is neutral),
        therefore [H+] =
Ö10-14 = 10-7 M

The pH (hydrogen ion potential) of a solution is defined as pH = -log10 (H+) , where (H+) is the hydrogen ion concentration. The pH scale can range between about -1 and +15; a neutral solution has pH 7.0.

For a weak acid, which dissociates as follows: HA [equilibrium arrows] H+ + A-

[Ka=(H+ x A- )/HA]

An interesting and extremely useful relationship between pH and pKa can be obtained simply by taking logarithms (to the base 10; see footnote 2) of the above:

log10Ka = log10[H+] + log10[A- ] - log10[HA]

Therefore             -log10[H+] = -log10Ka + log10[A-] - log10[HA]
giving the Henderson-Hasselbalch equation:         [Henderson-Hasselbalch equation; ph=pka+log10([A-/[HA)]    


The most convenient form of this Henderson-Hasselbalch equation, is

[Henderson-Hasselbalch equation; ph=pka+log10([proton acceptor]/[proton donor])]

By using pKa values, we are able to express the strength of an acid (i.e. its tendency to dissociate) with reference to the pH scale. If Ka, the dissociation constant, is large, then pKa will have a low numerical value. A strong acid is one which is largely, perhaps completely, dissociated, and which therefore has a high Ka value. A weak acid is one that is only slightly dissociated in solution, and has a low Ka value. There is no generally accepted dividing line between weak and strong acids, but as a rough guide it is suggested that a strong acid would be at least 25% dissociated in a 0.1 M solution; this corresponding to Ka of about 10-2. If Ka = 10-2, the corresponding pKa value is (+) 2.0. Lower values of pKa (e.g. 0.7) correspond to stronger acids (Ka = 0.2), while higher values (e.g. 4.7) correspond to weaker acids (Ka = 2 x 10-5). pKa values are used both for acids and for the conjugate acids of bases. Thus NH3 has pKa (for the dissociation NH4+ [equilibrium arrows] H+ + NH3) about 9.3. There is no real need for Kb (and pKb) values, given by Kb = [BH+][HO-]/[B][H2O]. Should such values ever be required, for dilute aqueous solutions Kb = 10-14/Ka and pKb = 14.0 - pKa where Kb and pKb refer to the conjugate acid (BH+ [equilibrium arrows] H+ + B).

Perhaps it is useful to look at this in another way: if we consider the situation where the acid is one half dissociated, in other words where [A-] is equal to [HA], then, substituting in the Henderson-Hasselbalch Equation 

                                                pH = pKaa + log10(1)
Therefore                               pH = pKa + 0
Therefore                               pH = pKa

This means that an acid is half dissociated when the pH of the solution is numerically equal to the pKa of the acid. Therefore acids with the lowest pKa values are able to dissociate in solutions of low pH, i.e. even where the hydrogen ion concentration is high. Acids with higher pKa values dissociate only in solutions of high (more alkaline) pH.

Acid

Ka

 

pKa

Trichloroacetic

2 x10-1

=10-0.7

0.7

Dichloroacetic

5 x10-2

=10-1.3

1.3

Monochloroacetic

1.6 x10-3

=10-2.8

2.8

Formic

2.1 x10-4

=10-3.7

3.7

Benzoic

7.8 x10-5

=10-4.1

4.1

Acetic

1.9 x10-5

=10-4.7

4.7

H2CO3

2.9 x10-7

=10-6.5

6.5

H2S

5.8 x10-8

=10-7.2

7.2

HCN

1.3 x10-9

=10-8.9

8.9

The table above gives some examples of Ka and pKa values for a number of acids, which are listed in order of decreasing strength.

Buffers

A buffer is a mixture of dissolved substances which tends to keep the pH of a solution constant when modest additions of acid or base are made to that solution. No buffer can keep the pH exactly constant in such situations. The Henderson-Hasselbalch equation is very useful for calculations involving buffers, and is fairly accurate under reasonable conditions.
Practical buffers consist of mixtures of weak acids and their salts. They will buffer solutions over a range of pH about 1 unit either side of the pKa value.

[titration]

Polybasic acids or mixtures of acids will have several pKa values, and (when mixed with their salts) can therefore buffer over several ranges of pH. Provided the individual pKa values differ by less than 2 units, the mixture will show more or less continuous buffering over a wide range of pH. For example, citric acid (pKa values 3.1, 4.8, and 6.4) mixed with one of its salts such as trisodium citrate will buffer continuously from pH 2 to 7.5, but not at higher pH values. A mixture of "Tris" (tris(hydroxymethyl)aminomethane, pKa 8.1), trimethylamine (pKa 9.8), and piperidine (pKa 11.1) together with (say) their hydrochlorides, will buffer from pH about 7 to 12. These three bases with citric acid would buffer at all pH values between about 2 and 12.

The Henderson-Hasselbalch equation is used for the calculation of the pH or composition of a buffer solution. With mixtures consisting of weak acids (only slightly dissociated) and their salts (nearly totally ionised) the convenient approximations [HA] = total acid concentration, and [A- ] = salt concentration, can often be made. This is adequate for most buffer design purposes, the exact pH required being obtained by adjusting the composition by adding a little strong acid or base. Hence an acetic acid/acetate buffer solution containing 0.1 M acetic acid and 0.05 M sodium acetate would have a pH of 4.4.

            pH = pKa + log10[conjugate base] = 4.7 + log10(0.05) = 4.7 + (-0.3) = 4.4
                                          [conjugate acid]                         (0.1)

Buffers function as follows: when a strong acid is added, the H+ from that acid combine with a portion of the anion to form undissociated acid, thereby removing most of the added H+ from the solution

                H+ + A- [equilibrium arrows] HA

When strong base is added part of the undissociated acid reacts to form anions

        Base + HA [equilibrium arrows] A- + BaseH+

For further information, do look at the 'pH/Titrations' PC program.

Notes

1.    Square brackets means “concentration of” whatever they contain; e.g. [H+] means the concentration of hydrogen ions in moles per litre. (Back)

2.    If this derivation presents any difficulty then you are reminded of the following relationships:

  1. log (x x y) = log (x) + log (y)
  2. log (x/y) = log (x) - log (y)
  3. log (1/x) = -log (x)
  4. log (1) = 0                             (Back)


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This page was last updated by Martin Chaplin
on 10 February, 2005

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