pH to pKa conversion


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               Equilibrium constant of a weak acid 

 In discussing the equilibrium involved when a weak acid is added to water, a heterogeneous equilibrium forms in which aqueous acid molecules, HB(aq), react with liquid water to form aqueous hydronium ions and aqueous anions, B-(aq). The latter are produced when the acid molecules lose H+ ions to water.   


               HB(aq) + H2O <----à H30+ (aq) + B- (aq)

When the equilibrium in question occurs in solution, the chemical formulas enclosed in brackets in the equilibrium constant expression represent the molarities of the substances

 It is convenient to represent the proton transfer as a simple ionisation. Remember that H+ can be used to represent H3O+ .

    HB (aq) <-------à  H+ (aq) + B- (aq)

In which case the expression for the equilibrium constant becomes

                        Ka = [H+]  x [B-]



The equilibrium constant Ka is called acid equilibrium constant of weak acid HB.


 The Ka value of some weak acid in the order of decreasing strength are listed in tabular form. The weaker the acid,the smaller the value of Ka.Forexample HCN with Ka=5.8 x 10^-10 is a weaker acid than HNO2 for which Ka = 6.0 x 10^-4        

                pKavalue of weak acid

We sometimes refer to the pKa value of a weak acid.

                     pKa= -log10Ka  

HNO2 Ka = 6.0 x 10^-4  pKa =3.22

HCN Ka=5.8 x 10^-10 pKa = 9.24

                        pH to pKa

There are several ways to determine Ka of a weak acid .A simplest approach involves measuring [H+] or pH in a solution prepared by dissolving a known amount of the weak acid to form a given volume of solution.

Lets see an example to understand this 

Aspirin is a weak organic acid whose molecular formula may be written as HC9H7O4.A water solution of aspirin is prepared by dissolving 3.60g per liter.The pH pfthis solution is found to be 2.60.Calculate Ka for aspirin.

Strategy :

•The mass is given. So, we can calculate original concentration of HC9H704 (MM=180.15 g/mol) . [ HC9H7O4] = 3.60g / 1L x 1 mol/180.15 g = 2.00 x 10^-2 M


•The equilibrium concentration of H+ can be calculated from the given pH =2.60


              [H+] eq=   10^-2.60 = 2.5 x 10^-3 M 


Now we have to calculate acid equilibrium constant Ka.



HC9H7O4 <-----à H+ (aq) + C9H704- (aq)


 Ka = [ H+] x [ C9H704- ] /  [HC9H7O4]


Solution: From chemical equation for the ionisation of the weak acid, it should be clear that 1 mol of C9H7O4- is produced and 1 mol of HC9H7O4 is consumed for every mole of H+ produced. It follows that


Delta[C9H7O4-] = delta[H+] = 0.0025    and     detla[HC9H7O4] = - delta [H+] = -0.0025  


 Originally there is essentially no H+ ( ignoring the slight ionisation of water).The same holds for the anion C9H7O4- , the only species present is the weak acid HC9H7O4 at a concentration of 0.0200 M.



Putting this information togather

     HC9H7O4   <-----à   H+ (aq)  +   C9H704- (aq)

I      0.0200                          0.0          0.0

C     -0.0025                    +0.0025      +0.0025


E      0.0175                     0.0025        0.0025


All information needed to calculate Ka is now available ,

    Ka = (2.5 x 10^-3)2 / 0.0175


  Ka  = 3.6 x 10^-4

 pKa=  3.45