Home > Weak Acids and Bases – Calculate the pH of a weak acid (a general equation) > Solving Weak Acid and Weak Base pH problems > Weak Acid Weak Base pH calculation solved example > Weak Acid Weak Base pH calculation solved example > Cubic Equation Calculator for Weak Acid-Base Equilibria > Polyprotic Acids / pH Calculation

Polyprotic Acids / pH Calculation

A large number of acids can give two or more protons on ionization (dissociation) and these are referred to as polyprotic acids. For example, with sulfurous acid (H₂SO₃) we have the successive ionizations:

Fig. I.1: Stepwise dissociation of sulphurous acid (H₂SO₃)

A polyprotic acid always dissociates in a stepwise manner, one proton at a time. Note that the acid dissociation constants are labelled k_a1 and k_a2. The numbers on the constants refer to the particular proton of the acid that is ionizing. Thus, k_a1 always refers to the equilibrium involving removal of the first proton of a polyprotic acid. Note also that k_a2 for sulfurous acid is much smaller than k_a1. This can be explained by the fact that the second H⁺ has to leave from a negatively charged species, HSO₃^- - electrostatic attraction has to be overcomed – while the first H⁺ from a neutral compound H₂SO₃.

The above observation is general: It is always easier to remove the first proton from a polyprotic acid than to remove the second and so on. The k_a values become successively smaller as successive protons are removed.

The acid dissociation constants for common polyprotic acids are given in Table I.1

Table I.1: Stepwise dissociation constants for several common polyprotic acids

Polyprotic Acids and Ionic Equilibria  

Depending on the pH of the solution, a polyprotic acid may exist predominantly as the undissociated acid or any one of its anionic forms. It is easy to calculate the fraction present of the species involved in the equilibrium as a function of [H⁺].

 As an example let us calculate the fraction of phosphoric acid present as a function of the pH. Phosphoric acid is typical of most weak polyprotic acids in that its successive k_a values are very different.

 

Fig. I.1: Stepwise dissociation of phosphoric acid

From the ionic equilibria shown above it is obvious that the relative acid strengths are:

H₃PO₄   >>  H₂PO₄^-   >> HPO₄^-2

This means that in a solution prepared by dissolving H₃PO₄ in water, only the first dissociation step makes an important contribution to [H⁺]. This greatly simplifies the pH calculations for phosphoric acid solutions as is illustrated in the Example I.1 below.

Let us though first derive  general expressions for [H₃PO₄], [H₂PO₄^-], [HPO₄^-2] as a function of [H⁺] in phosphoric acid solutions.

Polyprotic Acids – Phosphoric Acid - General Expression of the Equilibrium Concentrations (of the species involved as a function of  [H⁺])

Since there are six unknowns (the concentration of the acid, of the three conjugate bases and of H⁺ and OH^-] we need six equations to define the relations between these quantities. These are

The equilibria ( equilibrium constant expressions):

[H⁺] * [H₂PO₄^-] = k_a1 * [H₃PO₄]     (1)

[H⁺] * [HPO₄^-2 ] = k_a2 * [H₂PO₄^-]  (2)

[H⁺] * [PO₄^-3] = k_a3 * [HPO₄^-2]      (3)

[H⁺] * [OH^-] = k_w        (4)

The mass balance on phosphate:

[H₃PO₄] + [H₂PO₄^-] +[HPO₄^-2] + [PO₄^-3] = C            (4a)

where C is the initial concentration of phosphoric acid.

The fraction of the acid present as each species is the ratio of the concentration of that species to the initial concentration C of the acid.

a₃ = [H₃PO₄] / C     

Note: The index on a shows the number of protons attached to the molecule. It is more convenient to calculate the reciprocal of this fraction, since this is directly expressible in terms of the mass balance (4a).

1/a₃ = C/[H₃PO₄] = 1 + [H₂PO₄^-] / [H₃PO₄] + [HPO₄^-2] / [H₃PO₄] + [PO₄^-3] / [H₃PO₄]   (5)

From equation (1):  [H₂PO₄^-] / [H₃PO₄] = k_a1 / [H⁺]                (6)

Multiplying (1) by (2) gives the third term of (5):

[HPO₄^-2] / [H₃PO₄] = (k_a1* k_a2)/ [H⁺]²    (7)

Multiplying (1) by (2) by (3) gives the fourth term of (5):

[PO₄^-3] / [H₃PO₄]    = (k_a1* k_a2* k_a3)/ [H⁺]³    (8)

Substituting (6), (7) and (8) in (5) gives:

a₃ = [H₃PO₄] / C = [ 1 + k_a1 / [H⁺] + (k_a1* k_a2)/ [H⁺]² + (k_a1* k_a2* k_a3)/ [H⁺]³]^-1     (9)           

Combining (9) and (6):

a₂ = [H₂PO₄^-] / C = a₃ * k_a1 / [H⁺]    (10)

Combining (9) and (7):

a₁ = [HPO₄^-2] / C = a₃ * (k_a1* k_a2) / [H⁺]²    (11)

Combining (9) and (8):

a_o = [PO₄^-3] / C = a₃ * (k_a1* k_a2* k_a3) / [H⁺]³     (12)

There are very few situations where the above exact relations are used. The above formulas should not be memorized since equilibrium concentrations for the species involved can be derived from equilibrium constant expressions and the assumption that all [H⁺] in solution is generated by the dissociation of H₃PO₄ (Example I.1).

Example I.1

Calculate the pH of a 5.0 M H₃PO₄ solution and determine equilibrium concentrations of the species involved

Solution

DATA	[H3PO4]initial = C = 5.0 M  ka1 = 7.5 * 10-3 ka2 = 6.2 * 10-8 ka3 = 4.8 * 10-13
UNKNOWNS	[H3PO4] = ?  [H2PO4-] = ?  [HPO4-2] = ?  [PO4-3] = ?

Mathematical Approach (Exact Solution)

STEP 1: Write down the equilibrium reactions. Phosphoric acid H₃PO₄ is a weak triprotic acid – see the ka ‘s in Fig. I.1. It dissociates partially in water and most of the H⁺ is generated at the first dissociation reaction (notice that  k_a1 >> k_a2 >> k_a3)

STEP 2: Write down the equilibrium constant expression and the equilibrium constant.

The dominant equilibrium will be the dissociation of H₃PO₄ and the equilibrium constant expression is given by:

k_a1 = [H⁺] * [H₂PO₄^-] / [H₃PO₄] = 7.5 * 10^-3      (13)

STEP 3: Write down the equilibrium concentration of the species involved in the dominant equilibrium reaction.

Suppose that x M of H₃PO₄ dissociates in water

	H3PO4	H+	H2PO4-
Initial	5.0 Μ	0	0
Change	-x  M	+x M	+x M
Final (at equilibrium)	(5.0 – x) M	x M	x M

STEP 4: Substitute the equilibrium concentrations into the expression for k_a1 and solve for x.

Since the value of x is small (5.0 – x) ~ 5

k_a1 = [H⁺] * [H₂PO₄^-] / [H₃PO₄] = (x)*(x) / (5.0 – x) ~ x²/5 =  7.5 * 10^-3

and therefore x ~ 1.9*10^-1 M,  [H⁺]= [H₂PO₄^-] = x = 0.19 M,  pH = 0.72

The [H₃PO₄] = 5.0 – x = 4.8 M

The concentration of [HPO₄^-2] can be obtained from (11):

a₁ = [HPO₄^-2] / C = a₃ * (k_a1* k_a2) / [H⁺]²    (11)

The only unknown in the above equation is a₃. It can be calculated from (9):

a₃ = [H₃PO₄] / C = [ 1 + k_a1 / [H⁺] + (k_a1* k_a2)/ [H⁺]² + (k_a1* k_a2* k_a3)/ [H⁺]³]^-1 =

[ 1 + (7.5*10^-3/ 0.19) + (7.5 * 10^-3* 6.2 * 10^-8)/(0.19)² + (7.5 * 10^-3* 6.2 * 10^-8* 4.8 * 10^-13)/(0.19)³]^-1 = [ 1 + (0.039) + (1.288*10^-8) + (3.258*10^-20) ]^-1 ~ 1 / 1.039 = 0.96       

         

Therefore, a₃ ~ 0.96    (12)      

Substituting in (11) gives:

a₁ = [HPO₄^-2] / 5 = a₃ * (k_a1* k_a2) / [H⁺]² = 0.96 * (7.5 * 10^-3* 6.2 * 10^-8) / (0.19)² = 1.236 *10^-8

Therefore, [HPO₄^-2] = 5 * 1.236 * 10^-8 = 6.18 * 10^-8 M  ~  6.2 * 10^-8 M      (13)      

Similarly, [PO₄^-3] can be calculated. [PO₄^-3] = 1.6 *10^-19 M       (14) 

“Chemical Approach” (Approximations)

Proceed as above up to the calculation of [H⁺] = [H₂PO₄^-] = 0.19 M. The concentration of [HPO₄^-2] can be obtained from the equilibrium constant expression and k_a2 (dissociation of H₂PO₄^-).

H₂PO₄^- (aq)    =   H⁺(aq)   + HPO₄^-2   (aq)       k_a2 = 6.2 * 10^-8       (15)

k_a2 = [H⁺] * [HPO₄^-2 ] / [H₂PO₄^-] = 6.2 * 10^-8   (16)

Substituting the values for [H⁺] = [H₂PO₄^-] in (16):

[HPO₄^-2 ] = k_a2  = 6.2 * 10^-8 M           (13)

Similarly, [PO₄^-3] can be calculated from the equilibrium constant expression and k_a3 to give as above:

[PO₄^-3] = 1.6 *10^-19 M                        (14)

The results show that the dissociation of H₂PO₄^- and HPO₄^-2 is negligible relative to that of H₃PO₄. This is apparent from the fact that [HPO₄^-2] is 6.2 * 10^-8 M – showing that only 6.2 * 10^-8 mol/l of H₂PO₄^- has dissociated. The value of [PO₄^-3] shows that the dissociation of HPO₄^-2 is even smaller.        

For many polyprotic acids k_a1 is much larger than subsequent dissociation constants, in which case the H⁺ in the solution comes almost entirely from the first dissociation reaction. As long as successive k_a values differ by a factor of 10³ or more, it is possible to obtain a satisfactory estimate of the pH of polyprotic acid solution by treating the acids as if they were monoprotic, considering only k_a1.

In a manner analogous to a triprotic acid the following formulas can be derived for a diprotic acid of the general form H₂A:

a₂ =[H₂A] / C = ([ 1 + k_a1 / [H⁺] + (k_a1* k_a2)/ [H⁺]²)^-1         (17)

a₁= [HA^-] / C = a₂ * k_a1 / [H⁺]            (18)

a_o = [A^-] / C = a₂ * (k_{a1 *} k_a2)/ [H⁺]²                  (19)

Where a_o , a₁, a₂ are the fraction of a diprotic acid present as each of the three species H₂A, HA^- and A^-.

However, as for the triprotic acid there are very few situations where the above exact relations are used. The above formulas should not be memorized since equilibrium concentrations for the species involved can be derived from equilibrium constant expressions and the assumption that all [H⁺] in solution is generated by the dissociation of H₂A.

For another solved example regarding pH calculation of a polyprotic acid please see the video below:

 

---

Relevant Posts

Solving Weak Acid and Weak Base pH problems

Weak Acid Weak Base pH calculation solved example

Weak Acids and Bases - Calculate the pH of a weak acid

Cubic Equation Calculator for Weak Acid-Base Equilibria

Self-ionization of water - Autoionization of water - The ion product of water (kw)

 

---

References

J-L. Burgot “Ionic Equilibria in Analytical Chemistry”, Springer Science & Business Media, 2012

J.N. Butler  “Ionic Equilibrium – Solubility and pH calculations”, Wiley – Interscience, 1998

Clayden, Greeves, Waren and Wothers “Organic Chemistry”, Oxford,

D. Harvey,  “Modern Analytical Chemistry”, McGraw-Hill Companies Inc., 2000

Toratane Munegumi, World J. of Chem. Education, 1.1, 12 (2013)

J.N. Spencer et al., “Chemistry structure and dynamics”, 5th Edition, John Wiley & Sons, Inc., 2012

L. Cardellini, Chem. Educ. Res. Pract. Eur., 1, 151, (2000)

---

Key Terms

polyprotic acids, dissociation, dissociation constants, phosphoric acid, concentration of the species as a function of [H⁺], conjugate base, equilibrium constant expressions, equilibrium concentrations, calculate the pH, monoprotic, k_a,  triprotic acid, polyprotic acids and bases problems polyprotic acid and pka, polyprotic acid base equilibria, polyprotic acid concentration calculation, diprotic acid concentration calculation, polyprotic acids dissociation