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Chapter: Modern Analytical Chemistry: Equilibrium Chemistry

Solving Equilibrium Problems: pH of a Polyprotic Acid or Base

A more challenging problem is to find the pH of a solution prepared from a polyprotic acid or one of its conjugate species.

pH of a Polyprotic Acid or Base

A more challenging problem is to find the pH of a solution prepared from a polyprotic acid or one of its conjugate species. As an example, we will use the amino acid alanine whose structure and acid dissociation constants are shown in Figure 6.11.


pH of 0.10 M H2L+ 

Alanine hydrochloride is a salt consisting of the diprotic weak acid H2L+ and Cl. Because H2L+ has two acid dissociation reactions, a complete systematic solution to this problem will be more complicated than that for a mono- protic weak acid. Using a ladder diagram (Figure 6.12) can help us simplify the problem. Since the areas of predominance for H2L+ and L are widely separated, we can assume that any solution containing an appreciable quantity of H2L+ will con- tain essentially no L. In this case, HL  is such a weak acid that H2L+ behaves as if it were a monoprotic weak acid.


To find the pH of 0.10 M H2L+, we assume that


Because H2L+ is a relatively strong weak acid, we cannot simplify the problem fur- ther, leaving us with


Solving the resulting quadratic equation gives the [H3O+] as 1.91 x 10–2 M or a pH of 1.72. Our assumption that [H3O+] is significantly greater than [OH] is acceptable.

pH of 0.10 M L 

The alaninate ion is a diprotic weak base, but using the ladder dia- gram as a guide shows us that we can treat it as if it were a monoprotic weak base. Following the steps in Example 6.11 (which is left as an exercise), we find that the pH of 0.10 M alaninate is 11.42.


pH of 0.1 M HL 

Finding the pH of a solution of alanine is more complicated than that for H2L+ or L because we must consider two equilibrium reactions involving HL. Alanine is an amphiprotic species, behaving as an acid

HL(aq)+ H2O(l) < = = = = > H3O+(aq)+ L(aq)

and a base

HL(aq)+ H2O(l)  < ===   ==   >  OH(aq)+ H2L+(aq)

As always, we must also consider the dissociation of water

2H2O(l)     < == == >   H3O+(aq)+ OH(aq)


This leaves us with five unknowns ([H2L+], [HL], [L], [H3O+], and [OH]), for which we need five equations. These equations are Ka2 and Kb2 for HL,


From the ladder diagram it appears that we may safely assume that the concentra- tions of H2L+ and L are significantly smaller than that for HL, allowing us to sim- plify the mass balance equation to



Triprotic Acids and Bases, and Beyond 

The treatment of a diprotic acid or base is easily extended to acids and bases having three or more acid–base sites. For a tripro- tic weak acid such as H3PO4, for example, we can treat H3PO4 as if it was a mono- protic weak acid, H2PO4– and HPO 2– as if they were intermediate forms of diprotic weak acids, and PO 3– as if it was a monoprotic weak base.


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