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# Solving Equilibrium Problems: pH of a Monoprotic Weak Acid

To illustrate the systematic approach, let us calculate the pH of 1.0 M HF.

pH of a Monoprotic Weak Acid

To illustrate the systematic approach, let us calculate the pH of 1.0 M HF. Two equilbria affect the pH of this system. The first, and most obvious, is the acid disso- ciation reaction for HF

HF(aq)+ H2O(l)   < = = >   H3O+(aq)+ Fâ€“(aq)

for which the equilibrium constant expression is The second equilibrium reaction is the dissociation of water, which is an obvious yet easily disregarded reaction

2H2O(l)  < == == >   H3O+(aq)+ OHâ€“(aq)

Kw = [H3O+][OHâ€“] = 1.00 x 10â€“14            6.36

Counting unknowns, we find four ([HF], [Fâ€“], [H3O+], and [OHâ€“]). To solve this problem, therefore, we need to write two additional equations involving these un- knowns. These equations are a mass balance equation

CHF = [HF]+ [Fâ€“]            6.37

and a charge balance equation

H3O+]= [Fâ€“] + [OHâ€“]           6.38

We now have four equations (6.35, 6.36, 6.37, and 6.38) and four unknowns ([HF], [Fâ€“], [H3O+], and [OHâ€“]) and are ready to solve the problem. Before doing so, however, we will simplify the algebra by making two reasonable assumptions. First, since HF is a weak acid, we expect the solution to be acidic; thus it is reason- able to assume that

[H3O+] >> [OHâ€“]

simplifying the charge balance equation (6.38) to

[H3O+]= [Fâ€“]               6.39

Second, since HF is a weak acid we expect that very little dissociation occurs, and

[HF] >> [Fâ€“]

Thus, the mass balance equation (6.36) simplifies to

CHF = [HF]                       6.40

For this exercise we will accept our assumptions if the error introduced by each as- sumption is less than Â±5%.

Substituting equations 6.39 and 6.40 into the equilibrium constant expression for the dissociation of HF (equation 6.35) and solving for the concentration of H3O+ gives us Before accepting this answer, we must verify that our assumptions are acceptable. The first assumption was that the [OHâ€“] is significantly smaller than the [H3O+]. To calculate the concentration of OHâ€“ we use the Kw expression (6.36) Clearly this assumption is reasonable. The second assumption was that the [Fâ€“] is significantly smaller than the [HF]. From equation 6.39 we have

[Fâ€“] = 2.6 x 10â€“2 M

Since the [Fâ€“] is 2.6% of CHF, this assumption is also within our limit that the error be no more than Â±5%. Accepting our solution for the concentration of H3O+, we find that the pH of 1.0 M HF is 1.59.

How does the result of this calculation change if we require our assumptions to have an error of less than Â±1%. In this case we can no longer assume that [HF] >> [Fâ€“]. Solving the mass balance equation (6.37) for [HF]

[HF] = CHF â€“ [Fâ€“]

and substituting into the Ka expression along with equation 6.39 gives where a, b, and c are the coefficients in the quadratic equation ax2 + bx + c = 0. Solving the quadratic formula gives two roots, only one of which has any chemical significance. For our problem the quadratic formula gives roots of Only the positive root has any chemical significance since the negative root implies that the concentration of H3O+ is negative. Thus, the [H3O+] is 2.6 x 10â€“2 M, and the pH to two significant figures is still 1.59.

This same approach can be extended to find the pH of a monoprotic weak base, replacing Ka with Kb, CHF with the weak baseâ€™s concentration, and solving for the [OHâ€“] in place of [H3O+] Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
Modern Analytical Chemistry: Equilibrium Chemistry : Solving Equilibrium Problems: pH of a Monoprotic Weak Acid |