BIOCHEMISTRY TOPICS

Problems in acid-base chemistry

Weak acid dissociation and fraction of dissociation

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Problem 8-24. (ECA5, p.187) Calculate the pH and fraction of dissociation of (a) 10^−2.00 M and (b) 10^−10.00 M barbituric acid. ( K_a  =  9.8 × 10⁻⁵ )

Solution: This is a weak acid equilibrium calculation. We can write a table to help us define the equation we need to solve. We will adopt the convention used in the text denoting the initial concentration of the acid by F (or "formal" concentration).

		HA	=	H+	+	A−
	initial concentration	F		0*		0
	change in concentration
	final concentration	F − x		x*		x

The asterisk attached to the concentrations of H⁺ means that it is not, strictly speaking, zero initially, or derived solely from the dissociation of HA, but a small amount (~10⁻⁷ M) comes from the autodissociation of water. We will for the most part only deal with problems where this can be safely neglected, but we should keep in mind the need to verify that this assumption is valid.

The next step in this problem is to substitute the quantities in the final concentration row into the expression for K_a  and solve the resulting equation for x, plugging in numbers for F and K_a.

K_a  =  [H⁺][A⁻ ] / [HA]  =  x² / F − x

Solving the above for x yields a quadratic equation:

x²  +  K_ax  −  K_aF  =  0

We have kept the equation in general form, and we can write the general form of the solution for this type of problem using the quadratic formula. The solution ("roots") to the equation

ax²  +  bx  +  c  =  0

has the formula

x = { −b ± ( b² − 4ac)^1/2} / 2a .

Thus, with a = 1, b = K_a, and c = −K_aF, we can write the general formula for the solution as

x = { −K_a ± (K_a² + 4K_aF)^1/2} / 2 .

Plugging in the numbers K_a = 9.8 × 10⁻⁵ and F = 10^−2.00 = 0.010, and discarding the negative root, we obtain

x = 9.4₂₁₆ × 10⁻⁴,

from which we obtain [H⁺] = 9.4 × 10⁻⁴ and pH = 3.03.

The fraction of dissociation, α, can be directly computed as x / F.

α  = 9.4₂₁₆ × 10⁻⁴ / 10^−2.00 = 9.4₂₁₆ × 10⁻² = 0.094 or 9.4%

For the case where F = 10^−10.00 , we do not obtain the pH from solving a quadratic equation derived from the HA equilibrium. Rather, the pH is set to that of pure water, because H⁺(aq) from the acid HA at this concentration is insignificant in relation to the H⁺(aq) from autodissociation of water. So, the pH is 7, and the fraction of dissociation is found by

[A⁻ ] / ( [HA] + [A⁻ ] )  =  K_a / ( K_a + [H⁺] ) = 9.8 × 10⁻⁵ / ( 9.8 × 10⁻⁵ + 1.0 × 10⁻⁷) = 0.99898

The acid is 99.9% dissociated at this concentration!

Example Problem: Weak acid pH calculation

For a weak acid HA, with pK_a = 2.328, calculate the pH of a 0.0500 M solution.

Solution: This is a weak acid calculation. In this case, we start with a weak acid, made into a(n aqueous) solution of a given concentration, and calculate the pH using the acid dissociation equation and its associated Ka value. Why does the acid dissociation equation govern the pH? It is because we start with only HA, and the pH will change according to the extent of dissociation of HA:

HA  =  H⁺  +  A⁻       K_a  =  [H⁺][A⁻ ] / [HA].

Note that the Henderson-Hasselbalch equation does not apply here because we are not starting with a mixture of HA and A⁻. If we were to start with only A⁻, then we would be dealing with a weak base calculation, and the relevant chemical equation would be the base hydrolysis reaction, with equilibrium constant K_b.

One thing that we need to do to solve this problem - and we may as well do it first off - is to derive K_a from pK_a.

pK_a  =  − log₁₀K_a   iff   K_a   =  10^−pKa

Thus,

K_a  =  10^−2.328  =  4.70 × 10⁻³

where we have followed the rule for treating significant figures when we take the antilog of a number.

We can write a table to help us define the equation we need to solve. Using the convention denoting the initial concentration of the acid by F (or "formal" concentration), and letting x be the quantity we need to solve for, we have:

		HA	=	H+	+	A−
	initial concentration	F		(10−7 M)		0
	change	− x		+ x		+ x
	final concentration	F − x		+ x		+ x

(Note: In all such problems we will need to solve in this context, we will be able to safely ignore the small amount
(10⁻⁷ M) of H⁺ coming from the autodissociation of water, relative to that coming from the dissociation of HA. But it is important to understand that there are situations when this assumption breaks down.)