**BIOCHEMISTRY TOPICS**

### Problems in acid-base chemistry

Weak acid dissociation and fraction of dissociation

**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] =x^{2}/F−x

Solving the above for *x* yields a quadratic equation:

x^{2}+K_{a}x−K_{a}F= 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

ax^{2}+bx+c= 0

has the formula

x= { −b± (b^{2}− 4ac)^{1/2}} / 2a.

Thus, with *a* = 1, *b* = *K*_{a},
and *c* = −*K*_{a}*F*, we can write
the general formula for the solution as

x= { −K_{a}± (K_{a}^{2}+ 4K_{a}F)^{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= 9.4_{216}× 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 = 9.4

_{216}× 10^{−4}/ 10^{−2.00}= 9.4_{216}× 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 p*K*_{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 *K*a 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
p*K*_{a}.

pK_{a}= − log_{10}K_{a}iffK_{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.)