Question

Consider a solution of monoprotic weak acid having dissociation constant $$K_a$$ . What is the minimum concentration $$C$$ in terms of $$K_a$$ such that the concentration of the undissociated acid can be equated to $$C$$ within a $$10\mathrm{\%}$$ limit of error?
A.$$90K_a$$
B.$$50K_a$$
C.$$70K_a$$
D.$$100K_a$$

Answer 1

Hint: According to the Arrhenius acid-base theory, a weak monoprotic acid is an acid that partially dissociates into its ions. Thus, the unionized acid and the dissociated ions in the aqueous form are present in an ionic equilibrium.

Complete answer:
Let the weak monoprotic acid have a general formula $$HA$$ consisting of hydrogen ions and an anionic part $$A^{-}$$ .
On dissolving $$HA$$ in water, the acid dissociates into its ions giving one mole of hydrogen ions (as it is a monobasic acid) and one mole of the anionic species per mole of the acid. The dissociation is partial in nature as a weak acid is classified as a weak electrolyte.
The degree or extent to which the acid dissociates is represented by $$\alpha$$ .
Initially, only undissociated acid is present and the concentration of ions is zero. When an equilibrium is established then the concentration of the acid reduces by a factor of and the ions are formed in equal proportions. The extent of dissociation remains the same i.e. an equal amount of acid is consumed to give the same amount of ions. This can be represented as follows:
$$HA\rightleftharpoons H^{+}(aq)+A^{-}(aq)$$

t=0	$c\alpha$	0	0
$t_{eq}$	$c(1-\alpha )$	$c\alpha$	$c\alpha$

The dissociation constant (an equilibrium constant) can be represented as a ratio of product concentrations of the product and the concentration of the reactant at equilibrium.
$$K_a={\displaystyle \frac{(C\alpha )(C\alpha )}{C(1-\alpha )}}={\displaystyle \frac{C{\alpha }^2}{1-\alpha }}$$
Now, the error allowed is a $$10\mathrm{\%}$$ error. Which makes the lowest value possible for the degree of dissociation to be $$\alpha =0.1$$ which can be inserted into the expression of dissociation constant.
$$K_a={\displaystyle \frac{C{(0.1)}^2}{(1-0.1)}}$$
Upon solving the above equation we get the relationship between the initial concentration and the dissociation constant, which can be written as follows:
$$C=90K_a$$
Hence, the correct option is (A) $$C=90K_a$$

Note:
Even though weak acids act as weak electrolytes, that does not mean that salts made up of weak acids are also weak electrolytes. If a neutralization reaction takes place between a weak acid and a strong base then the salt formed is a strong electrolyte that dissociates completely in water.