Question

A monoprotic acid in a \[0.1M\] solution ionises to \[0.001\% \]. Its ionization constant is:
A. $1 \times {10^{ - 11}}$
B. $1 \times {10^{ - 3}}$
C. $1 \times {10^{ - 6}}$
D. $1 \times {10^{ - 8}}$

Answer 1

Hint: Monoprotic acid is an acid which can donate only one proton. For example, \[HCl\], \[HBr\], \[HN{O_3}\], \[C{H_3}COOH\] etc are all monoprotic acid. The monoprotic acid undergoes ionization to produce hydrogen ion and a conjugate base.

Complete step by step answer: The ionization of a monoprotic acid can be shown as
\[HA \to {H^ + } + {A^ - }\]
Let the initial concentration of the acid is \[C\]. Let the degree of dissociation for the acid \[HA\] be \[\alpha \]. Then the progress of the dissociation can be shown as
\[HA \to {H^ + } + {A^ - }\]
\[C\] \[0\] \[0\] (At the beginning)
\[C - C\alpha \] \[C\alpha \] \[C\alpha \]
In this case, the concentration of acid given is\[0.1M\],i.e. \[C = 0.1M\].
The degree of dissociation is \[0.001\% \], i.e. \[\alpha {\text{ }} = {\text{ }}0.001\% {\text{ }} = \dfrac{{0.001}}{{100}} = 1 \times {10^{ - 5}}\].
 Thus the ionization constant for the acid will be written as
${K_a} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}}$
Inserting the values of \[[{H^ + }]\], \[[{A^ - }]\] and \[[HA]\] in the equation,
\[{K_a} = \dfrac{{C\alpha \times C\alpha }}{{C - C\alpha }}\]
${K_a} = \dfrac{{[1 \times {{10}^{ - 5}}] \times [1 \times {{10}^{ - 5}}]}}{{0.1 - [0.1 \times 1 \times {{10}^{ - 5}}]}}$
${K_a} = 1 \times {10^{ - 11}}$.
So the option A is the correct answer, i.e. the ionization constant of the monoprotic acid is $1 \times {10^{ - 11}}$ .

Note: Unlike monoprotic acid the acid which can donate more than one proton during the ionization is called a polyprotic acid, i.e. diprotic for two, triprotic for three. In a similar way the monoprotic base is the base which can accept one proton. Normally diprotic acids are stronger than monoprotic acid, for example \[{H_2}S{O_4}\] is stronger than \[HCl\] due to the availability of two protons for donation. This in fact is clearer by comparing the \[pH\]values of the respective acids. The \[pH\] of \[{H_2}S{O_4}\] (2.75) is lower than \[pH\] of \[HCl\] (3.01) for a \[1mM\]solution. The acid acts as Brønsted acids.