Question

: If the degree of ionization is $0.01$ of decimolar solution of weak acid HA, then ${\text{p}}{{\text{K}}_{\text{a}}}$of acid is
A. $2$
B. $3$
C. $5$
D.$7$

Answer 1

Hint:The dissociation constant is determined as the product of the square of the degree of dissociation and concentration. We will use the dissociation constant formula to determine the dissociation constant. After determining the dissociation constant we will take its negative logarithm to determine the ${\text{p}}{{\text{K}}_{\text{a}}}$of acid.


Formula used: ${{\text{K}}_{\text{a}}}{\text{ = }}\,\dfrac{{{\text{C}}{{\alpha }^{\text{2}}}}}{{{\text{1}} - {\alpha }}}$


Complete solution:
The formula which relates the degree of dissociation with dissociation constant is as follows:
 ${{\text{K}}_{\text{a}}}{\text{ = }}\,\dfrac{{{\text{C}}{{\alpha}^{\text{2}}}}}{{{\text{1}} - {\alpha }}}$
Where,
${{\text{K}}_{\text{a}}}$is the acid dissociation constant.
C is the concentration.
${\alpha}$is the degree of dissociation
Decimolar solution means the concentration of the solution is \[0.1\]M.
On substituting $0.1$M for the concentration and $0.01$ for degree of dissociation.
${{\text{K}}_{\text{a}}}{\text{ = }}\,\dfrac{{0.1\, \times {{{\text{(0}}{\text{.01)}}}^{\text{2}}}}}{{{\text{1}} - 0.01}}$
${{\text{K}}_{\text{a}}}{\text{ = }}\,\dfrac{{{\text{1}} \times {\text{1}}{{\text{0}}^{- 5}}}}{{0.99}}$
${{\text{K}}_{\text{a}}}{\text{ = }}\,{\text{1}} \times {\text{1}}{{\text{0}}^{- 5}}$
The relation between ${{\text{K}}_{\text{a}}}$ and ${\text{p}}{{\text{K}}_{\text{a}}}$ is as follows:
${\text{p}}{{\text{K}}_{\text{a}}}\, = - {\text{log}}\,{{\text{K}}_{\text{a}}}$
Where,
${\text{p}}{{\text{K}}_{\text{a}}}$is the negative log of acid dissociation constant.
On substituting ${\text{1}} \times {\text{1}}{{\text{0}}^{ - 5}}$ for${{\text{K}}_{\text{a}}}$.
${\text{p}}{{\text{K}}_{\text{a}}}\, = - {\text{log}}\,\left( {{\text{1}} \times {\text{1}}{{\text{0}}^{ - 5}}} \right)$
$\,{\text{p}}{{\text{K}}_{\text{a}}}\,{\text{ = }}\,{\text{5}}$
So, the ${\text{p}}{{\text{K}}_{\text{a}}}$of acid is $5$.

Therefore, option (C) is correct.

Additional information: By using the acid dissociation constant we can determine the base dissociation constant. The relation between acid dissociation constant and base dissociation constant is,
${{\text{K}}_{\text{a}}}{\times }\,\,{{\text{K}}_{\text{b}}}\,{\text{ = }}\,{{\text{K}}_{\text{w}}}$
Where,
${{\text{K}}_{\text{a}}}$ is the acid dissociation constant
${{\text{K}}_{\text{b}}}$ is the base dissociation constant
${{\text{K}}_{\text{w}}}$ is the ionic product of water whose value is ${10^{ - 14}}$.
A similar formula is used for the determination of the degree of dissociation for a weak base only the${{\text{K}}_{\text{a}}}$is replaced with${{\text{K}}_{\text{b}}}$where, ${{\text{K}}_{\text{b}}}$is the base dissociation constant. The degree of dissociation can also be calculated by using equivalent conductance at a concentration and equivalent conductance at infinite dilution. The formula to calculate the degree of dissociation is $\alpha = \,\dfrac{{{\lambda _m}}}{{\lambda _m^ \circ }}$. Where, ${\lambda _m}$is the equivalent conductance at a concentration and $\lambda _m^ \circ $is the equivalent conductance at infinite dilution.


Note:Mostly weak electrolyte dissociates very less so, for weak electrolyte the degree of dissociation is very-very less than one,${\alpha < < 1}$ so, the value of \[{\text{1}} - {\alpha }\] can be taken to equal to $1$ so, the formula of the dissociation constant can be reduced as, ${{\text{K}}_{\text{a}}}{\text{ = }}\,{\text{C}}{{\alpha }^{\text{2}}}$.
${{\text{K}}_{\text{a}}}{\text{ = }}\,0.1\, \times {{\text{(0}}{\text{.01)}}^{\text{2}}}$
${{\text{K}}_{\text{a}}}{\text{ = }}\,{\text{1}} \times {\text{1}}{{\text{0}}^{ - 5}}$
The degree of dissociation tells the dissociated amount of the weak electrolyte. The strong electrolyte dissociates completely, so it is not calculated for the strong electrolyte.