Question

The $pH$ of 0.1 M monobasic acid is 4.50. The acidity constant $({{K}_{a}})$ of the monobasic acid is:
A. $1.0\text{ }\times \text{ 1}{{\text{0}}^{-7}}$
B. $1.0\text{ }\times \text{ 1}{{\text{0}}^{-5}}$
C. $1.0\text{ }\times \text{ 1}{{\text{0}}^{-4}}$
D. $1.0\text{ }\times \text{ 1}{{\text{0}}^{-8}}$

Answer 1

Hint: $pH$means negative logarithm of hydrogen ion concentration.
‘mono’ – suggests one
Monobasic acid means an acid which has only one hydrogen ion to donate to a base in an acid-base reaction.
Acidity constant means the quantitative measure of the strength of acid in solution in an acid-base reaction.

Formula used:
As we know, $pH$ means negative logarithm of hydrogen ion concentration in an aqueous solution.
$\text{pH = - log}\left[ {{\text{H}}^{\text{+}}} \right]$
$\Rightarrow \left[ {{\text{H}}^{\text{+}}} \right]\text{ = antilog }\left( \text{-pH} \right)$
Acidity constant ,${{K}_{a}}=\dfrac{\left[ \text{hydrogen ion } \right]\left[ \text{conjugate base} \right]}{\left[ \text{acid} \right]}$
$=\text{ }\dfrac{\left[ {{\text{H}}^{\text{+}}} \right]\left[ {{\text{A}}^{-}} \right]}{\left[ \text{HA} \right]}\text{ }\left[ \text{where HA}\rightleftharpoons \text{ }{{\text{H}}^{+}}\text{ + }{{\text{A}}^{-}} \right]$

Complete step by step answer:
Given, $pH$= 4.50
So, $\text{pH = - log}\left[ {{\text{H}}^{\text{+}}} \right]$
$\Rightarrow \left[ {{\text{H}}^{\text{+}}} \right]\text{ = antilog }\left( \text{-pH} \right)$
$\Rightarrow \left[ {{\text{H}}^{\text{+}}} \right]\text{ = antilog }\left( -4.5 \right)$
={{10}^{(-4.5)}}$$$$=3.16\text{ }\times \text{ 1}{{\text{0}}^{-5}}
Now, $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{HA}\;\;\;\;\rightleftharpoons \;\;\; \text{ }{{\text{H}}^{+}}\;\;\;\text{ + }\;\;\;\;{{\text{A}}^{-}}$
Initial Conc. $\;\;\;\;\;\;\;0.1\;\;\;\;\;\;\;\;\;\;\;\;\;\;0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0$
At Equilibrium $\;\;(0.1-x)\;\;\;\;\;\;x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x$
(where, x = concentration of hydrogen formed or amount of acid dissociated)
x = \left[ {{\text{H}}^{\text{+}}} \right]$$$$=3.16\text{ }\times \text{ 1}{{\text{0}}^{-5}}
${{K}_{a}}=\text{ }\dfrac{\left[ {{\text{H}}^{\text{+}}} \right]\left[ {{\text{A}}^{-}} \right]}{\left[ \text{HA} \right]}=\dfrac{\text{x}\text{.x}}{(0.1-x)}$
$\text{= }\dfrac{\text{x}\text{.x}}{0.1}\text{ (}\because \text{0}\text{.1}\gg \text{x)}$
$\text{ =}\dfrac{{{(3.16\text{ }\times \text{ 1}{{\text{0}}^{-5}})}^{2}}}{0.1}$
$\begin{aligned} & \text{=0}\text{.998 }\times \text{ 1}{{\text{0}}^{-8}} \\\ & \simeq \text{1 }\times \text{ 1}{{\text{0}}^{-8}} \\\ \end{aligned}$
So, acidity constant of the monobasic acid is $\text{1 }\times \text{ 1}{{\text{0}}^{-8}}\text{ }$
So, the correct option is D.

Additional information:
$pH$ scale ranges from 0 to 14. $pH$of less than 7 is acidic and greater than 7 is basic. $pH$of neutral medium is 7.
Equilibrium constant $\text{(}{{\text{K}}_{\text{eq}}}\text{)}$ is the ratio of product concentration to the reactant concentration at equilibrium condition while concentration coefficient ($\text{Q}$) is the ratio of product concentration to the reactant concentration at any concentration.

Note: While calculating constant value, we should not take concentration of any solid or liquid as their concentration does not change throughout the reaction. It is considered to be 1. Always give attention to the basicity of the acid. If it is a dibasic acid, upon dissociation it gives two hydrogen ions for donation and so on for tribasic acid and so.