Question

The dissociation constants of two acids $H{A_1}$ and $H{A_2}$ are $3.0 \times {10^{ - 4}}$ and $1.8 \times {10^{ - 5}}$ respectively. The relative strengths of acid will be:
A. $1:4$
B. $4:1$
C. $1:16$
D. $16:1$

Answer 1

${K_a}$ is the acid dissociation constant and it is the equilibrium constant for the dissociation of acid. Whenever the compound has the tendency to readily lose its proton by dissociation, then it indicates that the compound is highly acidic. $p{K_a}$is the negative logarithm of acid dissociation constant i.e.) $p{K_a} = - \log \left( {{K_a}} \right)$. Thus, $p{K_a}$ value decreases with increase in the acid dissociation constant. The strength of acid will be easily analysed by increase in ${K_a}$ value and decrease in $p{K_a}$ value.

Complete step by step answer:
It is known that dissociation of acid represents the release of a proton from the given acid. Therefore, the strength of the acid is easily determined by the dissociation constant.
The strength of an acid is directly proportional to the square root of their dissociation constants. The dissociation constants of two acids $H{A_1}$ and $H{A_2}$ are given as $3.0 \times {10^{ - 4}}$ and $1.8 \times {10^{ - 5}}$ respectively.
The ratio of relative strength of two acids can be calculated as,
$\dfrac{{Acid\,strength[H{A_1}]}}{{Acid\,strength[H{A_2}]}} = \dfrac{{\sqrt {K{a_{[H{A_1}]}}} }}{{\sqrt {K{a_{[H{A_2}]}}} }}$
$\Rightarrow \dfrac{{Acid\, strength[H{A_1}]}}{{Acid\,strength[H{A_2}]}} = \dfrac{{\sqrt {3.0 \times {{10}^{ - 4}}} }}{{\sqrt {1.8 \times {{10}^{ - 5}}} }}$
$\Rightarrow \dfrac{{1.73}}{{0.424}}$
$\Rightarrow 4.80$
Thus, the relative strength of two acids is in the ratio of $4.80:1 \simeq 4:1$

So, the correct answer is Option B.

Note: Similar to acid dissociation, the dissociation constant for the base can be written as ${K_b}$. $p{K_b}$ is the negative logarithm of the base dissociation constant. Higher the ${K_b}$ the greater will be the basicity of the compound. Also, the lower the greater will be the basicity of the compound. The equation which relates $p{K_a}$ and $p{K_b}$ can be written as,
$p{K_a} + p{k_b} = 14$
Thus, if one value of the dissociation constant is known, then other dissociation constants can be calculated. Ionic product of water ${K_W}$ is defined as the product of the molar concentration of hydroxyl ion and hydrogen ion concentration at constant temperature and its $p{K_W}$ is the sum of pH and pOH. Thus, if one of concentration {hydrogen ion/hydroxyl ion} is known, then ${K_W}$ of water is calculated since for pure water, hydroxyl ion and hydrogen ion concentration must be equal.