Question

$p{K_a}$ values of four acids are given below at ${25^ \circ }C$ . Indicate the strongest acid.
A. 2.0
B. 3.0
C. 2.5
D. 4.0

Answer 1

$p{K_a}$ value decides the concentration of the acid. As its value increases, ${K_a}$ decreases which means acid is less strong. If the value of $p{K_a}$ is low means acid is strong.

Complete step by step answer:
Here, we first need to know what $p{K_a}$ value is. Firstly, we need to know about ${K_a}$ value.
${K_a}$ is also called as acid dissociation constant or acidity constant or acid-ionization constant. It is a quantitative measure of the strength of an acid in solution. It is the equilibrium constant for a chemical reaction known as dissociation. The dissociation constant formula is given as follows.
$p{K_a} = - {\log _{10}}{K_a}$
It is the negative log of the acid dissociation constant or ${K_a}$ value. An acid is classified as strong acid when the concentration of its undissociated species is too low to be measured. Any aqueous acid with a $p{K_a}$ value of less than 0 is almost completely deprotonated and is considered a strong acid. A lower $p{K_a}$ value indicated a stronger acid. Lower value indicates that acid fully dissociates in water.
The value of $p{K_a}$ also depends on the molecular structure of the acid in many ways. For example, Pauling proposed two rules, one for successive $p{K_a}$ of polyprotic acids and one to estimate the $p{K_a}$ of oxyacids based on the number of oxygen groups. Other structural factors that influence the magnitude of the acid dissociation constant include inductive effects, mesomeric effects and hydrogen bonding.
Now, moving on to find the solution of the question. We need to find the ${K_a}$ values form given $p{K_a}$ values in the options by substituting it in the equation given above.
Substituting $p{K_a} = 2$
$2 = - \log ({K_a})$
$\Rightarrow {K_a} = 0.01$
Substituting $p{K_a} = 3$
$3 = - \log ({K_a})$
$\Rightarrow {K_a} = 0.001$
Substituting $p{K_a} = 2.5$
$2.5 = - \log ({K_a})$
$\Rightarrow {K_a} = 0.00316$
Substituting $p{K_a} = 4$
$4 = - \log ({K_a})$
$\Rightarrow {K_a} = 0.0001$

As $p{K_a}$ value increases, ${K_a}$ value decreases and the acid becomes less strong. Hence, smaller the $p{K_a}$ value larger the ${K_a}$ value and stronger the acid. So, according to the above substitutions we get to know that the smallest $p{K_a}$ value is 2 and it has higgest ${K_a}$ value is 0.01.

**Therefore, the correct answer is option A.

Note: **
$p{K_a}$ value decides the concentration of the acid. Higher the $pK_a$, lower is the dissociation of the acid to give $H^+$ ions, lower is its acidic strength.