Question

Value of dissociation constant of acetic acid is ${10}^{-6}$, whereas dissociation constant of formic acid is ${10}^{-5}$. Which of the following will be the value of $p{K}_a$(acetic acid) - $p{K}_a$(formic acid)?

Answer 1

Hint: Write down the dissociation of the base in water. Find the equilibrium constant for the dissociation process with the degree of dissociation. The formula for equilibrium constant is given below. $K_b$ will be equal to the equilibrium constant calculated.
Formula: ${\text{K}}_{\text{c}}\text{=}{\displaystyle \frac{\text{Concentration}{\text{n}}_{}\text{o}{\text{f}}_{}\text{products}}{\text{Concentration}{\text{n}}_{}\text{o}{\text{f}}_{}\text{reactants}}}$


Complete step-by-step answer:
We will write the dissociation of acetic acid as well as of formic acid.
Acetic acid : $\text{C}{\text{H}}_{\text{3}}\text{COOH }\to {\text{H}}^{\text{+}}\text{ + C}{\text{H}}_{\text{3}}\text{CO}{\text{O}}^{-}$
Formic acid : $\text{HCOOH }\to {\text{H}}^{\text{+}}\text{ + HCO}{\text{O}}^{-}$
The values of dissociation constant have already been given to us. The symbol for dissociation constant is $K_a$.
$p{K}_a$ stands for negative log of $K_a$.
We will now calculate the $p{K}_a$ of acetic acid and formic acid.
Acetic acid:
$K_a\text{ = 1}{\text{0}}^{-6}$
$p{K}_a\text{ = }-\text{log(}{K}_a)$ = 6
Formic acid:
$K_a\text{ = 1}{\text{0}}^{-5}$
$p{K}_a\text{ = }-\text{log(}{K}_a)$ = 5
The value of $p{K}_a$(acetic acid) - $p{K}_a$(formic acid) = 6 - 5 = 1.

Therefore, the correct answer is option (B).


Note: $p{K}_a$ numerically denotes the concentration of hydrogen ions in a solution. pH stands for potenz of hydrogen and indicates the amount of $H^{+}$ ions present. $p{K}_a$ is the pH of the solution when the concentration of acid is greater than the concentration of hydrogen ions in water i.e. ${10}^{-7}$. However, when the concentration of acid is lesser than that of acid, we have to include the concentration of hydrogen ions present in water as well in order to find the pH of the solution.