Question

What is the concentration of $C{H_3}COOH$ which can be added to $0.5\,\,M\,\,HCOOH$ solution so that dissociation of both is the same.
[Given that: ${K_{C{H_3}COOH}} = 1.8 \times {10^{ - 5}}$, ${K_{HCOOH}} = 2.4 \times {10^{ - 4}}$]
(as nearest integer)

Answer 1

Dissociation is a process in chemistry in which an ionic compound or complex split down into simpler particles such as atoms, radicals or ions. Dissociation constant is represented by $Kd$ and it is the ratio of dissociated compound to the undissociated compound.

Complete step by step answer:
Given, molarity concentration of $HCOOH = 0.5M$
${K_{C{H_3}COOH}} = 1.8 \times {10^{ - 5}}$
${K_{HCOOH}} = 2.4 \times {10^{ - 4}}$
Concentration of $C{H_3}COOH$ is to be found.
$C{H_3}COOH$ is acetic acid and $HCOOH$ is formic acid.
Dissociation of $C{H_3}COOH$:
$C{H_3}COOH \rightleftharpoons {H^ + } + C{H_3}CO{O^ - }$
Dissociation of $HCOOH$
$HCOOH \rightleftharpoons {H^ + } + HCO{O^ - }$
As given in the question, the dissociation constant of both $C{H_3}COOH$ and $HCOOH$ is the same.
Therefore,
$\left[ {C{H_3}CO{O^ - }} \right] = \left[ {HCO{O^ - }} \right]$
Also, ${C_1}\,{\alpha _1} = {C_2}\,{\alpha _2}$
As we know, $\alpha = \sqrt {\dfrac{{{K_a}}}{C}}$
Therefore, $\sqrt {{K_{a\,1}}{C_1}} = \sqrt {{K_{a\,2}}{C_2}}$
On squaring both sides, we get
{K_{a\,1}}{C_1}$$$$ = $$$${K_{a\,2}}{C_2}
Now putting the values,
$1.8 \times {10^{ - 5}} \times C = 2.4 \times {10^{ - 4}} \times 0.5$
On rearranging for concentration $C$,
$C = \dfrac{{2.4 \times {{10}^{ - 4}} \times 0.5}}{{1.8 \times {{10}^{ - 5}}}}$
$C = 6.67M$
Hence, concentration of $C{H_3}COOH$ required is $6.67M$

Additional Information:
-When an acid is dissociated in water, its covalent bond and hydrogen bonds breaks by heterolytic fission and give proton ($s$) i.e. ${H^ + }$and an anion.
-In heterolytic fission, the more electronegative element pulls or attracts the shared pair of electrons towards itself and thus gain electrons towards itself and thus gain electrons and become anion. Whereas, the electropositive elements form the cation. The energy required for this process is known as heterolytic bond dissociation energy.

Note: $\alpha$ is the degree of dissociation. It is the fraction of dissociated solute molecules into ions or radicals per mole. For strong acids and bases, the degree of dissociation is high, that is, close to $1$. Whereas for weak acids and bases degree of dissociation is low.
-${K_a}$ is acid ionization constant and it indicates the strength of the acid. Strong acids and weak acids have high and low values of ${K_a}$ respectively. ${K_a}$ is inverse of $p{K_a}$, therefore strong and weak acids have low and high value of $p{K_a}$ respectively.