Calculate the degree of ionization of $0.05 \,M$ acetic acid if its $p K_{ a }$ value is $4.74$ 

How is the degree of dissociation affected when its solution also contains $(a)$ $0.01 \,M$ $(b)$ $0.1 \,M$ in $HCl$ ?

Vedclass pdf generator app on play store
Vedclass iOS app on app store

$c=0.05 \,M$

$p K_{a}=4.74$

$p K_{a}=-\log \left(K_{a}\right)$

$K_{a}=1.82 \times 10^{-5}$

$K_{a}=c \alpha^{2}$ $\alpha=\sqrt{\frac{K_{a}}{c}}$

$\alpha=\sqrt{\frac{1.82 \times 10^{-5}}{5 \times 10^{-2}}}=1.908 \times 10^{-2}$

When $HCI$ is added to the solution, the concentration of $H ^{+}$ ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.

Case $I:$ When $0.01 \,M$ $HCl$ is taken.

Let $x$ be the amount of acetic acid dissociated after the addition of $HCl$.

                                    $C{H_3}COOH\quad  \leftrightarrow \quad {H^ + }\quad  + \quad C{H_3}CO{O^ - }$

Initial conc.                      $0.05\,M$                          $0$                       $0$

After dissociation           $0.05-x$                     $0.01+x$          $x$

As the dissociation of a very small amount of acetic acid will take place, the values i.e., $0.05-x$ and $0.01+x$ can be taken as $0.05$ and $0.01$ respectively.

$K_{a}=\frac{\left[ CH _{3} COO ^{-}\right]\left[ H ^{+}\right]}{\left[ CH _{3} COOH \right]}$

$\therefore K_{a}=\frac{(0.01) x}{0.05}$

$x=\frac{1.82 \times 10^{-5} \times 0.05}{0.01}$

$x=1.82 \times 10^{-3} \times 0.05 \,M$

Now, $\alpha=\frac{\text { Amount of acid dissociated }}{\text { Amount of acid taken }}$

$=\frac{1.82 \times 10^{-3} \times 0.05}{0.05}$

$=1.82 \times 10^{-3}$

Case $II:$ When $0.1 \,M$ $HCl$ is taken.

Let the amount of acetic acid dissociated in this case be $X$. As we have done in the first case, the concentrations of various species involved in the reaction are:

$\left[ CH _{3} COOH \right]=0.05-X ; 0.05\, M$

$\left[ CH _{3} COO ^{-}\right]=X$

$\left[ H ^{+}\right]=0.1+X ; 0.1 \,M$

$K_{a}=\frac{\left[ CH _{3} COO ^{-}\right]\left[ H ^{+}\right]}{\left[ CH _{3} COOH \right]}$

$\therefore K_{a}=\frac{(0.1) X}{0.05}$

$x=\frac{1.82 \times 10^{-5} \times 0.05}{0.1}$

$x=1.82 \times 10^{-4} \times 0.05 \,M$

Now,  $\alpha=\frac{\text { Amount of acid dissociated }}{\text { Amount of acid taken }}$

$=\frac{1.82 \times 10^{-4} \times 0.05}{0.05}$

$=1.82 \times 10^{-4}$

Similar Questions

Find $pH$ of $5 \times 10^{-3}\, M$ $H_2CO_3$ solution having $10\%$ dissociation

When $100 \ mL$ of $1.0 \ M \ HCl$ was mixed with $100 \ mL$ of $1.0 \ M \ NaOH$ in an insulated beaker at constant pressure, a temperature increase of $5.7^{\circ} C$ was measured for the beaker and its contents (Expt. $1$). Because the enthalpy of neutralization of a strong acid with a strong base is a constant $\left(-57.0 \ kJ \ mol ^{-1}\right)$, this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. $2$), $100 \ mL$ of $2.0 \ M$ acetic acid $\left(K_a=2.0 \times 10^{-5}\right)$ was mixed with $100 \ mL$ of $1.0 M \ NaOH$ (under identical conditions to Expt. $1$) where a temperature rise of $5.6^{\circ} C$ was measured.

(Consider heat capacity of all solutions as $4.2 J g ^{-1} K ^{-1}$ and density of all solutions as $1.0 \ g mL ^{-1}$ )

$1.$ Enthalpy of dissociation (in $kJ mol ^{-1}$ ) of acetic acid obtained from the Expt. $2$ is

$(A)$ $1.0$ $(B)$ $10.0$ $(C)$ $24.5$ $(D)$ $51.4$

$2.$ The $pH$ of the solution after Expt. $2$ is

$(A)$ $2.8$ $(B)$ $4.7$ $(C)$ $5.0$ $(D)$ $7.0$

Give the answer question $1$ and $2.$

  • [IIT 2015]

Which oxychloride has maximum $pH$

The $pH$ of $ 0.1 \,M$  solution of a weak monoprotic acid  $1\%$  ionized is

Derive ${K_w} = {K_a} \times {K_b}$ and ${K_w} = p{K_a} \times p{K_b}$ for weak base $B$ and its conjugate acid ${B{H^ + }}$.