Question

Calculate the degree of ionization of 0.05 M acetic acid if its $pK_{a}$ value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0. 01 M (b) 0. 1M in HCl?

Answer 1

Given

C =  0.05 M

$p{ K }_{ a }\quad =\quad 4.74$

$p{ K }_{ a }\quad =\quad -\log { \left( { K }_{ a } \right) }$

${ K }_{ a }\quad =\quad 1.82\quad \times \quad { 10 }^{ -5 }$

We know that,

${ K }_{ a }\quad =\quad c{ \alpha }^{ 2 }$

$\alpha \quad =\quad \sqrt { \frac { { K }_{ a } }{ c } }$

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

$\alpha \quad =\quad 1.908\quad \times \quad { 10 }^{ -2 }$

Where α is the dissociation constant. When HCl is added drop wise the concentration of H+ ions increases and shifts the equilibrium to backward direction.  

Therefore, the dissociation of acetic acid decreases.

i) When 0.01 M HCl is taken

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

$C{ H }_{ 3 }COOH\quad \leftrightarrow \quad { H }^{ + }\quad +\quad { CH }_{ 3 }{ COO }^{ - }$

$\begin{matrix} Initial\quad Concentration \\ \quad \\ After\quad Dissociation \end{matrix}\quad \begin{matrix} 0.05M \\ \quad \\ 0.05-x \end{matrix}\quad \begin{matrix} 0 \\ \quad \\ 0.01+x \end{matrix}\quad \begin{matrix} 0 \\ \quad \\ x \end{matrix}$

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.

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

$\therefore { K }_{ a }\quad =\quad \frac { \left( 0.01 \right) \left( x \right) }{ \left( 0.05 \right) }$

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

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

Now,

$\alpha \quad =\quad \frac { Amount\quad of\quad acid\quad dissociated }{ Amount\quad of\quad acid\quad taken }$

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

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

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:

$\left[ { CH }_{ 3 }COOH \right] \quad =\quad 0.05-x\quad :\quad 0.05M$

$\left[ { CH }_{ 3 }{ COO }^{ - } \right] \quad -\quad x$

$\left[ { H }^{ + } \right] \quad =\quad 0.1\quad +\quad x;\quad 0.1M$

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

$\therefore { K }_{ a }\quad =\quad \frac { \left( 0.1 \right) \left( x \right) }{ \left( 0.05 \right) }$

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

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

Now,

$\alpha \quad =\quad \frac { Amount\quad of\quad acid\quad dissociated }{ Amount\quad of\quad acid\quad taken }$

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

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

Answer 2

Case I: When 0.01 M HCl is taken.

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

                               CH3COOH     ↔     H+    +   CH3COO-

Initial Conc.             0.05M                   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.

Ka  =  [CH3COO-] [ H+]  / [CH3COOH]

∴ Ka  = (0.01)  (x) / (0.05)

x  =   1.82 x 10-5  x(multiply) 0.05 /  0.01

x  =  1.82 x 10-3  x(multiply) 0.05

Now

α =  Amount of acid dissociated /  amount of acid taken

= 1.82 x 10-3  x 0.05  / 0.05

= 1.82 x 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:

[CH3COOH]  = 0.05 - X  : 0.05M

[CH3COO-] =  X

[ H+]   =   0.1+X ; 0.1M

Ka  =  [CH3COO-] [ H+]  / [CH3COOH]

∴ Ka  = (0.1)  (X) / (0.05)

x  =   1.82 x 10-5  x(multiply) 0.05 /  0.1

x  =  1.82 x 10-4  x(multiply) 0.05

Now

α =  Amount of acid dissociated /  amount of acid taken

= 1.82 x 10-4  x 0.05  / 0.05

= 1.82 x 10-4