Question

Calculate the dissociation constant of ${ NH }_{ 4 }{ OH }$ at ${ 298K }$, if ${ \Delta H }^{ - }$ and ${ \Delta S }^{ - }$ for the given changes are as follows:
${ NH }_{ 3 }{ +H }^{ + }{ \rightarrow NH }_{ 4 }^{ + }$;
${ \Delta H }^{ - } = { -52.2kJmol }^{ -1 }; { \Delta S }^{ - } = { 1.67JK }^{ -1 }{ mol }^{ -1 }$
${ H }_{ 2 }{ O }{ \rightleftharpoons H }^{ + }{ +OH }^{ - }$
${ \Delta H }^{ - } = { 56.6kJmol }^{ -1 }; { \Delta S }^{ - } = { -76.53JK }^{ -1 }{ mol }^{ -1 }$

a.) ${ { K }_{ b }=1.7\times 10 }^{ -5 }$
b.) ${ { K }_{ b }=1.7\times 10 }^{ -3 }$
c.) ${ { K }_{ b }=1.7\times 10 }^{ -1 }$
d.) ${ { K }_{ b }=3.4\times 10 }^{ -5 }$

Answer 1

Hint: To solve this question first we have to understand the meaning of dissociation constant. A dissociation constant is a specific type of equilibrium constant that measures the propensity of a larger object to separate (dissociate) reversibly into smaller components, as when a complex falls apart into its component molecules, or when a salt splits up into its component ions. The dissociation constant is the inverse of the association constant.

Complete Solution :
Given in the question:
Temperature, T = ${ 298K }$
The following reactions are given;
${ NH }_{ 3 }{ +H }^{ + }{ \rightarrow NH }_{ 4 }^{ + }$.........(1)
and ${ H }_{ 2 }{ O }{ \rightleftharpoons H }^{ + }{ +OH }^{ - }$........(2)
When we add equation 1 and 2, we get
${ NH_{ 4 }OH\rightleftharpoons NH_{ 4 } }^{ + }{ +OH }^{ - }$........(3)

As we know that:
${ \Delta H }^{ - } = { -52.2kJmol }^{ -1 } for equation (1) and { \Delta H }^{ - } = { 56.6kJmol }^{ -1 }$ for dissociation of water
where ${ \Delta H }$ =change in enthalpy in a reaction
 so when we add these two equations, we will get
${ \Delta H }^{ \circ } = { (-52.2) +56.6 } ={ 4.4kJ mol }^{ -1 }$

For equation ${ 1 }$, ${ \Delta S }^{ - } = { 1.67JK }^{ -1 }{ mol }^{ -1 }$
and for equation ${ 2 }$, ${ \Delta S }^{ - } = { -78.2JK }^{ -1 }{ mol }^{ -1 }$
where, ${ \Delta S }^{ \circ }$ = change in entropy
So, when we add these values, we will get ${ \Delta S }$ for equation 3,
${ \Delta S }^{ \circ } = { 1.67 + (-78.2) } = { -76.53JK }^{ -1 }{ mol }^{ -1 }$
As we know that, ${ \Delta G }^{ \circ }{ =\Delta H }^{ \circ }{ -T\Delta S }^{ \circ } { =-2.303RTlogK }_{ b }$……(1)
where, ${ \Delta G }^{ \circ }$ = change in free energy
R = Universal gas constant
T = Temperature
K = Dissociation constant

- Now, put the values in equation ${ 1 }$, we get
 ${ \Delta G }^{ \circ } ={ 4.4 - 298K(-76.53) } = { -2.303RTlogK }_{ b }$
${ 4.4 } + { 298\times 76.3\times 10 }^{ -3 } = { -2.303\times 8.314\times 10^{ -3 } }{ \times 298\times }{ logK }_{ b }$
\[27.2=-2.303\times 8.314\times {{10}^{-3}}\times 298\times \log {{K}_{b}}\]
${ logK }_{ b } = { 27.2\div (-2303)\times 8.314\times 10^{ -3 } }$
${ K }_{ b }{ =1.7\times 10 }^{ -5 }$
So, the correct answer is “Option A”.

Note: The possibility to make a mistake is that we have to calculate the dissociation constant of ${ NH }_{ 4 }{ OH }$ so, we have to add the given equations then we can calculate the enthalpy and entropy of ${ NH }_{ 4 }{ OH }$.