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Question: Calculate the dissociation constant of \({ NH }_{ 4 }{ OH }\) at \({ 298K }\), if \({ \Delta H }^{ -...

Calculate the dissociation constant of NH4OH{ NH }_{ 4 }{ OH } at 298K{ 298K }, if ΔH−{ \Delta H }^{ - } and ΔS−{ \Delta S }^{ - } for the given changes are as follows:
NH3+H+→NH4+{ NH }_{ 3 }{ +H }^{ + }{ \rightarrow NH }_{ 4 }^{ + };
ΔH−=−52.2kJmol−1;ΔS−=1.67JK−1mol−1{ \Delta H }^{ - } = { -52.2kJmol }^{ -1 }; { \Delta S }^{ - } = { 1.67JK }^{ -1 }{ mol }^{ -1 }
H2O⇌H++OH−{ H }_{ 2 }{ O }{ \rightleftharpoons H }^{ + }{ +OH }^{ - }
ΔH−=56.6kJmol−1;ΔS−=−76.53JK−1mol−1{ \Delta H }^{ - } = { 56.6kJmol }^{ -1 }; { \Delta S }^{ - } = { -76.53JK }^{ -1 }{ mol }^{ -1 }

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

Explanation

Solution

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{ 298K }
The following reactions are given;
NH3+H+→NH4+{ NH }_{ 3 }{ +H }^{ + }{ \rightarrow NH }_{ 4 }^{ + }.........(1)
and H2O⇌H++OH−{ H }_{ 2 }{ O }{ \rightleftharpoons H }^{ + }{ +OH }^{ - }........(2)
When we add equation 1 and 2, we get
NH4OH⇌NH4++OH−{ NH_{ 4 }OH\rightleftharpoons NH_{ 4 } }^{ + }{ +OH }^{ - }........(3)

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

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

- Now, put the values in equation 1{ 1 }, we get
ΔG∘=4.4−298K(−76.53)=−2.303RTlogKb{ \Delta G }^{ \circ } ={ 4.4 - 298K(-76.53) } = { -2.303RTlogK }_{ b }
4.4+298×76.3×10−3=−2.303×8.314×10−3×298×logKb{ 4.4 } + { 298\times 76.3\times 10 }^{ -3 } = { -2.303\times 8.314\times 10^{ -3 } }{ \times 298\times }{ logK }_{ b }
27.2=−2.303×8.314×10−3×298×log⁡Kb27.2=-2.303\times 8.314\times {{10}^{-3}}\times 298\times \log {{K}_{b}}
logKb=27.2÷(−2303)×8.314×10−3{ logK }_{ b } = { 27.2\div (-2303)\times 8.314\times 10^{ -3 } }
Kb=1.7×10−5{ 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 NH4OH{ NH }_{ 4 }{ OH } so, we have to add the given equations then we can calculate the enthalpy and entropy of NH4OH{ NH }_{ 4 }{ OH }.