Question

Calculate the [OH] of ${\left[ {{\rm{N}}{{\rm{H}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right]^ + }$ and ${\left[ {{{\rm{H}}_{\rm{3}}}{\rm{N}} - {{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right]^{2 + }}$ in 0.15 M ethylene diamine (aq) if

$NH_{2}C_{2}H_{4}NH_{2}+H_2O\rightleftharpoons NH_{2}C_{2}H_{4}NH_{3}^{+}+OH^{-}\left ( K_{1}=8.5\times 10^{-5} \right )$

$NH_{2}C_{2}H_{4}NH_{3}^{+}+H_2O\rightleftharpoons NH_{3}C_{2}H_{4}NH_{3}^{2+}+OH^{-}\left ( K_{2}=2.7\times 10^{-8} \right )$

A. Case I: $3.57 \times {10^{ - 3}}\,{\rm{M}}$ and Case II: $2.7 \times {10^{ - 8}}{\rm{M}} = {K_{{b_2}}}$

B. Case I: $4.57 \times {10^{ - 3}}\,{\rm{M}}$ and Case II: $3.7 \times {10^{ - 8}}{\rm{M}} = {K_{{b_2}}}$

C. Case I: $5.57 \times {10^{ - 3}}\,{\rm{M}}$ and Case II: $4.7 \times {10^{ - 8}}{\rm{M}} = {K_{{b_2}}}$

D. Case I: $6.57 \times {10^{ - 3}}\,{\rm{M}}$ and Case II: $5.7 \times {10^{ - 8}}{\rm{M}} = {K_{{b_2}}}$

Answer 1

Hint: We know that, the law of chemical equilibrium states that at a given temperature, a chemical system might reach a state in which a particular ratio of reactant and product concentration has a constant value. This constant is termed as equilibrium constant $\left( {{K_c}} \right)$.

Complete step by step answer:
The reaction in case 1 is,

$NH_{2}C_{2}H_{4}NH_{2}+H_2O\rightleftharpoons NH_{2}C_{2}H_{4}NH_{3}^{+}+OH^{-}\left ( K_{1}=8.5\times 10^{-5} \right )$


So, the concentration of $\left[ {{\rm{N}}{{\rm{H}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\mathop {\rm{N}}\limits^ + {{\rm{H}}_{\rm{3}}}} \right]\,\,$ and $\left[ {{\rm{O}}{{\rm{H}}^ - }} \right]$ is equal.

Therefore, the equilibrium constant expression for the above reaction is,

\[{K_1} = \dfrac{{\left[ {{\rm{N}}{{\rm{H}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{N}}{{\rm{H}}_{\rm{3}}}^ + } \right]\left[ {{\rm{O}}{{\rm{H}}^ - }} \right]}}{{\left[ {{\rm{N}}{{\rm{H}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{N}}{{\rm{H}}_{\rm{2}}}} \right]}}\]…… (1)


As the concentration of $\left[ {{\rm{N}}{{\rm{H}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\mathop {\rm{N}}\limits^ + {{\rm{H}}_{\rm{3}}}} \right]\,\,$ and $\left[ {{\rm{O}}{{\rm{H}}^ - }} \right]$ is equal. So, we can write equation (1) as,
\[{K_1} = \dfrac{{\left[ {{\rm{O}}{{\rm{H}}^ - }} \right]\left[ {{\rm{O}}{{\rm{H}}^ - }} \right]}}{{\left[ {{\rm{N}}{{\rm{H}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{N}}{{\rm{H}}_{\rm{2}}}} \right]}}\]

\[ \Rightarrow {K_1} = \dfrac{{{{\left[ {{\rm{O}}{{\rm{H}}^ - }} \right]}^2}}}{{\left[ {{\rm{N}}{{\rm{H}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{N}}{{\rm{H}}_{\rm{2}}}} \right]}}\]

$ \Rightarrow \left[ {{\rm{O}}{{\rm{H}}^ - }} \right] = \sqrt {{K_1}\left[ {{\rm{N}}{{\rm{H}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{N}}{{\rm{H}}_{\rm{2}}}} \right]} $…… (2)

Given the value of ${K_1} = 8.5 \times {10^{ - 5}}$ and $\left[ {{\rm{N}}{{\rm{H}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{N}}{{\rm{H}}_{\rm{2}}}} \right] = 0.15\,{\rm{M}}$
.

Now, we have to put the above values in equation (2) to calculate hydroxide concentration.


$ \Rightarrow \left[ {{\rm{O}}{{\rm{H}}^ - }} \right] = \sqrt {8.5 \times {{10}^{ - 5}} \times 0.15} = 3.57 \times {10^{ - 3}}$


Now, for the second case reaction given is,

$NH_{2}C_{2}H_{4}NH_{3}^{+}+H_2O\rightleftharpoons NH_{3}C_{2}H_{4}NH_{3}^{2+}+OH^{-}\left ( K_{2}=2.7\times 10^{-8} \right )$

Now, we have to write an ICE Table for the above reaction.

	$NH_{2}C_{2}H_{4}NH_{3}^{+}+H_2O\rightleftharpoons NH_{3}C_{2}H_{4}NH_{3}^{2+}+OH^{-}\left ( K_{2}=2.7\times 10^{-8} \right )$
Initial	$3.57 \times {10^{ - 3}}$		0	$3.57 \times {10^{ - 3}}$
Equilibrium	$3.57 \times {10^{ - 3}} - x$		x	$3.57 \times {10^{ - 3}} + x$

Now, the equilibrium constant for the second case is,

${K_2} = \dfrac{{x\left( {3.57 \times {{10}^{ - 3}} + x} \right)}}{{3.57 \times {{10}^{ - 3}} - x}}$


Given $\,\,{K_2} = 2.7 \times {10^{ - 8}}$
.

$\,\,2.7 \times {10^{ - 8}} = \dfrac{{x\left( {3.57 \times {{10}^{ - 3}} + x} \right)}}{{3.57 \times {{10}^{ - 3}} - x}}$

$\,\,2.7 \times {10^{ - 8}} = \dfrac{{{x^2} + 3.57 \times {{10}^{ - 3}}x}}{{3.57 \times {{10}^{ - 3}} - x}}$


As x is very small, we have to neglect ${x^2}$ and $3.57 \times {10^{ - 3}} - x = 3.57 \times {10^{ - 3}}$.

$\,\,2.7 \times {10^{ - 8}} = \dfrac{{3.57 \times {{10}^{ - 3}}x}}{{3.57 \times {{10}^{ - 3}}}}$

$\, \Rightarrow x = 2.7 \times {10^{ - 8}}$


Therefore, the concentration of ${\left[ {{{\rm{H}}_{\rm{3}}}{\rm{N}} - {{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right]^{2 + }}$
is $2.7 \times {10^{ - 8}} = {K_{{b_2}}}$

So, the correct answer is Option A.

Additional Information:
Let’s learn about equilibrium constant in detail. Equilibrium constant helps in quantitative measurement of strength of bases and acids in water.

${K_a}$ is acid constant for acidic reaction.

$HA+H_2O\rightleftharpoons H_3O^{+}+A^{-}$

For this reaction, acidic constant is,

${K_a} = \dfrac{{\left[ {{{\rm{H}}_{\rm{3}}}{{\rm{O}}^ + }} \right]\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]}}$


${K_b}$ is the basic constant for equilibrium reaction.

$BOH+H_2O\rightleftharpoons B^{+}+OH^{-}$

${K_b} = \dfrac{{\left[ {{{\rm{B}}^ + }} \right]\left[ {{\rm{O}}{{\rm{H}}^ - }} \right]}}{{\left[ {{\rm{BOH}}} \right]}}$

Note: Always remember that, in equilibrium constant expression only gaseous reactants and products are considered. The value of equilibrium constant is dependent on temperature. Its value is independent of the actual amount of product and reactant, presence of catalyst and presence of inert material, pressure and volume of reactants and products.