Chapter 22, Problem 27QAP

When ethylamine, a weak base $\left(K_b=4.3\times {10}^{-4}\right)$ , reacts with formic acid, a weak acid $\left(K_a=1.8\times {10}^{-4}\right)$ , the following reaction takes place:
${\text{CH}}_3{\text{CH}}_2{\text{NH}}_2\left(aq\right)+\text{HCOOH}\left(aq\right)\rightleftharpoons {\text{CH}}_3{\text{CH}}_2{\text{NH}}_3{}^{+}\left(aq\right)+{\text{HCOO}}^{-}\left(aq\right)$ Calculate K for this reaction.

Interpretation Introduction

Interpretation:

The K of the given reaction is to be calculated.

Concept introduction:

The relationship between the concentration of products and reactants at equilibrium for a general reaction:

$\text{aA}\text{+}\text{bB}\rightleftarrows \text{cC}\text{+}\text{dD}$

Where A, B, C, and D represents chemical species and a, b, c, and d are the coefficients for balanced reaction.

The equilibrium expression, K_c for reversible reaction is determined by multiplying the concentrations of products together and divided by the concentrations of the reactants. Every concentration term is raised to the power that is equal to the coefficient in the balanced reaction. So, the expression is:

${\text{K}}_{\text{c}}\text{=}\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}}$

pH scale is a type of measurement of H+ ion concentration in the solution. It is taken as a negative log of H+ ion concentration as:

$pH=-{\log }_{1o}[H^{+}]$

For the equilibrium conditions:

$k_a\times k_b=k_w={10}^{-14}$

Where, k_a is acid dissociation constant, k_b is base dissociation constant and k_w is ionic product of water whose value is fixed as 10-14.

Answer to Problem 27QAP

The required value of $K=7.74\times {10}^{-7}$.

Explanation of Solution

Given Information:

The weak base ethylamine $C_2{H}_{5}N{H}_2(k_b=4.3\times {10}^{-3})$ reacts with a weak acid formic acid $HCOOH(k_a=1.8\times {10}^{-4})$.

The reaction is:

$C{H}_{3}C{H}_{2}N{H}_2(aq)+HCOOH(aq)\rightleftharpoons C{H}_{3}C{H}_{2}N{H}_3^{+}(aq)+HCO{O}^{-}(aq)$

The ethylamine dissociates as;

$C_2{H}_{5}N{H}_2(aq)+H_{2}O\rightleftharpoons C_2{H}_{5}N{H}_3^{+}(aq)+O{H}^{-}(aq)$

The equilibrium constant expression will be:

${\text{K}}_b\text{=}\frac{[C_2{H}_{5}N{H}_3^{+}][O{H}^{-}]}{[C_2{H}_{5}N{H}_2]}$

Further the formic acid dissociates as:

$HCOOH(aq)+H_{2}O(l)\rightleftharpoons H_3{O}^{+}(aq)+HCO{O}^{-}(aq)$

Hence, we have:

The equilibrium constant expression will be:

${\text{K}}_a\text{=}\frac{[H_3{O}^{+}][HCO{O}^{-}]}{[HCOOH]}$

Now both the reactions are combined as:

$C_2{H}_{5}N{H}_2(aq)+HCOOH(aq)\rightleftharpoons C_2{H}_{5}N{H}_3^{+}(aq)+HCO{O}^{-}(aq)$

The equilibrium constant expression will be:

$\text{K=}\frac{[C_2{H}_{5}N{H}_3^{+}][HCO{O}^{-}]}{[C_2{H}_{5}N{H}_2][HCOOH]}$

The expression can be rearranged as:

$\begin{array}{l}\text{K=}\frac{[C_2{H}_{5}N{H}_3^{+}][HCO{O}^{-}][H_3{O}^{+}]}{[C_2{H}_{5}N{H}_2][HCOOH][H_3{O}^{+}]}=\frac{[HCO{O}^{-}][H_3{O}^{+}]}{[HCOOH]}\times \frac{[C_2{H}_{5}N{H}_3^{+}]}{[C_2{H}_{5}N{H}_2][H_3{O}^{+}]}=k_a\frac{[C_2{H}_{5}N{H}_3^{+}]}{[C_2{H}_{5}N{H}_2][H_3{O}^{+}]}\\ \text{K=}{k}_a\times k_b\end{array}$

Putting the values we get:

$\begin{array}{c}\text{K=}{k}_a\times k_b=1.8\times {10}^{-4}\times 4.3\times {10}^{-3}\\ K=7.74\times {10}^{-7}\end{array}$

Conclusion

The required value of $K=7.74\times {10}^{-7}$.