Chapter 18, Problem 72A

Interpretation Introduction

Interpretation:

The ionization equations and base ionization expressions for the given reaction is to be written.

Concept introduction:

Base ionization constant tells about equilibrium that whether reactants are favored or products.

For a reaction,

$BOH\rightleftharpoons B^{+}+O{H}^{-}$

Base ionization constant is given by:

$\begin{array}{l}{k}_b=\frac{\left[B^{+}\right]\times \left[O{H}^{-}\right]}{\left[BOH\right]}\\ where,\\ \left[B^{+}\right]isconcentrationof{B}^{+}\\ \left[O{H}^{-}\right]isconcentrationofO{H}^{-}\\ \left[BOH\right]isconcentrationofBOH\end{array}$

Answer to Problem 72A

Equation:

$C_6{H}_{5}N{H}_2+H_{2}O\rightleftharpoons C_6{H}_{5}N{H}_3{}^{+}+O{H}^{-}$

Base ionization constant is given by:

$k_b=\frac{\left[C_6{H}_{5}N{H}_3{}^{+}\right]\times \left[O{H}^{-}\right]}{\left[C_6{H}_{5}N{H}_2\right]}$.

Explanation of Solution

Aniline is a base, it accepts proton from water. Therefore equation is written as:

$C_6{H}_{5}N{H}_2+H_{2}O\rightleftharpoons C_6{H}_{5}N{H}_3{}^{+}+O{H}^{-}$

Using the general expression, k_b is given by:

$k_b=\frac{\left[C_6{H}_{5}N{H}_3{}^{+}\right]\times \left[O{H}^{-}\right]}{\left[C_6{H}_{5}N{H}_2\right]}$

Due to weak acidic character, it is safer to use it in cleaning windows.

Conclusion

Balanced Equation is:

$C_6{H}_{5}N{H}_2+H_{2}O\rightleftharpoons C_6{H}_{5}N{H}_3{}^{+}+O{H}^{-}$

Base ionization constant is:

$k_b=\frac{\left[C_6{H}_{5}N{H}_3{}^{+}\right]\times \left[O{H}^{-}\right]}{\left[C_6{H}_{5}N{H}_2\right]}$