Chapter 18.2, Problem 20SSC

Interpretation Introduction

Interpretation:

Interpret the information from the statement:

The k_b for aniline is $4.3\times {10}^{-10}$

Concept introduction:

The k_b value tells about the formation of reactant and product in base equilibrium reaction.

For a reaction,

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

Base ionization constant is given by:

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

Higher value of k_b implies base is strong and vice-versa.

Answer to Problem 20SSC

The low value of $k_b$ for aniline implies aniline is weak base.

Explanation of Solution

Base ionization constant is given by:

$k_b=\frac{\left[B^{+}\right]\times \left[O{H}^{-}\right]}{\left[BOH\right]}$

Higher value of k_b implies base is strong and vice-versa.

Conclusion

Thus aniline is a weak base.