Chapter 16.7, Problem 1RC

Interpretation Introduction

Interpretation:

The stronger conjugate base has to be determined from the given conjugates bases.

Concept Information:

If equilibrium is established between hydronium ions, hydroxide ions and water molecules, an equilibrium law can be written as follows,

  $\begin{array}{c}{K}_a=\frac{\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]}{\left[H_{2}O\right]\left[H_{2}O\right]}\\ K_a\times \left[H_{2}O\right]\left[H_{2}O\right]=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]\\ K_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]\end{array}$

Ionic product water can also be written as,

  $K_w=\left[H^{+}\right]\left[O{H}^{-}\right]=1.0\times {10}^{-14}$

Acid ionization constant${\text{K}}_a$:

The ionization of a weak acid $\text{HA}$ can be given as follows,

  ${\text{HA}}_{(aq)}\rightleftarrows {\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$

The equilibrium expression for the above reaction is given below.

  $K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Base ionization constant${\text{K}}_{\text{b}}$:

It is the equilibrium constant for the ionization of a base.

The ionization of a weak base $B$ is,

  ${\text{B}}_{\left(\text{aq}\right)}\text{+}{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\rightleftarrows {\text{BH}}_{\left(\text{aq}\right)}^{\text{+}}\text{+}{\text{OH}}_{\left(\text{aq}\right)}^{\text{-}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{{\text{[BH}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}{\text{[B]}}$

Strength of a base is increases with increase in $K_b$ value.

  $K_w=K_a\times K_b$

Where,

  $\begin{array}{l}{\text{K}}_{\text{w}}\to \text{Ionic}\text{product}\text{of}\text{water}\\ {\text{K}}_{\text{a}}\to \text{Acid}\text{ionization}\text{constant}\\ {\text{K}}_{\text{b}}\to \text{Base}\text{ionization}\text{constant}\end{array}$

According to Bronsted- Lowry concepts of acids and bases,

An acid is $H^{+}$(proton) donor

A base is a proton acceptor.

  $Conjugateacid =Base+H^{+}$

  $Conjugatebase =Acid-H^{+}$