Chapter 18.2, Problem 15PP

Interpretation Introduction

(a)

Interpretation:

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 15PP

Equation:

$C_6{H}_{13}N{H}_2+H_{2}O\rightleftharpoons C_6{H}_{13}N{H}_3{}^{+}+O{H}^{-}$

Base ionization constant is given by:

$k_b=\frac{\left[C_6{H}_{13}N{H}_3{}^{+}\right]\times \left[O{H}^{-}\right]}{\left[C_6{H}_{13}N{H}_2\right]}$

Explanation of Solution

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

$C_6{H}_{13}N{H}_2+H_{2}O\rightleftharpoons C_6{H}_{13}N{H}_3{}^{+}+O{H}^{-}$

Using the general expression, k_b is given by:

$k_b=\frac{\left[C_6{H}_{13}N{H}_3{}^{+}\right]\times \left[O{H}^{-}\right]}{\left[C_6{H}_{13}N{H}_2\right]}$

Interpretation Introduction

(a)

Interpretation:

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 15PP

Equation:

$C_3{H}_{7}N{H}_2+H_{2}O\rightleftharpoons C_3{H}_{7}N{H}_3{}^{+}+O{H}^{-}$

Base ionization constant is given by:

$k_b=\frac{\left[C_3{H}_{7}N{H}_3{}^{+}\right]\times \left[O{H}^{-}\right]}{\left[C_3{H}_{7}N{H}_2\right]}$

Explanation of Solution

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

$C_3{H}_{7}N{H}_2+H_{2}O\rightleftharpoons C_3{H}_{7}N{H}_3{}^{+}+O{H}^{-}$

Using the general expression, k_b is given by:

$k_b=\frac{\left[C_3{H}_{7}N{H}_3{}^{+}\right]\times \left[O{H}^{-}\right]}{\left[C_3{H}_{7}N{H}_2\right]}$

Interpretation Introduction

(c)

Interpretation:

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 15PP

Equation:

$C{O}_3{}^{2-}+H_{2}O\rightleftharpoons HC{O}_3{}^{-}+O{H}^{-}$

Base ionization constant is given by:

$k_b=\frac{\left[HC{O}_3{}^{-}\right]\times \left[O{H}^{-}\right]}{\left[C{O}_3{}^{2-}\right]}$

Explanation of Solution

Carbonate ion can accept proton from water. Therefore equation is written as:

$C{O}_3{}^{2-}+H_{2}O\rightleftharpoons HC{O}_3{}^{-}+O{H}^{-}$

Using the general expression, k_b is given by:

$k_b=\frac{\left[HC{O}_3{}^{-}\right]\times \left[O{H}^{-}\right]}{\left[C{O}_3{}^{2-}\right]}$

Interpretation Introduction

(d)

Interpretation:

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 15PP

Equation:

$HS{O}_3{}^{-}+H_{2}O\rightleftharpoons H_{2}S{O}_3+O{H}^{-}$

Base ionization constant is given by:

$k_b=\frac{\left[H_{2}S{O}_3\right]\times \left[O{H}^{-}\right]}{\left[HS{O}_3{}^{-}\right]}$

Explanation of Solution

Hydrogen sulphite ion can accept proton from water. Therefore equation is written as:

$HS{O}_3{}^{-}+H_{2}O\rightleftharpoons H_{2}S{O}_3+O{H}^{-}$

Using the general expression, k_b is given by:

$k_b=\frac{\left[H_{2}S{O}_3\right]\times \left[O{H}^{-}\right]}{\left[HS{O}_3{}^{-}\right]}$