Chapter 16, Problem 94GQ

The butylammonium ion, C₄H₉NH₃⁺, has a K_a of 2.3 × 10⁻¹¹.

C₄H₉NH₃⁺(aq) + H₂O(ℓ) ⇄ H₃O⁺(aq) + C₄H₉NH₂(aq)

1. a) Calculate K_b for the conjugate base, C₄H₉NH₂ (butyl amine).
2. b) Place the butylammonium ion and its conjugate base in Table 16.2. Name an acid weaker than C₄H₉NH₃⁺ and a base stronger than C₄H₉NH₃.
3. c) What is the pH of a 0.015M solution of butylammonium chloride?

Interpretation Introduction

Interpretation:

${\text{K}}_{\text{b}}$ has to be calculated for the conjugate base (${\text{C}}_{\text{4}}{\text{H}}_{\text{9}}{\text{NH}}_{\text{2}}$, butyl amine)

Concept Introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

${\text{K}}_{\text{a}}$ is an acid constant for equilibrium reactions.

$\begin{array}{l}{\text{HA + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+ A}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{\text{[HA]}}\end{array}$

${\text{K}}_{\text{b}}$ is a base constant for equilibrium reaction.

$\begin{array}{l}{\text{BOH + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{B}}^{\text{+}}{\text{+ OH}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[B}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}{\text{[BOH]}}\end{array}$

$\begin{array}{l}\text{Ion product constant for wter}\\ {\text{ K}}_{\text{w}}{\text{= [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \text{ =1}{\text{.00×10}}^{\text{-14}}\\ {\text{ pH = -log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ {\text{pOH= -log[OH}}^{\text{-}}\text{]}\\ \end{array}$

Relation between pH and pOH

$\text{pH + pOH =14}$

Answer to Problem 94GQ

${\text{K}}_{\text{b}}$ of conjugate base is $\text{4}{\text{.34 × 10}}^{\text{-4}}$

Explanation of Solution

According to the Brown stead –Lowry theory the compound which is accepts a proton to be considered as a base.

A weak base accepts a proton from water molecules producing hydroxide ions in water.

The ${\text{K}}_{\text{a}}\text{ of 2}{\text{.3× 10}}^{\text{-11}}$

$\begin{array}{l}{\text{K}}_{\text{a}}{\text{×K}}_{\text{b}}{\text{ = K}}_{\text{w}}\\ {\text{K}}_{\text{b}}\text{= }\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ {\text{K}}_{\text{w}}\text{= 1}{\text{.0×10}}^{\text{-14}}\\ {\text{K}}_{\text{a}}\text{ = 2}{\text{.3×10}}^{\text{-11}}\end{array}$

Substitute the values,

$\begin{array}{l}{\text{K}}_{\text{b}}\text{= }\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{2}{\text{.3×10}}^{\text{-11}}}\\ \text{ = 4}{\text{.34×10}}^{\text{-4}}\end{array}$

Therefore, ${\text{K}}_{\text{b}}$ of conjugate base is $\text{4}{\text{.34 × 10}}^{\text{-4}}$

Interpretation Introduction

Interpretation:

Compound should be Named for an acid weaker than ${\text{C}}_{\text{4}}{\text{H}}_{\text{9}}{\text{NH}}_3^{+}$ and a base stronger than ${\text{C}}_{\text{4}}{\text{H}}_{\text{9}}{\text{NH}}_2$.

Concept Introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of  the strength of the acid and bases in the water .

${\text{K}}_{\text{a}}$ an acid constant for equilibrium reactions.

$\begin{array}{l}{\text{HA + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+ A}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{\text{[HA]}}\end{array}$

${\text{K}}_{\text{b}}$ is base constant for equilibrium reaction.

$\begin{array}{l}{\text{BOH + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{B}}^{\text{+}}{\text{+ OH}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[B}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}{\text{[BOH]}}\end{array}$

$\begin{array}{l}\text{Ion product constant for wter}\\ {\text{ K}}_{\text{w}}{\text{= [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \text{ =1}{\text{.00×10}}^{\text{-14}}\\ {\text{ pH = -log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ {\text{pOH= -log[OH}}^{\text{-}}\text{]}\\ \end{array}$

Relation between pH and pOH

$\text{pH + pOH =14}$

Answer to Problem 94GQ

An acid weaker than ${\text{C}}_{\text{4}}{\text{H}}_{\text{9}}{\text{NH}}_{\text{3}}^{\text{+}}$ is ${\text{HPO}}_{\text{4}}^{\text{2-}}$

A base stronger than ${\text{C}}_{\text{4}}{\text{H}}_{\text{9}}{\text{NH}}_{\text{2}}{\text{ is PO}}_{\text{4}}^{\text{3-}}$

Explanation of Solution

From the table,

${\text{K}}_{\text{a}}$ of ${\text{C}}_{\text{4}}{\text{H}}_{\text{9}}{\text{NH}}_{\text{3}}^{\text{+}}\text{ is 2}{\text{.3×10}}^{\text{-11}}$

${\text{K}}_{\text{b}}{\text{ of C}}_{\text{4}}{\text{H}}_{\text{9}}{\text{NH}}_{\text{2}}\text{ is 4}{\text{.4×10}}^{\text{-4}}$

The butyl ammonium ion has ${\text{K}}_{\text{a}}$ value is in between ${\text{HPO}}_{\text{4}}^{\text{2-}}{\text{ (K}}_{\text{a}}\text{=3}{\text{.6×10}}^{\text{-13}}\text{)}$ and ${{\text{[Ni(H}}_2{\text{O)}}_6\text{]}}^{2+}{\text{ (K}}_{\text{a}}\text{=3}{\text{.6×10}}^{\text{-13}}\text{)}$

The order of ${\text{K}}_{\text{a}}$ is as follows.

${\text{HPO}}_{\text{4}}^{\text{2-}}{\text{ < C}}_{\text{4}}{\text{H}}_{\text{9}}{\text{NH}}_{\text{3}}^{\text{+}}{\text{ < [Ni(H}}_{\text{2}}{\text{O)}}_{\text{6}}{\text{]}}^{\text{2+}}$

Butyl amine has a ${\text{K}}_{\text{b}}$ value in between ${{\text{[Ni(H}}_{\text{2}}{\text{O)}}_{\text{5}}\text{OH]}}^{\text{+}}{\text{ (K}}_{\text{b}}\text{=4}{\text{.0×10}}^{\text{-4}}\text{)}$ and ${\text{PO}}_{\text{4}}^{\text{3-}}{\text{ (K}}_{\text{b}}\text{=2}{\text{.8×10}}^{\text{-2}}\text{)}$.

The ${\text{K}}_{\text{b}}$ order is as follows.

${{\text{[Ni(H}}_{\text{2}}{\text{O)}}_{\text{5}}\text{OH]}}^{\text{+}}{\text{< C}}_{\text{4}}{\text{H}}_{\text{9}}{\text{NH}}_{\text{2}}{\text{< PO}}_{\text{4}}^{\text{3-}}$

Therefore, An acid weaker than ${\text{C}}_{\text{4}}{\text{H}}_{\text{9}}{\text{NH}}_{\text{3}}^{\text{+}}$ is ${\text{HPO}}_{\text{4}}^{\text{2-}}$ . A base stronger than ${\text{C}}_{\text{4}}{\text{H}}_{\text{9}}{\text{NH}}_{\text{2}}{\text{ is PO}}_{\text{4}}^{\text{3-}}$

Interpretation Introduction

Interpretation:

The pH has be calculated for 0.015M solution of butylammonium chloride?

Concept Introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of  the strength of the acid and bases in the water .

${\text{K}}_{\text{a}}$ is an acid constant for equilibrium reactions.

$\begin{array}{l}{\text{HA + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+ A}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{\text{[HA]}}\end{array}$

${\text{K}}_{\text{b}}$ is a base constant for equilibrium reaction.

$\begin{array}{l}{\text{BOH + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{B}}^{\text{+}}{\text{+ OH}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[B}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}{\text{[BOH]}}\end{array}$

$\begin{array}{l}\text{Ion product constant for wter}\\ {\text{ K}}_{\text{w}}{\text{= [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \text{ =1}{\text{.00×10}}^{\text{-14}}\\ {\text{ pH = -log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ {\text{pOH= -log[OH}}^{\text{-}}\text{]}\\ \end{array}$

Relation between pH and pOH

$\text{pH + pOH =14}$

Answer to Problem 94GQ

The pH of the 0.015M solution of butylammonium chloride is 6.2.

Explanation of Solution

Butylammonium chloride dissociates into butyl ammonium ion and chloride ion. Chloride ion does not affect the solution pH. Butylammonium donate a proton to water to form hydronium ions.

The equilibrium chemical reaction of butylammonium chloride is as follows.

${\text{C}}_{\text{4}}{\text{H}}_{\text{9}}{\text{NH}}_{\text{3}}^{\text{+}}{\text{(aq) + H}}_{\text{2}}\text{O(l) }\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq) + C}}_{\text{4}}{\text{H}}_{\text{9}}{\text{NH}}_{\text{2}}\text{(aq)}$

The equilibrium expression:

${\text{K}}_{\text{a}}\text{= }\frac{{\text{[C}}_{\text{4}}{\text{H}}_{\text{9}}{\text{NH}}_{\text{2}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[C}}_{\text{4}}{\text{H}}_{\text{9}}{\text{NH}}_{\text{3}}^{\text{+}}\text{]}}$

The initial concentration of ammonium chloride is 0.015M.

Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.

$\begin{array}{l}{\text{ C}}_{\text{4}}{\text{H}}_{\text{9}}{\text{NH}}_{\text{3}}^{\text{+}}{\text{(aq) + H}}_{\text{2}}\text{O(l) }\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq) + C}}_{\text{4}}{\text{H}}_{\text{9}}{\text{NH}}_{\text{2}}\text{(aq)}\\ \text{I 0}\text{.015M -- -- --}\\ \text{C -x -- + x + x}\\ \text{E (0}\text{.015- x) -- x x}\end{array}$

Hence,

${\text{K}}_{\text{a}}\text{= }\frac{{\text{[C}}_{\text{4}}{\text{H}}_{\text{9}}{\text{NH}}_{\text{2}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[C}}_{\text{4}}{\text{H}}_{\text{9}}{\text{NH}}_{\text{3}}^{\text{+}}\text{]}}$

${\text{K}}_{\text{a}}\text{= 2}{\text{.3×10}}^{\text{-11}}$

Substitute the values form the ICE table.

$\begin{array}{l}\text{2}{\text{.3×10}}^{\text{-11}}\text{ = }\frac{\text{(x)(x)}}{\text{0}\text{.015- x}}\\ \text{2}{\text{.3×10}}^{\text{-11}}\text{ = }\frac{{\text{(x)}}^{\text{2}}}{\text{0}\text{.015- x}}\\ \text{(0}\text{.015-x) approximately equals to 0}\text{.015}\\ \text{2}{\text{.3×10}}^{-11}\text{ = }\frac{{\text{(x)}}^{\text{2}}}{\text{(0}\text{.015)}}\\ {\text{ x}}^{\text{2}}\text{= (0}\text{.015)(2}{\text{.3×10}}^{\text{-11}}\text{)}\\ \end{array}$

Therefore,

$\begin{array}{l}\text{x = }\sqrt{\text{(0}\text{.015)(2}{\text{.3×10}}^{\text{-11}}\text{)}}\\ \text{ = 5}{\text{.9×10}}^{\text{-7}}\\ {\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] =5}{\text{.9×10}}^{-7}\text{M}\end{array}$

Let’s calculate the pH of the solution:

$\begin{array}{l}{\text{pH = -log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{pH = -log[5}{\text{.9×10}}^{\text{-7}}\text{]}\\ \text{pH = 6}\text{.2}\end{array}$

Therefore, pH of butylammonium chloride solution is 6.2.