Chapter 7, Problem 169MP

(a)

Interpretation Introduction

Interpretation: Thelowest pH of solution that is prepared by mixing of solution from group I and group II needs to be determined.

Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. A weak acid ionizes partially and reaches to equilibrium.

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

The acid dissociation constant is the ratio of concentration of product and reactant for the equilibrium reaction of weak acid in solution. For the given weak acid HA, it can be written as:

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

Answer to Problem 169MP

  $\text{pH = 2}\text{.2}$

Explanation of Solution

To get the lowest pH, the strong acid and weak base must be added. Hence 0.20 M HCl and 0.20 M ${\text{NaNO}}_{\text{2}}$ must be mixed to get the solution. Here HCl is strong acid and ${\text{NaNO}}_{\text{2}}$ will form ${\text{HNO}}_{\text{2}}$ that is weak acid and will ionize as given:

  ${\text{HNO}}_{\text{2}}{}_{\text{(aq)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+NO}}_{\text{2}}{{}^{\text{-}}}_{\text{(aq)}}$

• ${\text{K}}_{\text{a}}$ = $\text{4}\text{.0 }\times {\text{10}}^{\text{-4 }}$
• Concentration = 0.20

The equilibrium and ${\text{K}}_{\text{a}}$ expression:

  $\begin{array}{l}{\text{HNO}}_{\text{2}}{}_{\text{(aq)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+NO}}_{\text{2}}{{}^{\text{-}}}_{\text{(aq)}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] [NO}}_{\text{2}}{{}^{\text{-}}}_{\text{(aq)}}\text{]}}{{\text{[HNO}}_{\text{2}}{}_{\text{(aq)}}\text{]}}\end{array}$

Make ICE table:

Concentration	${\text{HNO}}_{\text{2}}$	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$	${\text{NO}}_{\text{2}}{{}^{\text{-}}}_{\text{(aq)}}$
Initial	0.20 M	0	0
Change	-x	x	x
Equilibrium	0.20 −x	x	x

Substitute the values to calculate ‘x’:

• ${\text{K}}_{\text{a}}$ = $\text{4}\text{.0 }\times {\text{10}}^{\text{-4 }}$

  $\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] [NO}}_{\text{2}}{{}^{\text{-}}}_{\text{(aq)}}\text{]}}{{\text{[HNO}}_{\text{2}}{}_{\text{(aq)}}\text{]}}\\ \text{4}\text{.0 }\times {\text{10}}^{\text{-4 }}\text{ =}\frac{\text{[x] [x]}}{\text{[0}\text{.20 -x]}}\text{(0}\text{.20>>x)}\\ {\text{[x]}}^{\text{2}}\text{ =4}\text{.0 }\times {\text{10}}^{\text{-4 }}\text{ ×0}\text{.20}\\ \text{[x]= 6}{\text{.1 ×10}}^{\text{-3}}{\text{M=[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\end{array}$

Calculate pH:

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}\text{]}\\ \text{pH = - log [6}{\text{.1 ×10}}^{\text{-3}}\text{M]}\\ \text{pH = 2}\text{.2}\end{array}$

(b)

Interpretation Introduction

Interpretation: Thehighest pH of solution that is prepared by mixing of solution from group I and group II needs to be determined.

Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. A weak acid ionizes partially and reaches to equilibrium.

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

The acid dissociation constant is the ratio of concentration of product and reactant for the equilibrium reaction of weak acid in solution. For the given weak acid HA, it can be written as:

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

Answer to Problem 169MP

  $\text{pH = 10}\text{.6}$

Explanation of Solution

To get the highest pH, the strong base and weak acid must be added. Hence 0.20 M ${{\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{)}}_{\text{3}}\text{NHCl}$ and 0.20 M $\text{KOI}$ must be mixed to get the solution. Here ${{\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{)}}_{\text{3}}\text{NHCl}$ is weak acid and $\text{KOI}$ will form ${\text{OI}}^{\text{-}}$. Hence the equilibrium must be written as:

  ${\text{OI}}^{\text{-}}{}_{\text{(aq)}}{\text{+ (C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{)}}_{\text{3}}{\text{NH}}^{\text{+}}{}_{\text{(l)}}\rightleftharpoons {\text{HOI}}_{\text{(aq)}}{\text{+(C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{)}}_{\text{3}}{\text{N}}_{\text{(aq)}}$

• $\text{K =}\frac{{\text{ K}}_{{\text{a(C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{)}}_{\text{3}}\text{N+}}}{{\text{K}}_{\text{aOI-}}}\text{=}\frac{\text{ 1}\text{.0 }\times {\text{10}}^{\text{-14 }}}{\text{4}\text{.0 }\times {\text{10}}^{\text{-4 }}}\times \frac{\text{ 1}\text{.0 }}{\text{2}\text{.0 }\times {\text{10}}^{\text{-11 }}}=1.25$
• Concentration = 0.20M

Make ICE table:

Concentration	${\text{OI}}^{\text{-}}$	${{\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{)}}_{\text{3}}{\text{NH}}^{\text{+}}$	$\text{HOI}$	${{\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{)}}_{\text{3}}\text{N}$
Initial	0.20 M	0.20	0	0
Change	-x	-x	x	x
Equilibrium	0.20 −x	0.20-x	x	x

Substitute the values to calculate ‘x’:

  $\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[HOI] [(C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{)}}_{\text{3}}\text{N]}}{{\text{[(C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{)}}_{\text{3}}{\text{NH}}^{\text{+}}{\text{][OI}}^{\text{-}}\text{]}}\\ \text{1}\text{.25 =}\frac{\text{[x] [x]}}{\text{[0}\text{.20 -x][0}\text{.20 -x]}}\\ \text{1}\text{.12 =}\frac{\text{[x]}}{\text{[0}\text{.20 -x]}}\\ \text{[x]= 0}\text{.22 - 1}\text{.12x}\\ \text{[x]= 0}{\text{.105 M =[HOI] [(C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{)}}_{\text{3}}\text{N]}\\ {\text{[OI}}^{\text{-}}\text{]=0}\text{.20-0}\text{.053=0}\text{.094 M}\end{array}$

Calculate concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ with the help of ${\text{K}}_{\text{a}}=2.0\times {10}^{-11}$ of HOI:

  $\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[OI}}^{\text{-}}{\text{] [H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{\text{[HOI]}}\\ 2.0\times {10}^{-11}\text{=}\frac{\text{[x] [0}\text{.094]}}{\text{[0}\text{.105]}}\\ \text{[x]= }2.2\times {10}^{-11}{\text{ =[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\end{array}$

Calculate pH:

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}\text{]}\\ \text{pH = - log [}2.2\times {10}^{-11}\text{]}\\ \text{pH = 10}\text{.6}\end{array}$

(c)

Interpretation Introduction

Interpretation: The solution with pH 7 that is prepared by mixing of solution from group I and group II needs to be determined.

Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. A weak acid ionizes partially and reaches to equilibrium.

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

The acid dissociation constant is the ratio of concentration of product and reactant for the equilibrium reaction of weak acid in solution. For the given weak acid HA, it can be written as:

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

Answer to Problem 169MP

Since ${\text{K}}_{\text{a}}{\text{ = K}}_{\text{b}}$ therefore the solution will have pH 7.00.

Explanation of Solution

To get the neutral solution, the strong base and strong acid must be added. Hence 0.20 M ${{\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{)}}_{\text{3}}\text{NHCl}$ and 0.20 M ${\text{NaNO}}_{\text{2}}$ must be mixed to get the solution. Here ${{\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{)}}_{\text{3}}\text{NHCl}$ is weak acid and ${\text{NaNO}}_{\text{2}}$ will form ${\text{HNO}}_{\text{2}}$ that is weak acid and will ionize as given:

  ${\text{HNO}}_{\text{2}}{}_{\text{(aq)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+NO}}_{\text{2}}{{}^{\text{-}}}_{\text{(aq)}}$

• ${\text{K}}_{\text{a}}{\text{ for (C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{)}}_{\text{3}}{\text{NH}}^{\text{+}}\text{ =}\frac{\text{ 1}\text{.0 }\times {\text{10}}^{\text{-14 }}}{\text{4}\text{.0 }\times {\text{10}}^{\text{-4 }}}=1.25\times {\text{10}}^{\text{-11 }}$
• ${\text{K}}_b{\text{ for NO}}_{\text{2}}{}^{\text{-}}\text{ =}\frac{\text{ 1}\text{.0 }\times {\text{10}}^{\text{-14 }}}{\text{4}\text{.0 }\times {\text{10}}^{\text{-4 }}}=1.25\times {\text{10}}^{\text{-11 }}$
• Since ${\text{K}}_{\text{a}}{\text{ = K}}_{\text{b}}$ therefore the solution will have pH 7.00.