Chapter 7, Problem 135AE

(a)

Interpretation Introduction

Interpretation: The pH of 0.10 M solution of acrylic acid needs to be determined if ${\text{K}}_{\text{a}}$ for acrylic acid is $\text{5}\text{.6 }\times {\text{10}}^{\text{-5}}$.

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 ionized 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 135AE

  $\text{pH = 2}\text{.62}$

Explanation of Solution

• ${\text{K}}_{\text{a}}$ = $\text{5}\text{.6 }\times {\text{10}}^{\text{-5}}$
• Concentration = 0.10

Consider Hacr as the abbreviation of acrylic acid.

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

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

Make ICE table:

Concentration	$\text{Hacr}$	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$	${\text{acr}}^{\text{-}}$
Initial	0.10 M	0	0
Change	-x	x	x
Equilibrium	0.10 −x	x	x

Substitute the values to calculate ‘x’:

• ${\text{K}}_{\text{a}}$ = $\text{5}\text{.6 }\times {\text{10}}^{\text{-5}}$

  $\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] [acr}}_{{}_{\text{(aq)}}}^{\text{-}}\text{]}}{{\text{[Hacr}}_{\text{(aq)}}\text{]}}\\ \text{5}\text{.6 }\times {\text{10}}^{\text{-5}}\text{ =}\frac{\text{[x] [x]}}{\text{[0}\text{.10 -x]}}\text{(0}\text{.10>>x)}\\ {\text{[x]}}^{\text{2}}\text{ =5}\text{.6 }\times {\text{10}}^{\text{-5}}\text{×0}\text{.10}\\ \text{[x]= 2}{\text{.4 ×10}}^{\text{-3}}{\text{M=[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\end{array}$

Calculate pH:

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}\text{]}\\ \text{pH = - log [2}{\text{.4 ×10}}^{\text{-3}}\text{M]}\\ \text{pH = 2}\text{.62}\end{array}$

(b)

Interpretation Introduction

Interpretation: The percent dissociation of 0.10 M solution of acrylic acid needs to be determined if the ${\text{K}}_{\text{a}}$ for acrylic acid is $\text{5}\text{.6 }\times {\text{10}}^{\text{-5}}$.

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 ionized 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 135AE

Percent dissociation = $\text{2}\text{.4\%}$

Explanation of Solution

• ${\text{K}}_{\text{a}}$ = $\text{5}\text{.6 }\times {\text{10}}^{\text{-5}}$
• Concentration = 0.10

Consider Hacr as the abbreviation of acrylic acid.

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

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

Make ICE table:

Concentration	$\text{Hacr}$	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$	${\text{acr}}^{\text{-}}$
Initial	0.10 M	0	0
Change	-x	x	x
Equilibrium	0.10 −x	x	x

Substitute the values to calculate ‘x’:

• ${\text{K}}_{\text{a}}$ = $\text{5}\text{.6 }\times {\text{10}}^{\text{-5}}$

  $\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] [acr}}_{{}_{\text{(aq)}}}^{\text{-}}\text{]}}{{\text{[Hacr}}_{\text{(aq)}}\text{]}}\\ \text{5}\text{.6 }\times {\text{10}}^{\text{-5}}\text{ =}\frac{\text{[x] [x]}}{\text{[0}\text{.10 -x]}}\text{(0}\text{.10>>x)}\\ {\text{[x]}}^{\text{2}}\text{ =5}\text{.6 }\times {\text{10}}^{\text{-5}}\text{×0}\text{.10}\\ \text{[x]= 2}{\text{.4 ×10}}^{\text{-3}}{\text{M=[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\end{array}$

Calculate percent dissociation:

  $\begin{array}{l}\text{Percent dissociation = }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{\text{[Hacr]}}\text{×100}\\ \text{Percent dissociation = }\frac{\text{[2}{\text{.4 ×10}}^{\text{-3}}\text{M]}}{\text{[0}\text{.10]}}\text{×100= 2}\text{.4\%}\end{array}$

(c)

Interpretation Introduction

Interpretation: The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions needs to be determined, if the percent dissociation is less than 0.010% for 0.10 M solution of acrylic acid. Also, the ${\text{K}}_{\text{a}}$ for acrylic acid is $\text{5}\text{.6 }\times {\text{10}}^{\text{-5}}$.

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 ionized 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 135AE

  ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]= 0}\text{.56M}$

Explanation of Solution

• ${\text{K}}_{\text{a}}$ = $\text{5}\text{.6 }\times {\text{10}}^{\text{-5}}$
• Concentration = 0.10M

Consider Hacr as the abbreviation of acrylic acid. The equilibrium and ${\text{K}}_{\text{a}}$ expression for $\text{Hacr}$ are:

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

Calculate ${\text{acr}}^{\text{-}}$ for 0.010 % percent dissociation:

  $\begin{array}{l}\text{Percent dissociation = }\frac{{\text{[acr}}^{\text{-}}\text{]}}{\text{[Hacr]}}\text{×100}\\ \text{0}\text{.010 = }\frac{{\text{[acr}}^{\text{-}}\text{]}}{\text{[0}\text{.10]}}\text{×100 }\\ {\text{[acr}}^{\text{-}}\text{]=}\frac{\text{0}\text{.010×0}\text{.10}}{\text{100 }}\\ {\text{[acr}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-5}}\text{M }\end{array}$

Calculate H₃O⁺:

  $\begin{array}{l}{\text{Hacr}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+ acr}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] [acr}}_{{}_{\text{(aq)}}}^{\text{-}}\text{]}}{{\text{[Hacr}}_{\text{(aq)}}\text{]}}\\ \text{5}{\text{.6×10}}^{\text{-5}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] [1}{\text{.0×10}}^{\text{-5}}\text{]}}{\text{[0}\text{.10-1}{\text{.0×10}}^{\text{-5}}\text{ ]}}\text{(0}\text{.10>>>1}{\text{.0×10}}^{\text{-5}}\text{)}\\ \text{5}{\text{.6×10}}^{\text{-5}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] [1}{\text{.0×10}}^{\text{-5}}\text{]}}{\text{[0}\text{.10-1}{\text{.0×10}}^{\text{-5}}\text{ ]}}\\ {\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]=}\frac{\text{5}{\text{.6×10}}^{\text{-5}}\text{×0}\text{.10}}{\text{1}{\text{.0×10}}^{\text{-5}}}\\ {\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]= 0}\text{.56M}\end{array}$

(d)

Interpretation Introduction

Interpretation: The value of pH of 0.050 M solution of sodium acrylate needs to be determined. The ${\text{K}}_{\text{a}}$ for acrylic acid is $\text{5}\text{.6 }\times {\text{10}}^{\text{-5}}$.

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 ionized 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 135AE

  $\text{pH = 8}\text{.42}$

Explanation of Solution

• ${\text{K}}_{\text{a}}$ = $\text{5}\text{.6 }\times {\text{10}}^{\text{-5}}$
• Concentration of sodium acrylate = 0.050M

Consider Hacr as the abbreviation of acrylic acid. Sodium acrylate will give ${\text{acr}}^{\text{-}}$ which is a weak base. The equilibrium and ${\text{K}}_b$ expression for ${\text{acr}}^{\text{-}}$ are:

  $\begin{array}{l}{\text{acr}}^{\text{-}}{}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{OH}}^{-}{}_{\text{(aq)}}{\text{+ Hacr}}_{\text{(aq)}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[OH}}^{-}{\text{] [Hacr}}_{{}_{\text{(aq)}}}^{}\text{]}}{{\text{[acr}}^{\text{-}}{}_{\text{(aq)}}\text{]}}\end{array}$

Make ICE table:

Concentration	${\text{acr}}^{\text{-}}$	${\text{OH}}^{-}$	$\text{ Hacr}$
Initial	0.050M	0	0
Change	-x	x	x
Equilibrium	0.050−x	x	x

Substitute the values to calculate ‘x’:

• ${\text{K}}_{\text{a}}$ = $\text{5}\text{.6 }\times {\text{10}}^{\text{-5}}$

Calculate ${\text{K}}_b$ :

  $\begin{array}{l}{\text{K}}_{\text{b}}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ {\text{K}}_{\text{b}}\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{5}{\text{.6 ×10}}^{\text{-5}}}\\ {\text{K}}_{\text{b}}\text{=1}{\text{.8×10}}^{\text{-10}}\end{array}$

  $\begin{array}{l}{\text{K}}_{\text{b}}\text{=}\frac{{\text{[OH}}^{-}{\text{] [Hacr}}_{{}_{\text{(aq)}}}^{}\text{]}}{{\text{[acr}}^{\text{-}}{}_{\text{(aq)}}\text{]}}\\ \text{1}{\text{.8×10}}^{\text{-10}}\text{ =}\frac{\text{[x] [x]}}{\text{[0}\text{.050 -x]}}\text{(0}\text{.050>>x)}\\ {\text{[x]}}^{\text{2}}\text{ =1}{\text{.8×10}}^{\text{-10}}\text{ ×0}\text{.050}\\ \text{[x]= 3}{\text{.0 ×10}}^{\text{-6}}{\text{M=[OH}}^{-}\text{]}\end{array}$

Calculate pH:

  $\begin{array}{l}{\text{pOH = - log [OH}}^{\text{-}}\text{]}\\ \text{pOH = - log [ 3}{\text{.0 ×10}}^{\text{-6}}\text{]}\\ \text{pOH = 5}\text{.52}\\ \text{pH=14-pOH=14-5}\text{.52=8}\text{.42}\end{array}$