Chapter 13, Problem 151AE

Acrylic acid (CH₂9CHCO₂H) is a precursor for many important plastics. K_a for acrylic acid is 5.6 × 10⁻⁵.

a. Calculate the pH of a 0.10-M solution of acrylic acid.

b. Calculate the percent dissociation of a 0.10-M solution of acrylic acid.

c. Calculate the pH of a 0.050-M solution of sodium acrylate (NaC₃H₃O₂).

Interpretation Introduction

Interpretation: The acid dissociation constant $K_{\text{a}}$  and molarity of acrylic acid and sodium acrylate are given. By using these values, the $\text{pH}$ value of solution of acrylic acid and sodium acrylate and percent dissociation of solution of acrylic acid is to be calculated.

Concept introduction: The equilibrium constant $K_{\text{b}}$ is the ratio of product of equilibrium concentrations of product to the equilibrium concentration of reactant base.

The equilibrium constant $K_{\text{a}}$ is the ratio of product of equilibrium concentrations of product to the equilibrium concentration of reactant acid.

To determine: The change in concentration of ${\text{H}}^{+}$ and the $\text{pH}$ value of acrylic acid.

Answer to Problem 151AE

The $\text{pH}$ value of acrylic acid is $\underset{\_}{2.63}$.

Explanation of Solution

Explanation

The dissociation reaction of acrylic acid is,

${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{H}}_{\left(\text{aq}\right)}\to {\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+H}}^{\text{+}}{}_{\left(\text{aq}\right)}$

The concentration of ${\text{H}}^{+}$ is $\underset{\_}{2.33\times 10^{-3}M}$.

Given

The equilibrium constant is $\text{5}\text{.6}\times {\text{10}}^{-\text{5}}$.

The initial concentration of acrylic acid is $\text{0}\text{.10}\text{M}$.

It is assumed that the change in concentration of $\left[{\text{H}}^{+}\right]$ is $x$.

Make the ICE table for the dissociation reaction of acrylic acid.

$\begin{array}{ccccc}& {\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_2{\text{H}}_{\left(\text{aq}\right)}& \to & {\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_2{{}^{-}}_{\left(\text{aq}\right)}& {\text{H}}^{+}{}_{\left(\text{aq}\right)}\\ \text{Initial}\left(\text{M}\right):& 0.10& & 0& 0\\ \text{Chang}\left(\text{M}\right):& -x& & x& x\\ \text{Equilibrium}\left(\text{M}\right):& 0.10-x& & x& x\end{array}$

The expression for the acid dissociation constant for the given reaction is,

$K_{\text{a}}=\frac{\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_2{}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_2\text{H}\right]}$

Substitute the value of $K_{\text{a}}$ and equilibrium concentrations in the above equation.

$\begin{array}{c}{K}_{\text{a}}=\frac{\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_2{}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_2\text{H}\right]}\\ \text{5}\text{.6}\times {\text{10}}^{-\text{5}}=\frac{\left(x\right)\left(x\right)}{\left(0.10-x\right)}\\ x^2=\text{5}\text{.6}\times {\text{10}}^{-\text{5}}\left(0.10-x\right)\end{array}$

Simplify the above equation,

$\begin{array}{c}\text{5}\text{.6}\times {\text{10}}^{-\text{5}}\left(0.10-x\right)=x^2\\ x=2.33\times {10}^{-3}\end{array}$

Thus, the concentration $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{2.33\times 10^{-3}M}$.

The dissociation reaction of acrylic acid is,

${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{H}}_{\left(\text{aq}\right)}\to {\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+H}}^{\text{+}}{}_{\left(\text{aq}\right)}$

The $\text{pH}$ value of acrylic acid is $\underset{\_}{2.63}$.

The change in concentration of $\left[{\text{H}}^{+}\right]$ is $\text{2}\text{.33}\times {\text{10}}^{-\text{3}}\text{M}$.

The $\text{pH}$ is calculated using the formula,

$\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]$

Substitute the value of concentration of ${\text{H}}^{+}$ in the above equation.

$\begin{array}{c}\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]\\ =-{\log }_{\text{10}}\left[\text{2}\text{.33}\times {\text{10}}^{-\text{3}}\right]\\ =\underset{\_}{2.63}\end{array}$

Interpretation Introduction

Interpretation: The acid dissociation constant $K_{\text{a}}$  and molarity of acrylic acid and sodium acrylate are given. By using these values, the $\text{pH}$ value of solution of acrylic acid and sodium acrylate and percent dissociation of solution of acrylic acid is to be calculated.

Concept introduction: The equilibrium constant $K_{\text{b}}$ is the ratio of product of equilibrium concentrations of product to the equilibrium concentration of reactant base.

The equilibrium constant $K_{\text{a}}$ is the ratio of product of equilibrium concentrations of product to the equilibrium concentration of reactant acid.

To determine: The percent dissociation of solution of acrylic acid.

Answer to Problem 151AE

Answer

The percent dissociation of acrylic acid is $\underset{\_}{2.38\%}$.

Explanation of Solution

Explanation

The dissociation reaction of acrylic acid is,

${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_2\text{H}\left(aq\right)\stackrel{}{\rightleftharpoons }{\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_2{}^{-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)$

The percent dissociation of solution of acrylic acid is $\underset{\_}{2.38\%}$.

The concentration of $\left[{\text{H}}^{+}\right]$ is $\text{2}\text{.33}\times {\text{10}}^{-\text{3}}\text{M}$. It is the amount of acrylic acid dissociated.

Formula

The percent dissociation is calculated using the formula,

$\begin{array}{c}\text{Percent dissociation}=\frac{\text{Amount}\text{of}\text{acrylic}\text{acid}\text{dissociated}}{\text{Concentration}\text{of}\text{acrylic}\text{acid}\text{at}\text{equilibrium}}\times \text{100}\\ \text{=}\frac{x}{\left(0.10-x\right)}\times 100\%\end{array}$

Substitute the value of $x$ from part (a) in the above equation.

$\begin{array}{c}\text{Percent dissociation}=\frac{x}{\left(0.10-x\right)}\times 100\%\\ \text{=}\frac{\text{2}\text{.33}\times {\text{10}}^{-\text{3}}\text{M}}{\left(0.10-2.33\times {10}^{-3}\right)\text{M}}\times \text{100\%}\\ =\underset{\_}{2.38\%}\end{array}$

Interpretation Introduction

Interpretation: The acid dissociation constant $K_{\text{a}}$  and molarity of acrylic acid and sodium acrylate are given. By using these values, the $\text{pH}$ value of solution of acrylic acid and sodium acrylate and percent dissociation of solution of acrylic acid is to be calculated.

Concept introduction: The equilibrium constant $K_{\text{b}}$ is the ratio of product of equilibrium concentrations of product to the equilibrium concentration of reactant base.

The equilibrium constant $K_{\text{a}}$ is the ratio of product of equilibrium concentrations of product to the equilibrium concentration of reactant acid.

To determine: The $\text{pH}$ value of solution of sodium acrylate.

Answer to Problem 151AE

Answer

The $\text{pH}$ value of sodium acrylate is $\underset{\_}{8.48}$.

Explanation of Solution

Explanation

The base dissociation reaction of sodium acrylate is,

${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_2{}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\rightleftharpoons }{\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_2\text{H}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

The base dissociation constant of sodium acrylate is $\underset{\_}{1.786\times 10^{-10}}$.

Given

The initial concentration of sodium acrylate is $\text{0}\text{.05}\text{M}$.

It is assumed that the change in concentration of $\left[{\text{OH}}^{-}\right]$ is $x$.

Make the ICE table for the above reaction.

$\begin{array}{cccccccc}& {\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_2{}^{-}\left(aq\right)& +& {\text{H}}_{\text{2}}\text{O}\left(l\right)& \rightleftharpoons & {\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_2\text{H}& +& {\text{OH}}^{-}\left(aq\right)\\ \text{Initial}\left(\text{M}\right)\text{:}& 0.05& & & & 0& & 0\\ \text{Change}\left(\text{M}\right)\text{:}& -x& & & & x& & x\\ \text{Equilibrium}\left(\text{M}\right)\text{:}& 0.05-x& & & & x& & x\end{array}$

The expression for dissociation constant for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_2\text{H}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_2{}^{-}\right]}$

Substitute the equilibrium concentrations in the above expression.

$K_{\text{b}}=\frac{\left(x\right)\left(x\right)}{\left(0.050-x\right)\text{M}}$ (1)

The acid dissociation constant of acrylic acid is $\text{5}\text{.6}\times {\text{10}}^{-\text{5}}$.

The relation between $K_{\text{b}}$ and $K_{\text{a}}$ is,

$K_{\text{a}}\times K_{\text{b}}=K_{\text{w}}$

Where,

• $K_{\text{w}}$ is ion- product constant of water $(\text{1}\text{.0}\times {\text{10}}^{-\text{14}})$.

Substitute the values of $K_{\text{a}}$ and $K_{\text{w}}$ in the above equation.

$\begin{array}{c}{K}_{\text{a}}\times K_{\text{b}}=K_{\text{w}}\\ \text{5}\text{.6}\times {\text{10}}^{-\text{5}}\times K_{\text{b}}=\text{1}\text{.0}\times {\text{10}}^{-\text{14}}\\ K_{\text{b}}=\underset{\_}{1.786\times 10^{-10}}\end{array}$

The stated reaction is,

${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_2{}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\rightleftharpoons }{\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{O}}_2{\text{H}}^{-}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

The concentration of $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{2.987\times 10^{-6}M}$.

The base dissociation constant is $\text{1}\text{.786}\times {\text{10}}^{-\text{10}}$.

Substitute the value of $K_{\text{b}}$ in equation (1).

$\begin{array}{c}{K}_{\text{b}}=\frac{\left(x\right)\left(x\right)}{\left(0.050-x\right)}\\ \text{1}\text{.786}\times {\text{10}}^{-\text{10}}=\frac{x^2}{\left(0.050-x\right)}\\ x^2=\text{1}\text{.786}\times {\text{10}}^{-\text{10}}\left(0.050-x\right)\end{array}$

Simplify the above equation,

$\begin{array}{c}{x}^2=\text{1}\text{.786}\times {\text{10}}^{-\text{10}}\left(0.050-x\right)\\ x=2.987\times {10}^{-6}\end{array}$

Thus, the concentration $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{2.987\times 10^{-6}M}$.

The $\text{pH}$ value of sodium acrylate is $\underset{\_}{8.48}$.

The concentration of $\left[{\text{OH}}^{-}\right]$ is $\text{2}\text{.987}\times {\text{10}}^{-\text{6}}\text{M}$.

Formula

The $\text{pOH}$ is calculated using the formula,

$\text{pOH}=-{\log }_{\text{10}}\left[{\text{OH}}^{-}\right]$

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above equation.

$\begin{array}{c}\text{pOH}=-{\log }_{\text{10}}\left[{\text{OH}}^{-}\right]\\ =-{\log }_{\text{10}}\left[\text{2}\text{.987}\times {\text{10}}^{-\text{6}}\right]\\ =5.52\end{array}$

The relation between $\text{pOH}$ and $\text{pH}$ is,

$\text{pH}+\text{pOH}=\text{14}$

Substitute the value of $\text{pOH}$ in the above equation.

$\begin{array}{c}\text{pH}+\text{pOH}=\text{14}\\ \text{pH}=\text{14}-\text{5}\text{.52}\\ =\underset{\_}{8.48}\end{array}$