Chapter 13, Problem 64E

(a)

Interpretation Introduction

Interpretation: The major species present in $0.250\text{M}$ solutions of the given acids and the $\text{pH}$ value of these solutions is to be calculated.

Concept introduction: The $\text{pH}$ of a solution is define as a figure that expresses the acidity of the alkalinity of a given solution. A logarithmic scale is used on which, the value $7$ corresponds to a neutral species, a value less than $7$ corresponds to an acid and a value greater than $7$ corresponds to a base.

The $\text{pH}$ of a solution is calculated by the formula, $\text{pH}=-\log \left[{\text{H}}^{+}\right]$

To determine: The major species present in $0.250\text{M}$ solution of ${\text{C}}_6{\text{H}}_5\text{OH}$ and the $\text{pH}$ of this solution.

Answer to Problem 64E

Answer

The major species present in $0.250\text{M}$ solution of ${\text{C}}_6{\text{H}}_5\text{OH}$ are ${\text{C}}_6{\text{H}}_5\text{OH}$ and ${\text{H}}_{\text{2}}\text{O}$. The $\text{pH}$ of this solution is $\underset{\_}{5.2}$.

Explanation of Solution

Explanation

The major species present in $0.250\text{M}$ solution of ${\text{C}}_6{\text{H}}_5\text{OH}$ are ${\text{C}}_6{\text{H}}_5\text{OH}$ and ${\text{H}}_{\text{2}}\text{O}$.

${\text{C}}_6{\text{H}}_5\text{OH}$ is a weak acid. Hence, it does not completely dissociate in water.

The major species present in the given solution of ${\text{C}}_6{\text{H}}_5\text{OH}$ are ${\text{C}}_6{\text{H}}_5\text{OH}$ and ${\text{H}}_{\text{2}}\text{O}$.

The equilibrium constant expression for the stated dissociation reaction is,

$K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]}{\left[{\text{C}}_6{\text{H}}_5\text{OH}\right]}$

${\text{C}}_6{\text{H}}_5\text{OH}$ is a comparatively stronger acid than ${\text{H}}_{\text{2}}\text{O}$.

The dominant equilibrium reaction for the given case is,

${\text{C}}_6{\text{H}}_5\text{OH}\left(aq\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\left(aq\right)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$K_{\text{a}}=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Where,

• $K_{\text{a}}$ is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

$K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]}{\left[{\text{C}}_6{\text{H}}_5\text{OH}\right]}$ (1)

The $\left[{\text{H}}^{\text{+}}\right]$ is $\underset{\_}{6.3\times 10^{-6}M}$.

The change in concentration of ${\text{HNO}}_{\text{2}}$ is assumed to be $x$.

The ICE table for the stated reaction is,

$\begin{array}{cccccc}& {\text{C}}_6{\text{H}}_5\text{OH}\left(aq\right)& \rightleftharpoons & {\text{H}}^{+}\left(aq\right)& +& {\text{C}}_6{\text{H}}_5{\text{O}}^{-}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.250& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.250-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\text{C}}_6{\text{H}}_5\text{OH}\right]$ is $\left(0.250-x\right)\text{M}$.

The equilibrium concentration of $\left[{\text{H}}^{+}\right]$ is $x\text{M}$.

The equilibrium concentration of $\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]$ is $x\text{M}$.

The $K_{\text{a}}$ for ${\text{C}}_6{\text{H}}_5\text{OH}$ is $1.6\times {10}^{-10}$.

Substitute the value of $K_{\text{a}}$, $\left[{\text{C}}_6{\text{H}}_5\text{OH}\right]$, $\left[{\text{H}}^{+}\right]$ and $\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]$ in equation (1).

$\begin{array}{c}{K}_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]}{\left[{\text{C}}_6{\text{H}}_5\text{OH}\right]}\\ 1.6\times {10}^{-10}=\frac{\left[x\right]\left[x\right]}{\left[0.250-x\right]}\\ 1.6\times {10}^{-10}=\frac{{\left[x\right]}^2}{\left[0.250-x\right]}\end{array}$

The value of $x$ will be very small as compared to $0.250$. Hence, it is ignored from the term $\left[0.250-x\right]$.

Simplify the above expression.

$\begin{array}{c}1.6\times {10}^{-10}=\frac{{\left[x\right]}^2}{\left[0.250\right]}\\ {\left[x\right]}^2=\left(4.0\times {10}^{-11}\right)\\ \left[x\right]=\underset{\_}{6.3\times 10^{-6}M}\end{array}$

Therefore, the $\left[{\text{H}}^{\text{+}}\right]$ that is equal to $x$ is $\underset{\_}{6.3\times 10^{-6}M}$.

The $\text{pH}$ of the solution is $\underset{\_}{5.2}$.

The calculated value of $\left[{\text{H}}^{\text{+}}\right]$ is $\text{6}{\text{.3×10}}^{-\text{6}}\text{M}$.

The $\text{pH}$ of a solution is calculated by the formula,

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

Substitute the value of $\left[{\text{H}}^{\text{+}}\right]$ in the above expression.

$\begin{array}{c}\text{pH}=-\log \left[\text{6}{\text{.3×10}}^{-\text{6}}\right]\\ =\underset{\_}{5.2}\end{array}$

(b)

Interpretation Introduction

Interpretation: The major species present in $0.250\text{M}$ solutions of the given acids and the $\text{pH}$ value of these solutions is to be calculated.

Concept introduction: The $\text{pH}$ of a solution is define as a figure that expresses the acidity of the alkalinity of a given solution. A logarithmic scale is used on which, the value $7$ corresponds to a neutral species, a value less than $7$ corresponds to an acid and a value greater than $7$ corresponds to a base.

The $\text{pH}$ of a solution is calculated by the formula, $\text{pH}=-\log \left[{\text{H}}^{+}\right]$

To determine: The major species present in $0.250\text{M}$ solution of $\text{HCN}$ and the $\text{pH}$ of this solution.

Answer to Problem 64E

Answer

The major species present in $0.250\text{M}$ solution of $\text{HCN}$ are $\text{HCN}$ and ${\text{H}}_{\text{2}}\text{O}$. The $\text{pH}$ of this solution is $\underset{\_}{4.9}$.

Explanation of Solution

Explanation

The major species present in $0.250\text{M}$ solution of $\text{HCN}$ are $\text{HCN}$ and ${\text{H}}_{\text{2}}\text{O}$.

$\text{HCN}$ is a weak acid. Hence, it does not completely dissociate in water.

The major species present in the given solution of $\text{HCN}$ are $\text{HCN}$ and ${\text{H}}_{\text{2}}\text{O}$.

The equilibrium constant expression for the given reaction is, $K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{CN}}^{-}\right]}{\left[\text{HCN}\right]}$

$\text{HCN}$ is a comparatively stronger acid than ${\text{H}}_{\text{2}}\text{O}$.

The dominant equilibrium reaction for the given case is,

$\text{HCN}\left(aq\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{CN}}^{-}\left(aq\right)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$K_{\text{a}}=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Where,

• $K_{\text{a}}$ is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

$K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{CN}}^{-}\right]}{\left[\text{HCN}\right]}$ (1)

The $\left[{\text{H}}^{\text{+}}\right]$ is $\underset{\_}{1.2\times 10^{-5}M}$.

The change in concentration of $\text{HCN}$ is assumed to be $x$.

The ICE table for the stated reaction is,

$\begin{array}{cccccc}& \text{HCN}\left(aq\right)& \rightleftharpoons & {\text{H}}^{+}\left(aq\right)& +& {\text{CN}}^{-}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.250& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.250-x& & x& & x\end{array}$

The equilibrium concentration of $\left[\text{HCN}\right]$ is $\left(0.250-x\right)\text{M}$.

The equilibrium concentration of $\left[{\text{H}}^{+}\right]$ is $x\text{M}$.

The equilibrium concentration of $\left[{\text{CN}}^{-}\right]$ is $x\text{M}$.

The $K_{\text{a}}$ for $\text{HCN}$ is $6.2\times {10}^{-10}$.

Substitute the value of $K_{\text{a}}$, $\left[\text{HCN}\right]$, $\left[{\text{H}}^{+}\right]$ and $\left[{\text{CN}}^{-}\right]$ in equation (1).

$\begin{array}{c}{K}_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{CH}}_3{\text{COO}}^{-}\right]}{\left[{\text{CH}}_3\text{COOH}\right]}\\ 6.2\times {10}^{-10}=\frac{\left[x\right]\left[x\right]}{\left[0.250-x\right]}\\ 6.2\times {10}^{-10}=\frac{{\left[x\right]}^2}{\left[0.250-x\right]}\end{array}$

The value of $x$ will be very small as compared to $0.250$. Hence, it is ignored from the term $\left[0.250-x\right]$.

Simplify the above expression.

$\begin{array}{c}6.2\times {10}^{-10}=\frac{{\left[x\right]}^2}{\left[0.250\right]}\\ {\left[x\right]}^2=\left(1.55\times {10}^{-10}\right)\\ \left[x\right]=\underset{\_}{1.2\times 10^{-5}M}\end{array}$

Therefore, the $\left[{\text{H}}^{\text{+}}\right]$ that is equal to $x$ is $\underset{\_}{1.2\times 10^{-5}M}$.

The $\text{pH}$ of the solution is $\underset{\_}{4.9}$.

The calculated value of $\left[{\text{H}}^{\text{+}}\right]$ is $1.25\times {10}^{-5}\text{M}$.

The $\text{pH}$ of a solution is calculated by the formula,

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

Substitute the value of $\left[{\text{H}}^{\text{+}}\right]$ in the above expression.

$\begin{array}{c}\text{pH}=-\log \left[1.25\times {10}^{-5}\right]\\ =\underset{\_}{4.9}\end{array}$