Chapter 15, Problem 15.61QE

(a)

Interpretation Introduction

Interpretation:

The ICE table and the equation needed to solve for the concentration of the hydrogen ion for $0.20M$${\text{C}}_6{\text{H}}_5\text{COOH}$ solution has to be determined.

Concept Introduction:

A weak acid in water produces a hydrogen ion and conjugate base. When weak acid dissolves in water, some acid molecules transfer proton to water.

In solution of weak acid, the actual concentration of the acid molecules becomes less because partial dissociation of acid has occurred and lost protons to form hydrogen ions.

The reaction is as follows:

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

The reaction is as follows:

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

The expression for $K_{\text{a}}$ is as follows:

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

For value of $K_{\text{a}}$ of different acids. (Refer to 15.6 in the book).

Answer to Problem 15.61QE

The ICE table for $0.20M$${\text{C}}_6{\text{H}}_5\text{COOH}$ is as follows:

  $\begin{array}{cccccccc}\text{Equation}& {\text{C}}_6{\text{H}}_5\text{COOH}& +& {\text{H}}_{\text{2}}\text{O}& \to & {\text{H}}_{\text{3}}{\text{O}}^{+}& +& {\text{C}}_6{\text{H}}_5{\text{COO}}^{-}\\ \text{Initial}\left(M\right)& 0.20& & 0& & 0& & 0\\ \text{Change}\left(M\right)& -x& & & & +x& & +x\\ \text{Equilibrium}\left(M\right)& 0.20-x& & & & x& & x\end{array}$

The equation needed to solve for the concentration of the hydrogen ion for $0.20M$${\text{C}}_6{\text{H}}_5\text{COOH}$ solution is as follows:

  $6.3\times {10}^{-5}=\frac{\left(x\right)\left(x\right)}{0.20-x}$

Explanation of Solution

The chemical equation for ionization of ${\text{C}}_6{\text{H}}_5\text{COOH}$ is as follows:

    ${\text{C}}_6{\text{H}}_5\text{COOH}+{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{3}}{\text{O}}^{+}+{\text{C}}_6{\text{H}}_5{\text{COO}}^{-}$

The concentration of ${\text{C}}_6{\text{H}}_5\text{COOH}$ is $0.20M$.

Also, ${\text{C}}_6{\text{H}}_5\text{COOH}$ is ionized into ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and ${\text{C}}_6{\text{H}}_5{\text{COO}}^{-}$. Therefore, concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is equal to ${\text{C}}_6{\text{H}}_5{\text{COO}}^{-}$.

Consider the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and ${\text{C}}_6{\text{H}}_5{\text{COO}}^{-}$ be x.

The ICE table for the above reaction is as follows:

  $\begin{array}{cccccccc}\text{Equation}& {\text{C}}_6{\text{H}}_5\text{COOH}& +& {\text{H}}_{\text{2}}\text{O}& \to & {\text{H}}_{\text{3}}{\text{O}}^{+}& +& {\text{C}}_6{\text{H}}_5{\text{COO}}^{-}\\ \text{Initial}\left(M\right)& 0.20& & 0& & 0& & 0\\ \text{Change}\left(M\right)& -x& & & & +x& & +x\\ \text{Equilibrium}\left(M\right)& 0.20-x& & & & x& & x\end{array}$

The expression for $K_{\text{a}}$ for ionization of ${\text{C}}_6{\text{H}}_5\text{COOH}$  is as follows:

  $K_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{C}}_6{\text{H}}_5{\text{COO}}^{-}\right]}{\left[{\text{C}}_6{\text{H}}_5\text{COOH}\right]}$        (1)

Substitute $0.20M$ for $\left[{\text{C}}_6{\text{H}}_5\text{COOH}\right]$, x for $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$, x for $\left[{\text{C}}_6{\text{H}}_5{\text{COO}}^{-}\right]$ and $6.3\times {10}^{-5}$ for $K_{\text{a}}$ in equation (1), the equation needed to solve for the concentration of the hydrogen ion is as follows:

  $6.3\times {10}^{-5}=\frac{\left(x\right)\left(x\right)}{0.20-x}$

(b)

Interpretation Introduction

Interpretation:

The ICE table and the equation needed to solve for the concentration of the hydrogen ion for $1.50M$$\text{HCOOH}$ solution has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 15.61QE

The ICE table for $1.50M$$\text{HCOOH}$ is as follows:

  $\begin{array}{cccccccc}\text{Equation}& \text{HCOOH}& +& {\text{H}}_{\text{2}}\text{O}& \to & {\text{H}}_{\text{3}}{\text{O}}^{+}& +& {\text{HCOO}}^{-}\\ \text{Initial}\left(M\right)& 1.50& & 0& & 0& & 0\\ \text{Change}\left(M\right)& -x& & & & +x& & +x\\ \text{Equilibrium}\left(M\right)& 1.50-x& & & & x& & x\end{array}$

The equation needed to solve for the concentration of the hydrogen ion for $1.50M$$\text{HCOOH}$ solution is as follows:

  $1.8\times {10}^{-4}=\frac{\left(x\right)\left(x\right)}{1.50-x}$

Explanation of Solution

The chemical equation for ionization of $\text{HCOOH}$ is as follows:

    $\text{HCOOH}+{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{3}}{\text{O}}^{+}+{\text{HCOO}}^{-}$

The concentration of $\text{HCOOH}$ is $1.50M$.

Also, $\text{HCOOH}$ is ionized into ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and ${\text{HCOO}}^{-}$. Therefore, concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is equal to ${\text{HCOO}}^{-}$.

Consider the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and ${\text{HCOO}}^{-}$ be x.

The ICE table for the above reaction is as follows:

  $\begin{array}{cccccccc}\text{Equation}& \text{HCOOH}& +& {\text{H}}_{\text{2}}\text{O}& \to & {\text{H}}_{\text{3}}{\text{O}}^{+}& +& {\text{HCOO}}^{-}\\ \text{Initial}\left(M\right)& 1.50& & 0& & 0& & 0\\ \text{Change}\left(M\right)& -x& & & & +x& & +x\\ \text{Equilibrium}\left(M\right)& 1.50-x& & & & x& & x\end{array}$

The expression for $K_{\text{a}}$ for ionization of $\text{HCOOH}$ is as follows:

  $K_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{HCOO}}^{-}\right]}{\left[\text{HCOOH}\right]}$        (2)

Substitute $1.50M$ for $\left[\text{HCOOH}\right]$, x for $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$, x for $\left[{\text{HCOO}}^{-}\right]$ and $1.8\times {10}^{-4}$ for $K_{\text{a}}$ in equation (2), the equation needed to solve for the concentration of the hydrogen ion is as follows:

  $1.8\times {10}^{-4}=\frac{\left(x\right)\left(x\right)}{1.50-x}$

(c)

Interpretation Introduction

Interpretation:

The ICE table and the equation needed to solve for the concentration of the hydrogen ion for $0.0055M$$\text{HCN}$ solution has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 15.61QE

The ICE table for $0.0055M$$\text{HCN}$ is as follows:

  $\begin{array}{cccccccc}\text{Equation}& \text{HCN}& +& {\text{H}}_{\text{2}}\text{O}& \to & {\text{H}}_{\text{3}}{\text{O}}^{+}& +& {\text{CN}}^{-}\\ \text{Initial}\left(M\right)& 0.0055& & 0& & 0& & 0\\ \text{Change}\left(M\right)& -x& & & & +x& & +x\\ \text{Equilibrium}\left(M\right)& 0.055-x& & & & x& & x\end{array}$

The equation needed to solve for the concentration of the hydrogen ion for $0.0055M$$\text{HCN}$ solution is as follows:

  $6.2\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{0.0055-x}$

Explanation of Solution

The chemical equation for ionization of $\text{HCN}$ is as follows:

    $\text{HCN}+{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{3}}{\text{O}}^{+}+{\text{CN}}^{-}$

The concentration of $\text{HCN}$ is $0.0055M$.

Also, $\text{HCN}$ is ionized into ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and ${\text{CN}}^{-}$. Therefore, concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is equal to ${\text{CN}}^{-}$.

Consider the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and ${\text{CN}}^{-}$ be $x$.

The ICE table for the above reaction is as follows:

  $\begin{array}{cccccccc}\text{Equation}& \text{HCN}& +& {\text{H}}_{\text{2}}\text{O}& \to & {\text{H}}_{\text{3}}{\text{O}}^{+}& +& {\text{CN}}^{-}\\ \text{Initial}\left(M\right)& 0.0055& & 0& & 0& & 0\\ \text{Change}\left(M\right)& -x& & & & +x& & +x\\ \text{Equilibrium}\left(M\right)& 0.055-x& & & & x& & x\end{array}$

The expression for $K_{\text{a}}$ for ionization of $\text{HCN}$ is as follows:

  $K_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{CN}}^{-}\right]}{\left[\text{HCN}\right]}$        (3)

Substitute $0.0055M$ for $\left[\text{HCN}\right]$, $x$ for $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$, $x$ for and ${\text{CN}}^{-}$ and $6.2\times {10}^{-5}$ for $K_{\text{a}}$ in equation (3), the equation needed to solve for the concentration of the hydrogen ion is as follows:

  $6.2\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{0.0055-x}$

(d)

Interpretation Introduction

Interpretation:

The ICE table and the equation needed to solve for the concentration of the hydrogen ion for $0.075M$${\text{HNO}}_2$ solution has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 15.61QE

The ICE table for $0.075M$${\text{HNO}}_2$ is as follows:

  $\begin{array}{cccccccc}\text{Equation}& {\text{HNO}}_2& +& {\text{H}}_{\text{2}}\text{O}& \to & {\text{H}}_{\text{3}}{\text{O}}^{+}& +& {\text{NO}}_2{}^{-}\\ \text{Initial}\left(M\right)& 0.075& & 0& & 0& & 0\\ \text{Change}\left(M\right)& -x& & & & +x& & +x\\ \text{Equilibrium}\left(M\right)& 0.075-x& & & & x& & x\end{array}$

The equation needed to solve for the concentration of the hydrogen ion for $0.075M$${\text{HNO}}_2$ solution is as follows:

  $5.6\times {10}^{-4}=\frac{\left(x\right)\left(x\right)}{0.075-x}$

Explanation of Solution

The chemical equation for ionization of ${\text{HNO}}_2$ is as follows:

    ${\text{HNO}}_2+{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{3}}{\text{O}}^{+}+{\text{NO}}_2{}^{-}$

The concentration of ${\text{HNO}}_2$ is $0.075M$.

Also, ${\text{HNO}}_2$ is ionized into ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and ${\text{NO}}_2{}^{-}$. Therefore, concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is equal to ${\text{NO}}_2{}^{-}$.

Consider the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and ${\text{NO}}_2{}^{-}$ be x.

The ICE table for the above reaction is as follows:

  $\begin{array}{cccccccc}\text{Equation}& {\text{HNO}}_2& +& {\text{H}}_{\text{2}}\text{O}& \to & {\text{H}}_{\text{3}}{\text{O}}^{+}& +& {\text{NO}}_2{}^{-}\\ \text{Initial}\left(M\right)& 0.075& & 0& & 0& & 0\\ \text{Change}\left(M\right)& -x& & & & +x& & +x\\ \text{Equilibrium}\left(M\right)& 0.075-x& & & & x& & x\end{array}$

The expression for $K_{\text{a}}$ for ionization of ${\text{HNO}}_2$ is as follows:

  $K_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{NO}}_2{}^{-}\right]}{\left[{\text{HNO}}_2\right]}$        (4)

Substitute $0.075M$ for $\left[{\text{HNO}}_2\right]$, $x$ for $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$, $x$ for $\left[{\text{NO}}_2{}^{-}\right]$ and $5.6\times {10}^{-4}$ for $K_{\text{a}}$ in equation (4), the equation needed to solve for the concentration of the hydrogen ion is as follows:

  $5.6\times {10}^{-4}=\frac{\left(x\right)\left(x\right)}{0.075-x}$