Chapter 15, Problem 34QRT

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the solution before start of titration has to be calculated.

Concept Introduction:

Molarity is defined as the ratio of number of moles of solute to the volume of solution is liters.  The S.I. unit if molarity is molar and it is represented by $\text{M}$.

Mathematical formulation of molarity is shown as follows:

  $\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution }\left(\text{L}\right)}$

The $\text{pH}$ of a solution is the negative logarithmic of hydrogen ion concentration in the solution.  Mathematically it can be represented as follows:

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

Answer to Problem 34QRT

The $\text{pH}$ of the solution is $\underset{\_}{1.00}$.

Explanation of Solution

Let the strong acid be hydrochloric acid and weak base be ammonia.

The concentration and volume of strong acid is $0.100\text{M}$ and $20.0\text{mL}$ respectively.

The concentration of weak base is $0.100\text{M}$.

As before the titration of strong acid and weak base, there is no base added to the solution of strong acid.

Thus, the $\text{pH}$ of strong acid is due to the original concentration of strong acid itself.

The relation between $\text{pH}$ and hydrogen ion concentration, $\left[{\text{H}}^{+}\right]$ in the solution is represented as follows:

  $\text{pH}=-\log \left[{\text{H}}^{+}\right]$        (1)

The value of $\left[{\text{H}}^{+}\right]$ is $0.100\text{M}$.

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the equation (1).

  $\begin{array}{c}\text{pH}=-\log \left(0.100\right)\\ =1.00\end{array}$

Therefore, the $\text{pH}$ of acid solution before titration is $\underset{\_}{1.00}$.

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the solution when half the volume of the titrant has been added has to be calculated.

Concept Introduction:

Same as part (a).

Answer to Problem 34QRT

The $\text{pH}$ of the solution is $\underset{\_}{1.5}$.

Explanation of Solution

It is assumed that the volume of base used for complete titration is ${\text{V}}_2\text{mL}$.

The volume of base can be calculated using molarity equation as shown below.

  ${\text{M}}_1{\text{V}}_1={\text{M}}_2{\text{V}}_2$        (2)

Where,

• ${\text{M}}_1$ is the molarity of acid.
• ${\text{V}}_1$ is the volume of acid.
• ${\text{M}}_2$ is the molarity of base.
• ${\text{V}}_2$ is the volume of base.

The value of ${\text{M}}_1$ is $0.100\text{M}$.

The value of ${\text{V}}_1$ is $20.0\text{mL}$.

The value of ${\text{M}}_2$ is $0.100\text{M}$.

Substitute the value of ${\text{M}}_1$, ${\text{V}}_1$ and ${\text{M}}_2$ in the equation (2).

  $\begin{array}{c}0.100\text{M}\times 20.0\text{mL}=0.100\text{M}\times {\text{V}}_2\\ {\text{V}}_2=\frac{0.100\times 20.0}{0.100}\text{mL}\\ =20.0\text{mL}\end{array}$

The number of moles of acid in the solution is calculated by using the relation shown below.

  $n=M\times V$        (3)

Where,

• $n$ is the number of moles.
• $M$ is the molarity.
• $V$ is the volume.

The value of $M$ for acid is $0.100\text{M}$.

The value of $V$ foracid is $20.0\text{mL}$.

Substitute the values of $M$ and $V$ for acid in the equation (3).

  $\begin{array}{c}{n}_{\text{acid}}=0.100\text{M}\times 20.0\text{mL}\times \frac{1\text{L}}{1000\text{mL}}\\ =2.00\times {10}^{-3}\text{mol}\end{array}$

The value of $M$ for baseis $0.100\text{M}$.

The value of $V$ for base is $20.0\text{mL}$.

Substitute the values of $M$ and $V$ for base in the equation (3).

  $\begin{array}{c}{n}_{\text{base}}=0.100\text{M}\times 20.0\text{mL}\times \frac{1\text{L}}{1000\text{mL}}\\ =2.00\times {10}^{-3}\text{mol}\end{array}$

Since when half of the volume of base is added, it neutralizes half of the acid solution.

Therefore, the number of moles of acid left in the solution is $1.00\times {10}^{-3}\text{mol}$.

The total volume of the solution after addition base to the acid solutionis calculated as shown below.

  $\begin{array}{c}{V}_{\text{total}}=20.0\text{mL}+10.0\text{mL}\\ =30.0\text{mL}\end{array}$

The final concentration of hydrogen ions in the solution is calculated by using the relation shown below.

  $M=\frac{n}{V}$        (4)

Where,

• $n$ is the number of moles.
• $M$ is the molarity.
• $V$ is the volume.

The value of $n$ for ${\text{H}}^{+}$ ions is $1.00\times {10}^{-3}\text{mol}$.

The value of $V$ for ${\text{H}}^{+}$ ions is $30.0\text{mL}$.

Substitute the values of $n$ and $V$ for ${\text{H}}^{+}$ ions in the equation (4).

  $\begin{array}{c}\left[{\text{H}}^{+}\right]=\frac{1.00\times {10}^{-3}\text{mol}}{\left(30.0\text{mL}\times \frac{1\text{L}}{1000\text{mL}}\right)}\\ =0.033\text{M}\end{array}$

Thus, the final concentration of hydrogen ions in the solution is $0.033\text{M}$.

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the equation (1).

  $\begin{array}{c}\text{pH}=-\log \left(0.033\text{M}\right)\\ =1.5\end{array}$

Hence, the $\text{pH}$ of the solution is $\underset{\_}{1.5}$.

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the solution at the equivalence point has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 34QRT

The $\text{pH}$ of the solution is $\underset{\_}{11.9}$.

Explanation of Solution

The equivalence point during titration of acid and base is that point at which the number of moles of acid completely neutralizes the number of moles of base.

At the equivalence point of hydrochloric acid and ammonia, the $\text{pH}$ of the solution is governed by hydrolysis of ammonium salt which is formed in the solution.

The chemical reaction for the hydrolysis of ammonium salt is shown below.

${\text{NH}}_4{}^{+}+{\text{H}}_2{\text{ONH}}_3+{\text{H}}_3{\text{O}}^{+}$

The concentration of ammonium ion at this point is $5.0\times {10}^{-5}\text{M}$.

The standard value of $K_{\text{a}}$ for strong acid that is hydrochloric acid is $1.3\times {10}^6$.

The base dissociation constant is calculated using the formula shown below.

  $K_{\text{b}}=\frac{K_{\text{w}}}{K_{\text{a}}}$        (5)

Where,

• $K_{\text{b}}$ is the base dissociation constant.
• $K_{\text{a}}$ is the acid dissociation constant.
• $K_{\text{w}}$ is the hydrolysis constant.

The value of $K_{\text{a}}$ is $1.3\times {10}^6$.

The value of $K_{\text{w}}$ is $1.0\times {10}^{-14}$.

Substitute the value of $K_{\text{a}}$ and $K_{\text{w}}$ in the equation (5).

  $\begin{array}{c}{K}_{\text{b}}=\frac{1.0\times {10}^{-14}}{1.3\times {10}^6}\\ =7.69\times {10}^{-21}\end{array}$

Let the number of moles of hydronium ions at equilibrium be $x\text{mol}$.

Therefore the concentration of ammonium ion at equilibrium is $\left(5.0\times {10}^{-5}-x\right)\text{M}$.

Thus, the concentration of hydronium ions is calculated as shown below.

$\begin{array}{c}\frac{x^2}{\left(5.0\times {10}^{-5}-x\right)}=7.69\times {10}^{-21}\\ x^2+7.69\times {10}^{-21}x-3.8\times {10}^{-25}=0\\ x=\frac{\left(-7.69\times {10}^{-21}\right)\pm \sqrt{{\left(7.69\times {10}^{-21}\right)}^2-4\left(1\right)\left(-3.8\times {10}^{-25}\right)}}{2}\\ =1.23\times {10}^{-12}\end{array}$

Therefore, the final concentration of hydrogen ions in the solution is $1.23\times {10}^{-12}$.

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the equation (1).

  $\begin{array}{c}\text{pH}=-\log \left(1.23\times {10}^{-12}\right)\\ =11.9\end{array}$

Hence, the $\text{pH}$ of the solution is $\underset{\_}{11.9}$.

(d)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the solution when $10.0\text{mL}$ of the base added to the solution after the equivalence point has to be calculated.

Concept Introduction:

Same as part (a).

Answer to Problem 34QRT

The $\text{pH}$ of the solution is $\underset{\_}{12.31}$.

Explanation of Solution

The initial volume of the solution containing strong acid and weak base is $40.0\text{mL}$.

As per the given data, $10.0\text{mL}$ of the base is added to the solution.  Therefore, the total volume of the solution after addition of base is $50.0\text{mL}$.

After the equivalence point, there is the salt solution.  Thus after addition of extra base, the number of moles of base in the solution is $1.00\times {10}^{-3}\text{mol}$.

The final concentration of hydroxide ions in the solution is calculated by using the relation shown below.

  $M=\frac{n}{V}$        (6)

Where,

• $n$ is the number of moles.
• $M$ is the molarity.
• $V$ is the volume.

The value of $n$ for ${\text{OH}}^{-}$ ions is $1.00\times {10}^{-3}\text{mol}$.

The value of $V$ for ${\text{OH}}^{-}$ ions is $30.0\text{mL}$.

Substitute the values of $n$ and $V$ for ${\text{OH}}^{-}$ ions in the equation (6).

  $\begin{array}{c}\left[{\text{OH}}^{-}\right]=\frac{1.00\times {10}^{-3}\text{mol}}{\left(50.0\text{mL}\times \frac{1\text{L}}{1000\text{mL}}\right)}\\ =2.0\times {10}^{-2}\text{M}\end{array}$

Thus, the final concentration of ${\text{OH}}^{-}$ ions in the solution is $2.0\times {10}^{-2}\text{M}$.

The relation between $\text{pOH}$ and hydrogen ion concentration, $\left[{\text{OH}}^{-}\right]$ in the solution is represented as shown below.

  $\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$        (7)

The value of $\left[{\text{OH}}^{-}\right]$ is $2.0\times {10}^{-2}\text{M}$.

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the equation (7).

  $\begin{array}{c}\text{pOH}=-\log \left(2.0\times {10}^{-2}\right)\\ =1.69\end{array}$

The relation between $\text{pOH}$ and $\text{pH}$ is shown below.

  $\text{pH}+\text{pOH}=14$        (8)

Substitute the value of $\text{pOH}$ in the equation (8).

  $\begin{array}{c}\text{pH}+\text{1}\text{.69}=14\\ \text{pH}=14-1.69\\ =12.31\end{array}$

Hence, the $\text{pH}$ of the solution is $\underset{\_}{12.31}$.

The titration curve is shown below.

Figure 1