Chapter 19.4, Problem 41LC

(a)

Interpretation Introduction

Interpretation: The number of moles of HCl required to neutralize 0.03 mol of KOH needs to be determined.

Concept Introduction: A reaction between an acid and a base is known as a neutralization reaction. Here, the acid reacts with the base to form salt and water. The general neutralization reaction is represented as follows:

  $\text{HA}+\text{BOH}\to \text{BA}+{\text{H}}_{\text{2}}\text{O}$

Here, HA is an acid, BOH is a base and BA is salt.

Explanation of Solution

The given solution is 0.03 mol KOH. To neutralize the given base, the number of moles of HCl required can be calculated by determining the number of moles of hydroxide ions present in the solution.

In 0.03 mol of KOH, the number of moles of hydroxide ions or ${\text{OH}}^{-}$ ions will be 0.03 mol as KOH is a strong base.

The number of moles of hydrogen ions required to neutralize the hydroxide ion will be 0.03 moles.

Similarly, HCl is a strong acid; thus, the number of moles of hydrogen ion will be equal to the number of moles of acid. Thus, the number of moles of HCl required will be 0.03 moles.

(b)

Interpretation Introduction

Interpretation: The number of moles of HCl required to neutralize 2 mol of ${\text{NH}}_{\text{3}}$ needs to be determined.

Concept Introduction: A reaction between an acid and a base is known as a neutralization reaction. Here, the acid reacts with the base to form salt and water. The general neutralization reaction is represented as follows:

  $\text{HA}+\text{BOH}\to \text{BA}+{\text{H}}_{\text{2}}\text{O}$

Here, HA is an acid, BOH is a base and BA is salt.

Explanation of Solution

The given solution is 2 mol of ${\text{NH}}_{\text{3}}$ . Here, ammonia is a weak base with a base dissociation constant $1.8\times {10}^{-5}$ . Thus, the hydroxide ion concentration can be calculated as follows:

  ${\text{NH}}_3+{\text{H}}_{\text{2}}\text{O}\rightleftharpoons {\mathrm{NH}}_4{}^{+}+{\text{OH}}^{-}$

The ICE table can be represented as follows:

  $\begin{array}{l}{\text{ NH}}_3+{\text{H}}_{\text{2}}\text{O}\rightleftharpoons {\mathrm{NH}}_4{}^{+}+{\text{OH}}^{-}\\ I\text{ C - - -}\\ C-\text{C}\alpha \text{ - +C}\alpha \text{ +C}\alpha \\ E\text{ C}-\text{C}\alpha \text{ - C}\alpha \text{ C}\alpha \end{array}$

The expression for the base dissociation constant will be:

  $K_b=\frac{\left[{\text{NH}}_4^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\mathrm{NH}}_3\right]}$

Substitute the values,

  $K_b=\frac{\left(C\alpha \right)\left(C\alpha \right)}{\left(C-C\alpha \right)}$

Or,

  $K_b=\frac{\left(\alpha \right)\left(C\alpha \right)}{\left(1-\alpha \right)}$

Considering 1 L solution thus, the value of C will be 2 M.

Or,

  $\begin{array}{l}1.8\times {10}^{-5}=\frac{2{\alpha }^2}{\left(1-\alpha \right)}\\ 2{\alpha }^2=1.8\times {10}^{-5}-1.8\times {10}^{-5}\alpha \\ 2{\alpha }^2+1.8\times {10}^{-5}\alpha -1.8\times {10}^{-5}=0\end{array}$

The value of $\alpha$ will be:

  $\begin{array}{l}x=\frac{-1.8\times {10}^{-5}\pm \sqrt{{\left(1.8\times {10}^{-5}\right)}^2-4\left(2\right)\left(-1.8\times {10}^{-5}\right)}}{2\left(2\right)}\\ =\frac{-1.8\times {10}^{-5}\pm \sqrt{{\left(1.8\times {10}^{-5}\right)}^2-4\left(2\right)\left(-1.8\times {10}^{-5}\right)}}{2\left(2\right)}\\ =\frac{-1.8\times {10}^{-5}\pm 0.012}{4}\\ =-0.003,+0.003\end{array}$

The value of x cannot be negative; thus, $x=\alpha =0.003$

The hydroxide ion concentration will be:

  $\begin{array}{c}\left[{\text{OH}}^{-}\right]=C\alpha \\ =2\left(0.003\right)\\ =0.006\text{ M}\end{array}$

For 1 L solution, the number of moles of hydroxide ion will be 0.006 mol.

The number of hydrogen ions required to neutralize will be 0.006 mol. Again, HCl is a strong acid thus, the number of moles of hydrogen ion is equal to the number of moles of HCl which is 0.006 mol.

(c)

Interpretation Introduction

Interpretation: The number of moles of HCl required to neutralize 0.1 mol of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ needs to be determined.

Concept Introduction: A reaction between an acid and a base is known as a neutralization reaction. Here, the acid reacts with the base to form salt and water. The general neutralization reaction is represented as follows:

  $\text{HA}+\text{BOH}\to \text{BA}+{\text{H}}_{\text{2}}\text{O}$

Here, HA is an acid, BOH is a base and BA is salt.

Explanation of Solution

The given solution is 0.1 mol of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ .

Here, calcium hydroxide is a strong base and completely dissociates into calcium and hydroxide ions. The reaction is represented as follows:

  $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\to {\text{Ca}}^{2+}+2{\text{OH}}^{-}$

Thus, the number of moles of hydroxide ions in the solution will be:

  $\begin{array}{c}\left[{\text{OH}}^{-}\right]=2\times 0.1\text{ mol}\\ =0.2\text{ mol}\end{array}$

Now, the concentration of hydrogen ions required to neutralize it will be 0.2 mol. Since HCl is a strong acid it gets completely dissociate into hydrogen ion and conjugate base. The number of moles of HCl required will be 0.2 mol.