Chapter 8, Problem 168AE

(a)

Interpretation Introduction

Interpretation:

The four titrations need to be ranked in order of increasing pH at the halfway point to equivalence.

Concept Introduction: For weak acid and base titrations, the pH at equivalence point is equal to its $p{K}_a$ value. If any strong acid or base is present in the solution, the pH at halfway equivalence will be always lower than the pH with strong species.

Explanation of Solution

(i)

At halfway point equivalence, the pH of the solution is equal to $p{K}_a$ . Now, the value for ammonium ion can be calculated as follows:

  $p{K}_a=-\log K_a=-\log \left(5.6\times {10}^{-10}\right)=9.25$

Thus, the pH of the solution at halfway point is 9.25.

(ii)

For a strong acid and strong base titration, the halfway point to the equivalence occurs when the solution is most acidic.

Thus, the titrations with weak species will always have higher pH at halfway equivalence than a titration of strong acid and strong base.

(iii)

At halfway point, for weak acid and strong base the pH of the solution is equal to $p{K}_a$ values.

The acid dissociation constant is $3.5\times {10}^{-8}$ .

Thus, $p{K}_a$ value will be:

  $p{K}_a=-\log K_a=-\log \left(3.5\times {10}^{-8}\right)=7.45$

The pH of solution at halfway point to the equivalence is 7.45.

(iv)

For weak acid-strong base titration at halfway point of the equivalence, the pH is equal to $p{K}_a$ Here,

  $p{K}_a=-\log K_a$

Thus, pH of the solution is:

  $\begin{array}{l}pH=-\log K_a\\ =-\log \left(7.2\times {10}^{-4}\right)\\ =3.14\end{array}$

Thus, the pH of the solution at halfway point is 3.14.

Thus, the order of pH at halfway point to equivalence is as follows:

  $ii<iv<iii<i$

(b)

Interpretation Introduction

Interpretation:

The four titrations need to be ranked in order of increasing pH at the equivalence point.

Concept Introduction: The relation between molarity, number of moles and volume of solution is as follows:

  $M=\frac{n}{V}$

Here, n is number of moles and V is volume of the solution.

Explanation of Solution

(i)

The molarity and volume of ${\text{NH}}_{\text{3}}$ solution is 0.2 M and 150 mL respectively. Similarly, the molarity of HCl is 0.2 M. The titration of HCl and ${\text{NH}}_{\text{3}}$ is a strong acid-weak base titration.

Calculate the number of moles of each species reacting as follows:

  $n_{{\text{NH}}_{\text{3}}}=\left(0.2\text{ M}\right)\left(150\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)=30\times {10}^{-3}\text{ mol}$

Now, at equivalence point the number of moles of acid and base will be equal thus, number of moles of HCl will also be $30\times {10}^{-3}\text{ mol}$ .

The molarity of HCl solution is 0.2 M thus, volume of HCl can be calculated as follows:

  $V_{HCl}=\frac{n}{M}=\frac{30\times {10}^{-3}\text{ mol}}{0.2\text{ M}}=150\text{ mL}$

The reaction between ammonia and HCl results in the complete neutralization and formation of $30\times {10}^{-3}\text{ mol}$ ammonium ion takes place.

The total volume of the solution will be sum of volume of ammonia and HCl which is 150 mL +150 mL that is 300 mL.

The molarity of ammonium ion in this solution will be:

  $M=\frac{30\times {10}^{-3}\text{ mol}}{300\text{ mL}\left[\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right]}=0.10\text{ M}$

The base dissociation constant of ammonia is $1.8\times {10}^{-5}$ . Thus, ammonium ion being its conjugate acid will have acid dissociation constant equal to:

  $K_a=\frac{K_w}{K_b}=\frac{{10}^{-14}}{1.8\times {10}^{-5}}=5.6\times {10}^{-10}$

The dissociation of ammonium ion can be represented as follows:

  ${\text{NH}}_4{}^{+}\to {\text{NH}}_{\text{3}}+{\text{H}}^{+}$

The ICE table can be prepared for the above reaction as follows:

  $\begin{array}{l}{\text{ NH}}_4{}^{+}\to {\text{NH}}_{\text{3}}+{\text{H}}^{+}\\ I\text{ 0}\text{.1 - -}\\ C\text{ -x x x}\\ E\text{ 0}\text{.1-x x x}\end{array}$

The expression for equilibrium constant will be:

  $\begin{array}{l}{K}_a=\frac{\left[N{H}_3\right]\left[H^{+}\right]}{\left[N{H}_4^{+}\right]}=\frac{x^2}{0.1}=5.6\times {10}^{-10}\\ x=7.5\times {10}^{-6}\text{ M}\end{array}$

The value of x is concentration of hydrogen ion. The pH of solution can be calculated as follows:

  $pH=-\log \left[{\text{H}}^{+}\right]=-\log \left(7.5\times {10}^{-6}\right)=5.12$

Thus, the pH of solution at equivalence point is 5.12.

(ii)

The given volume and molarity of HCl is 150 mL and 0.2 M respectively. Also, the molarity of NaOH is 0.2 M.

The number of moles of HCl can be calculated as follows:

  $n=M\times V$

Putting the values,

  $n=\left(0.2\text{ M}\right)\left(150\text{ mL}\right)=30\text{ mmol}$

At equivalence point, the number of moles of NaOH will be equal to HCl thus,

  $n_{NaOH}=30\text{ mmol}$

Now, molarity of NaOH is 0.2 M thus, volume of NaOH can be calculated as follows:

  $V=\frac{n}{M}$

Putting the values,

  $V=\frac{30\text{ mmol}}{0.2\text{ M}}=150\text{ mL}$

Thus, volume of NaOH added is 150 mL.

The dissociation of a strong acid and strong base is a complete dissociation. Thus, the pH of the titration at equivalence point will be 7.

(iii)

The volume and molarity of HOCl is 150 mL and 0.2 M respectively. The molarity of NaOH is 0.2 M.

The titration is a weak acid strong base titration.

The number of moles of HOCl can be calculated as follows:

  $n=M\times V=\left(0.2\text{ M}\right)\left(150\text{ mL}\right)=30\text{ mmol}$

The number of moles of HOCl will be equal to number of moles of NaOH at equivalent point. Thus, the number of moles of NaOH will be 30 mmol.

The volume of NaOH added at equivalence point can be calculated as follows:

  $V=\frac{n}{M}=\frac{30\text{ mmol}}{0.2\text{ M}}=150\text{ mL}$

Thus, the volume of NaOH added will be 150 mL.

Now, the reaction of HOCl and NaOH results in the formation of 30 mmol ${\text{OCl}}^{-}$ . The total volume is sum of volume of HOCl and NaOH which is 300 mL.

Molarity of solution can be calculated as follows:

  $M=\frac{30\text{ mmol}}{300\text{ mL}}=0.10\text{ M}$

Here, HOCl is a weak acid and ${\text{OCl}}^{-}$ is strong conjugate base. The equilibrium reaction is represented as follows:

  $OC{l}^{-}\left(aq\right)+H_{2}O\left(aq\right)\rightleftharpoons HOCl\left(aq\right)+O{H}^{-}\left(aq\right)$

The ICE table can be represented as follows:

  $\begin{array}{l}OC{l}^{-}\left(aq\right)+H_{2}O\left(aq\right)\rightleftharpoons HOCl\left(aq\right)+O{H}^{-}\left(aq\right)\\ I\text{ 0}\text{.1 - - -}\\ C\text{ -x - +x +x}\\ E\text{ 0}\text{.10-x x x}\end{array}$

The base dissociation constant for ${\text{OCl}}^{-}$ can be calculated as follows:

  $k_b=\frac{k_w}{k_a}=\frac{{10}^{-14}}{3.5\times {10}^{-8}}=2.86\times {10}^{-7}$

The expression for base dissociation constant can be represented as follows:

  $k_b=\frac{\left[HOCl\right]\left[O{H}^{-}\right]}{\left[OC{l}^{-}\right]}$

Putting all the values,

  $\begin{array}{l}2.86\times {10}^{-7}=\frac{x^2}{0.10}\\ x=1.7\times {10}^{-4}\text{ M}\end{array}$

Thus, the equilibrium concentration of hydroxide ion is $1.7\times {10}^{-4}\text{ M}$ .

The pH of solution can be calculated as follows:

  $\begin{array}{l}pH=14-pOH\\ =14-\left(-\log \left(1.7\times {10}^{-4}\text{ M}\right)\right)\\ =10.2\end{array}$

Here, the pH at equivalence point is 10.2.

(iv)

The molarity and volume HF is 0.2 M and 150 mL respectively. Molarity of NaOH is 0.2 M.

The number of moles of HF can be calculated as follows:

  $n=\left(0.2\text{ M}\right)\left(150\text{ mL}\right)=30\text{ mmol}$

At equivalence point, the number of moles of HF and NaOH is same thus, number of moles of NaOH will be 30 mmol.

Volume of NaOH can be calculated as follows:

  $V=\frac{n}{M}=\frac{30\text{ mmol}}{0.2\text{ M}}=150\text{ mL}$

Thus, the volume of NaOH added at equivalent point is 150 mL.

The reaction of HF and hydroxide ion results in the formation of F- ion.

The number of moles of F- ion will be 30 mmol.

The total volume of the solution is 300 mL, the molarity of solution can be calculated as follows:

  $M=\frac{30\text{ mmol}}{300\text{ mL}}=0.10\text{ M}$

HF is weak acid and F- is its strong conjugate base that abstract proton from water. The equilibrium reaction is represented as follows:

  $\begin{array}{l}{F}^{-}+H_{2}O\rightleftharpoons HF+O{H}^{-}\\ I\text{ 0}\text{.1 - 0 0}\\ C\text{ -x +x +x}\\ E\text{ 0}\text{.1-x x x}\end{array}$

The base dissociation constant can be represented as follows:

  $K_b=\frac{K_w}{K_a}=\frac{{10}^{-14}}{7.2\times {10}^{-4}}=1.4\times {10}^{-11}$

The expression for base dissociation constant is represented as follows:

  $K_b=\frac{\left[HF\right]\left[O{H}^{-}\right]}{\left[F^{-}\right]}$

Putting all the values,

  $\begin{array}{l}1.4\times {10}^{-11}=\frac{x^2}{0.10}\\ x=1.2\times {10}^{-6}\text{ M}\end{array}$

The pH of the solution can be calculated as follows:

  $\begin{array}{l}pH=14-\left(-\log \left[{\text{OH}}^{-}\right]\right)\\ =14-\left(-\log \left(1.2\times {10}^{-6}\right)\right)\\ =14-5.92\\ =8.08\end{array}$

Thus, the pH at equivalence point is 8.08.

Also, the increasing order of pH at the equivalence point is as follows:

  $i<ii<iv<iii$

(c)

Interpretation Introduction

Interpretation:

The titration that requires the largest volume of the titrant to reach the equivalence point needs to be determined.

Concept Introduction: The relation between molarity, number of moles and volume of solution is as follows:

  $M=\frac{n}{V}$

Here, n is number of moles and V is volume of the solution.

Explanation of Solution

(i)

The molarity and volume of ${\text{NH}}_{\text{3}}$ solution is 0.2 M and 150 mL respectively. Similarly, the molarity of HCl is 0.2 M. The titration of HCl and ${\text{NH}}_{\text{3}}$ is a strong acid-weak base titration.

Calculate the number of moles of each species reacting as follows:

  $n_{{\text{NH}}_{\text{3}}}=\left(0.2\text{ M}\right)\left(150\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)=30\times {10}^{-3}\text{ mol}$

Now, at equivalence point the number of moles of acid and base will be equal thus, number of moles of HCl will also be $30\times {10}^{-3}\text{ mol}$ .

The molarity of HCl solution is 0.2 M thus, volume of HCl can be calculated as follows:

  $V_{HCl}=\frac{n}{M}=\frac{30\times {10}^{-3}\text{ mol}}{0.2\text{ M}}=150\text{ mL}$

(ii)

The given volume and molarity of HCl is 150 mL and 0.2 M respectively. Also, the molarity of NaOH is 0.2 M.

The number of moles of HCl can be calculated as follows:

  $n=M\times V$

Putting the values,

  $n=\left(0.2\text{ M}\right)\left(150\text{ mL}\right)=30\text{ mmol}$

At equivalence point, the number of moles of NaOH will be equal to HCl thus,

  $n_{NaOH}=30\text{ mmol}$

Now, molarity of NaOH is 0.2 M thus, volume of NaOH can be calculated as follows:

  $V=\frac{n}{M}$

Putting the values,

  $V=\frac{30\text{ mmol}}{0.2\text{ M}}=150\text{ mL}$

Thus, volume of NaOH added is 150 mL.

(iii)

The volume and molarity of HOCl is 150 mL and 0.2 M respectively. The molarity of NaOH is 0.2 M.

The titration is a weak acid strong base titration.

The number of moles of HOCl can be calculated as follows:

  $n=M\times V=\left(0.2\text{ M}\right)\left(150\text{ mL}\right)=30\text{ mmol}$

The number of moles of HOCl will be equal to number of moles of NaOH at equivalent point. Thus, the number of moles of NaOH will be 30 mmol.

The volume of NaOH added at equivalence point can be calculated as follows:

  $V=\frac{n}{M}=\frac{30\text{ mmol}}{0.2\text{ M}}=150\text{ mL}$

Thus, the volume of NaOH added will be 150 mL.

(iv)

The molarity and volume HF is 0.2 M and 150 mL respectively. Molarity of NaOH is 0.2 M.

The number of moles of HF can be calculated as follows:

  $n=\left(0.2\text{ M}\right)\left(150\text{ mL}\right)=30\text{ mmol}$

At equivalence point, the number of moles of HF and NaOH is same thus, number of moles of NaOH will be 30 mmol.

Volume of NaOH can be calculated as follows:

  $V=\frac{n}{M}=\frac{30\text{ mmol}}{0.2\text{ M}}=150\text{ mL}$

Thus, the volume of NaOH added at equivalent point is 150 mL.

From the above calculations, the four titrations require same volume at equivalence point of the titrant.