Chapter 8, Problem 166AE

(a)

Interpretation Introduction

Interpretation: The pH of the solution needs to be calculated when 0 mL of KOH is added to given solution.

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

In the given solution, HCN is a weak acid and KOH is strong base. Molarity of HCN is 0.1 M and volume is 100 mL.

The acid dissociation constant of HCN is $6.2\times {10}^{-10}$ .

The number of moles of HCN can be calculated using molarity and volume as follows:

  $\begin{array}{l}n=M\times V\\ =\left(0.100\text{ M}\right)\left(100\text{ mL}\right)\\ =10\text{ mmol}\end{array}$

When 0 mL of KOH is added, the solution only contains a weak acid.

The equilibrium reaction for the dissociation of HCN is represented as follows:

  $HCN\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+C{N}^{-}\left(aq\right)$

The expression for equilibrium constant will be:

  $K_a=\frac{\left[C{N}^{-}\right]\left[H^{+}\right]}{\left[HCN\right]}$

The ICE table for the reaction can be represented as follows:

  $\begin{array}{l}HCN\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+C{N}^{-}\left(aq\right)\\ I\text{ 0}\text{.1 0 0}\\ C\text{ -x +x +x}\\ E\text{ 0}\text{.1-x x x}\end{array}$

The value of x can be neglected from equilibrium constant of HCN as the dissociation constant is very small.

Thus,

  $\begin{array}{l}6.2\times {10}^{-10}=\frac{x^2}{0.1}\\ x=7.9\times {10}^{-6}\text{ M}\end{array}$

The equilibrium concentration of hydrogen ion is $7.9\times {10}^{-6}\text{ M}$ .

Now, the pH of solution can be calculated as follows:

  $pH=-\log \left(7.9\times {10}^{-6}\right)=5.1$

Thus, the pH of the solution when 0.0 mL of KOH added is 5.1.

(b)

Interpretation Introduction

Interpretation: The pH of the solution needs to be calculated when 50 mL of KOH is added to given solution.

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

When 50 mL of KOH is added the number of moles of hydroxide ion can be calculated as follows:

  $n_{{\text{OH}}^{-}}=M\times V=0.1\text{ M}\times 50.0\text{ mL=5 mmol}$

Here, 5.00 mmol of KOH reacts completely with 5.00 mmol of HCN. The ICE table can be represented as follows:

  $\begin{array}{l}HCN\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\rightleftharpoons C{N}^{-}\left(aq\right)+H_{2}O\left(aq\right)\\ I\text{ 10 5 0 0}\\ C\text{ -5 -5 +5}\\ E\text{ 5 0 5}\end{array}$

The remaining solution will become a buffer solution. The $p{K}_a$ of HCN is 9.2. The pH of solution can be calculated as follows:

  $pH=p{K}_a+\log \frac{\left[C{N}^{-}\right]}{\left[HCN\right]}$

Since, volume is same for both thus,

  $\begin{array}{l}pH=9.2+\log \frac{5.00}{5.00}\\ =9.2\end{array}$

Thus, the pH of the solution when 50.0 mL of KOH is added to 100 0 mL of HCN is 9.2.

(c)

Interpretation Introduction

Interpretation: The pH of the solution needs to be calculated when 75 mL of KOH is added to given solution.

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

The number of moles of HCN initially present is 10 mmol. Now, number of moles of 75 mL hydroxide ion can be calculated as follows:

  $n=M\times V=\left(0.1\text{ M}\right)\left(75\text{ mL}\right)=7.5\text{ mmol}$

Now, 7.50 mmol of KOH reacts with 7.50 mmol of HCN. The number of moles of HCl remaining will be 10-7.5=2.5 mmol and the number of moles of CN- formed will be 7.50 mmol.

The pH of solution can be calculated as follows:

  $\begin{array}{l}pH=p{K}_a+\log \frac{\left[C{N}^{-}\right]}{\left[HCN\right]}\\ =9.2+\log \left(\frac{7.50}{2.50}\right)\\ =9.2+0.477\\ =9.7\end{array}$

Thus, pH of the solution after addition of 75.0 mL KOH is 9.7.

(d)

Interpretation Introduction

Interpretation: The pH of the solution needs to be calculated at 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

At the equivalence point, the number of moles of HCN added and KOH is same. Thus, the number of moles of KOH added will be 10 mmol. The volume of KOH added can be calculated from number of moles and molarity as follows:

  $V=\frac{n}{M}=\frac{10\text{ mmol}}{0.1}=100\text{ mL}$

If equal moles react thus, the number of moles of ${\text{CN}}^{-}$ ions formed will be 10.0 mmol. It can undergo hydrolysis.

The total volume will be sum of volume of HCl and KOH as follows:

  $V=100+100=200\text{ mL}$

For 10.0 mmol, the molarity can be calculated as follows:

  $M=\frac{n}{V}=\frac{10\text{ mmol}}{200\text{ mL}}=0.05\text{ M}$

The ICE table can be represented as follows:

  $\begin{array}{l}C{N}^{-}+H_{2}O\rightleftharpoons HCN+O{H}^{-}\\ I\text{ 0}\text{.05 - - -}\\ C\text{ -x x x}\\ E\text{ 0}\text{.05-x x x}\end{array}$

The hydrolysis of ${\text{CN}}^{-}$ can be represented as follows:

  $C{N}^{-}\left(aq\right)+H_{2}O\left(aq\right)\rightleftharpoons HCN\left(aq\right)+O{H}^{-}\left(aq\right)$

The hydrolysis constant value for ${\text{CN}}^{-}$ will be:

  $K_h=\frac{K_w}{K_a}=\frac{{10}^{-14}}{6.2\times {10}^{-10}}=1.6\times {10}^{-5}$

For the above reaction, the expression can be represented as follows:

  $\begin{array}{l}1.6\times {10}^{-5}=\frac{\left[HCN\right]\left[O{H}^{-}\right]}{\left[C{N}^{-}\right]}=\frac{x^2}{0.0500}\\ x=8.9\times {10}^{-4}\end{array}$

The above value of x is equal to the concentration of hydroxide ion in the solution. Thus, the pH of the solution can be calculated as follows:

  $pH=14-pOH=14-\left(-\log \left[{\text{OH}}^{-}\right]\right)=14-\left(-\log \left(8.9\times {10}^{-4}\right)\right)=10.95$

Thus, the pH of solution at equivalence point is 10.95.

(e)

Interpretation Introduction

Interpretation: The pH of the solution needs to be calculated when 125 mL of KOH is added to given solution.

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

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

  $n=M\times V=0.1\text{ M}\times \text{125 mL=12}\text{.5 mmol}$

Here, 10.0 mmol of KOH reacts with 10.0 mmol of HCN thus, the remaining number of moles of KOH will be 2.5 mmol.

Also, 10.0 mmol of ${\text{CN}}^{-}$ ion is formed. It is a weak acid thus, its hydrolysis can be neglected in the presence of KOH. The total volume of the solution will be sum of volume of HCN and KOH that is 100 mL+125 mL=225 mL.

The molarity of KOH can be calculated as follows:

  $M=\frac{n}{V}=\frac{2.50\text{ mmol}}{225\text{ mL}}=0.0111\text{ M}$

The pOH of the solution can be calculated by taking negative log of hydroxide ion conecntartion.

  $pOH=-\log \left[O{H}^{-}\right]=-\log \left(0.0111\right)=1.96$

The pH of the solution can be calculated as follows:

  $pH=14-pOH=14-1.96=12.04$

Thus, the pH of the solution is 12.04.