Chapter 16, Problem 16.91QP

A 50.0-mL sample of a 0.100 M solution of NaCN is titrated by 0.200 M HCl. K_b for CN⁻ is 2.0 × 10⁻⁵. Calculate the pH of the solution: a prior to the start of the titration; b after the addition of 15.0 mL of 0.200 M HCl; c at the equivalence point; d after the addition of 30.0 mL of 0.200 M HCl.

Interpretation Introduction

Interpretation:

The pH of the given points of the titration of $\text{NaCN}$  with hydrochloric acid has to be calculated.

1. (a) Prior to the start of the titration
2. (b) After the addition of 15 mL of 0.200 M $\text{HCl}$
3. (c) At the equivalence point
4. (d) After the addition of 30.0 mL of 0.200 M $\text{HCl}$

Concept Introduction:

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion ${\text{[OH}}^{-}\text{]}$ concentration.

$\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{]}$

Relationship between pH  and pOH:

$\text{pH + pOH = 14}$

Answer to Problem 16.91QP

The pH prior to the start of the titration is 11.15

Explanation of Solution

To Calculate: The pH prior to the start of the titration

Given data:

The volume of $\text{NaCN}$ = 50.0 mL

The concentration of $\text{NaCN}$ = 0.100 M

The concentration of hydrochloric acid = 0.200 M

The $K_b$ value for ${\text{CN}}^{-}$ is $2.0\times {10}^{-5}$

pH prior to the start of the titration

Construct an equilibrium table for the hydrolysis of cyanide ion from $\text{NaCN}$ as follows,

	${\text{ CN}}^{\text{-}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ HCN + OH}}^{-}$
Initial $(M)$	0.100 $-x$ 0.100-x	0.00	0.00
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		x	x

The $K_b$ value for ${\text{CN}}^{-}$ is $2.0\times {10}^{-5}$

Now substitute equilibrium concentrations into the equilibrium-constant expression.

$\begin{array}{l}{\text{ K}}_b=\frac{{\text{(}x\text{)}}^2}{(0.100-x)}\\ \text{Assume 0}\text{.100 is very small than }x\text{ and neglect it in the denominator}\\ \text{ 2}\text{.0}\times {10}^{-5}\approx \frac{{\text{(}x\text{)}}^2}{(0.100)}\\ x=1.414\times {10}^{-3}\text{ M}\end{array}$

Here, x gives the concentration of hydroxide ion $1.414\times {10}^{-3}\text{ M}$

Finally calculate pOH and then the pH as follows,

$\begin{array}{l}\text{pOH}{\text{=-log[OH}}^{\text{-}}\text{]}\\ \text{=-log(}1.414\times {10}^{-3})\\ =2.849\end{array}$

The pH is calculated as follows,

$\begin{array}{l}\text{pH + pOH }\text{= 14}\\ \text{pH}\text{=14 - pOH}\\ \text{=14 - 2}\text{.849}\\ \text{=11}\text{.15}\end{array}$

Conclusion

The pH prior to the start of the titration is 11.15

Interpretation Introduction

Interpretation:

The pH of the given points of the titration of $\text{NaCN}$  with hydrochloric acid has to be calculated.

1. (a) Prior to the start of the titration
2. (b) After the addition of 15 mL of 0.200 M $\text{HCl}$
3. (c) At the equivalence point
4. (d) After the addition of 30.0 mL of 0.200 M $\text{HCl}$

Concept Introduction:

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion ${\text{[OH}}^{-}\text{]}$ concentration.

$\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{]}$

Relationship between pH  and pOH:

$\text{pH + pOH = 14}$

Answer to Problem 16.91QP

The pH after the addition of 15 mL of 0.200 M $\text{HCl}$ is 9.12

Explanation of Solution

To Calculate: The pH after the addition of 15 mL of 0.200 M $\text{HCl}$

After the addition of 15.0 mL of 0.200 M $\text{HCl}$, the reaction will be as follows

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

At this point, the total volume is, $50.0\text{ mL + 15}\text{.0 mL = 65}\text{.0 mL}$

Now, the moles of ${\text{CN}}^{\text{-}}$ present at the initial and the moles of $\text{HCl}$ added are:

$\begin{array}{l}{\text{mol CN}}^{\text{-}}=\text{ M}\times \text{V}\\ \text{=0}\text{.100 M}\times \text{50}\times {\text{10}}^{-3}\text{ L}\\ \text{=5}\text{.00}\times {\text{10}}^{-3}\text{ mol}\\ \text{mol HCl }=\text{ M}\times \text{V}\\ \text{=0}\text{.200 M}\times 15\times {\text{10}}^{-3}\text{ L}\\ \text{=3}\text{.00}\times {\text{10}}^{-3}\text{ mol}\end{array}$

After the reaction, the moles of $\text{HCN}$ will be equal to the moles of $\text{HCl}$ added.

The moles of ${\text{CN}}^{\text{-}}$ present are:

$\begin{array}{l}{\text{mol CN}}^{\text{-}}\text{= 5}\text{.000}\times {\text{10}}^{-3}\text{ mol - 3}\text{.000}\times {\text{10}}^{-3}\\ =2.000\times {10}^{-3}\text{ mol}\\ \end{array}$

The concentrations are:

$\begin{array}{l}{\text{[CN}}^{\text{-}}\text{]}\text{=}\frac{2.000\times {10}^{-3}{\text{ mol CN}}^{\text{-}}}{65.0\times {10}^{-3}\text{ L}}\\ =0.03076\text{ M}\\ \text{[HCN]}\text{=}\frac{3.000\times {10}^{-3}\text{ mol HCN}}{65.0\times {10}^{-3}\text{ L}}\\ =0.04615\text{ M}\end{array}$

Solve for ${\text{[OH}}^{\text{-}}\text{]}$ by Substituting the concentrations into the ${\text{K}}_{\text{b}}$ expression

$\begin{array}{l}{\text{ K}}_{\text{b}}=\frac{[{\text{HCN][OH}}^{\text{-}}\text{]}}{[{\text{CN}}^{\text{-}}\text{]}}\\ 2.0\times {10}^{-5}=\frac{(0.04615+x)(x)}{(0.03076-x)}\\ \approx \frac{(0.04615)(x)}{(0.03076)}\\ =\frac{(0.04615)(x)}{(0.03076)}\\ x=\frac{(0.03076)(2.0\times {10}^{-5})}{(0.04615)}\\ =1.33\times {10}^{-5}\text{ M}\end{array}$

The pH is calculated from pOH as follows,

$\begin{array}{l}\text{pOH}\text{=-log(1}\text{.33}\times {\text{10}}^{\text{-5}})\\ =4.875\\ \text{pH}\text{=14 - pOH}\\ \text{=14-4}\text{.875}\\ \text{=9}\text{.12}\end{array}$

Conclusion

The pH after the addition of 15 mL of 0.200 M $\text{HCl}$ is 9.12

Interpretation Introduction

Interpretation:

The pH of the given points of the titration of $\text{NaCN}$  with hydrochloric acid has to be calculated.

1. (a) Prior to the start of the titration
2. (b) After the addition of 15 mL of 0.200 M $\text{HCl}$
3. (c) At the equivalence point
4. (d) After the addition of 30.0 mL of 0.200 M $\text{HCl}$

Concept Introduction:

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion ${\text{[OH}}^{-}\text{]}$ concentration.

$\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{]}$

Relationship between pH  and pOH:

$\text{pH + pOH = 14}$

Answer to Problem 16.91QP

The pH at the equivalence point is 5.24

Explanation of Solution

To Calculate: The pH at the equivalence point

Calculate the volume of $\text{HCl}$ added to reach the equivalence point

The volume of $\text{HCl}$ added is calculated as follows,

$\begin{array}{l}\text{Volume of HCl}\text{= }\frac{{\text{M}}_{base}{\text{ V}}_{base}}{{\text{M}}_{acid}}\\ =\frac{(0.100\text{ M)(50}\text{.0 mL)}}{0.200\text{ M}}\\ =\text{ 25}\text{.0 mL}\end{array}$

Hence, the total volume is as follows,

$\begin{array}{l}\text{Total volume}\text{= 50}\text{.0 mL + 25}\text{.0 mL}\\ =\text{75}\text{.0 mL}\\ \text{= 0}\text{.075 L}\end{array}$

The equilibrium reaction is,

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

$\begin{array}{l}\text{[HCN]}\text{=}\frac{5.000\times {10}^{-3}\text{ mol HCN}}{75.0\times {10}^{-3}\text{ L}}\\ =0.0666\text{ M}\end{array}$

The value of ${\text{K}}_{\text{a}}$ is calculated from ${\text{K}}_{\text{b}}$ as follows,

$\begin{array}{l}{\text{K}}_{\text{a}}=\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{b}}}\\ =\frac{\text{1}\text{.00}\times {\text{10}}^{\text{-14}}}{\text{2}\text{.0}\times {\text{10}}^{\text{-5}}}\\ =5.00\times {10}^{-10}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{a}}=\frac{[{\text{CN}}^{\text{-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{[\text{HCN]}}\\ 5.00\times {10}^{-10}=\frac{x^2}{0.0666-x}\\ x=5.773\times {10}^{-6}\text{ M}\end{array}$

Here, x gives the hydronium ion concentration.

$\begin{array}{l}\text{pH}{\text{=-log[H}}^{\text{+}}\text{]}\\ \text{=-log(5}\text{.773}\times {\text{10}}^{\text{-6}})\\ =5.24\end{array}$

Conclusion

The pH at the equivalence point is 5.24

Interpretation Introduction

Interpretation:

The pH of the given points of the titration of $\text{NaCN}$  with hydrochloric acid has to be calculated.

1. (a) Prior to the start of the titration
2. (b) After the addition of 15 mL of 0.200 M $\text{HCl}$
3. (c) At the equivalence point
4. (d) After the addition of 30.0 mL of 0.200 M $\text{HCl}$

Concept Introduction:

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion ${\text{[OH}}^{-}\text{]}$ concentration.

$\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{]}$

Relationship between pH  and pOH:

$\text{pH + pOH = 14}$

Answer to Problem 16.91QP

The pH after the addition of 30.0 mL of 0.200 M $\text{HCl}$ is 1.903

Explanation of Solution

To Calculate: The pH after the addition of 30 mL of 0.200 M $\text{HCl}$

Calculate the moles of acid added.

$(0.200\text{ M})\times 0.030\text{ L = 6}\text{.000}\times {\text{10}}^{\text{-3}}\text{ mol}$

$\begin{array}{l}\text{Moles of acid remaining }\text{= 6}\text{.000}\times {\text{10}}^{\text{-3}}\text{ mol - 5}\text{.000}\times {\text{10}}^{\text{-3}}\text{ mol}\\ \text{= 1}\text{.000}\times {\text{10}}^{\text{-3}}\text{ mol}\end{array}$

The total volume after the addition of 30.0 mL of $\text{HCl}$ is: $\text{50}\text{.0 mL + 30}\text{.0 mL = 80}\text{.0 mL}$

The hydronium ion concentration and the pH are:

$\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\text{=}\frac{\text{1}\text{.000}\times {\text{10}}^{\text{-3}}\text{ mol}}{80.0\times {\text{10}}^{\text{-3}}\text{ L}}\\ =0.01250\text{ M}\end{array}$

The pH is calculated as follows,

$\begin{array}{l}\text{pH}\text{=}-\log {\text{[H}}^{+}]\\ =-\log (0.01250)\\ =1.903\end{array}$

Conclusion

The pH after the addition of 30.0 mL of 0.200 M $\text{HCl}$ is 1.903