Chapter 16, Problem 16.106QP

A volume of 25.0 mL of 0.100 M HCl is titrated against a 0.100 M CH₃NH₂ solution added to it from a buret. Calculate the pH values of the solution (a) after 10.0 mL of CH₃NH₂ solution have been added, (b) after 25.0 mL of CH₃NH₂ solution have been added, (c) after 35.0 mL of CH₃NH₂ solution have been added.

Interpretation Introduction

Interpretation:

The $\text{pH}$ of titration $\text{HCl}\text{Vs}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ has to be calculated.

Concept introduction:

• Titration is used to find out the strength of an unknown solution from a concentration of known solution.  A standard solution whose concentration is known, is dropped slowly to an unknown solution and wait for the chemical reaction finish.
• The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$  in a solution.
• $\text{pH}$ is used to determine the acidity of an aqueous solution.

To calculate: $\text{pH}$ of $\text{25}\text{.0 mL of 0}\text{.100MHCl Vs}\text{10}\text{.0 mL of 0}\text{.100 M}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$

Answer to Problem 16.106QP

pH = 1.37

Explanation of Solution

The given data is recorded as such.

$\begin{array}{l}\text{0}\text{.100 M HCl}\\ \text{0}{\text{.100 M CH}}_{\text{3}}{\text{NH}}_{\text{2}}\end{array}$

$\begin{array}{l}{\text{HCl}}_{\left(\text{aq}\right)}\text{+}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}{}_{\left(\text{aq}\right)}\to {\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}{\text{Cl}}_{\text{(aq)}}\\ \text{Initial}\text{concentration }\left(\text{M}\right)\text{: }\text{2}\text{.50 }\times {10}^{-3}1.\text{00 }\times {10}^{-3}\text{0}\\ \text{Change}\text{in}\text{concentration }\left(\text{M}\right)\text{: }-1.\text{00 }\times {10}^{-3}-1.\text{00 }\times {10}^{-3}+1.\text{00 }\times {10}^{-3}\\ \text{Final concentration}\left(\text{M}\right)\text{: }1.\text{50 }\times {10}^{-3}01.\text{00 }\times {10}^{-3}\end{array}$

$\begin{array}{l}{\text{[H}}^{\text{+}}]=\frac{1.50\times {10}^{-3}mol}{0.0350L}\\ \\ \text{pH =}\text{-log[H+] }\\ \text{= -log[0}\text{.0429] }\\ \text{= 1}\text{.37}\end{array}$

The concentration of hydrogen ion is calculated from equilibrium table.  Using the concentration of hydrogen ion, the $\text{pH}$ is determined by taking negative logarithm of concentration of hydrogen ion.

Interpretation Introduction

Interpretation:

The $\text{pH}$ of titration $\text{HCl}\text{Vs}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ has to be calculated.

Concept introduction:

• Titration is used to find out the strength of an unknown solution from a concentration of known solution.  A standard solution whose concentration is known, is dropped slowly to an unknown solution and wait for the chemical reaction finish.
• The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$  in a solution.
• $\text{pH}$ is used to determine the acidity of an aqueous solution.

To calculate: $\text{pH}$ of $\text{25}\text{.0 mL of 0}\text{.100MHCl Vs}25.\text{0 mL of 0}\text{.100 M}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$

Answer to Problem 16.106QP

pH = 5.97

Explanation of Solution

$\begin{array}{l}{\text{[CH}}_{\text{3}}{\text{NH}}_{\text{3}}{}^{\text{+}}\text{]}\text{=}\frac{\text{2}\text{.50}\text{×}{\text{10}}^{\text{-3}}\text{mol}}{\text{0}\text{.0500L}}\text{=}\text{5}\text{.00}\text{×}{\text{10}}^{\text{-2}}\text{M}\\ \\ \text{The}\text{hydrolysis}\text{of}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}{}^{\text{+}}\text{is}\\ \\ {\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}{{}^{\text{+}}}_{\left(\text{aq}\right)}\to \text{H}{}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}{\text{Cl}}_{\text{(aq)}}\\ \text{I}\text{nitial}\text{concentration }\left(\text{M}\right)\text{: }\text{5}\text{.000 ×}{\text{10}}^{\text{-2}}\text{0}\text{0}\\ \text{Change}\text{in}\text{concentration }\left(\text{M}\right)\text{: }\text{-x }\text{+x }\text{+x}\\ \text{Equilibrium}\text{concentration}\left(\text{M}\right)\text{: }\text{(5}\text{.000 ×}{\text{10}}^{\text{-2}}\text{)}\text{-x}\text{x}\text{x}\\ \\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[CH}}_{\text{3}}{\text{NH}}_{\text{2}}{\text{][H}}^{\text{+}}\text{]}}{{\text{[CH}}_{\text{3}}{\text{NH}}_{\text{3}}{}^{\text{+}}\text{]}}\\ \text{2}\text{.3}\text{×}{\text{10}}^{\text{-11}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{(5}\text{.00}\text{×}{\text{10}}^{\text{-2}}\text{)}\text{-}\text{x}}\text{»}\frac{{\text{x}}^{\text{2}}}{\text{5}\text{.00}\text{×}{\text{10}}^{\text{-2}}}\\ \begin{array}{l}\begin{array}{l}\\ {\text{ x}}^{\text{2}}\text{=}\text{1}\text{.15}\text{×}{\text{10}}^{\text{-}}{}^{\text{12}}\end{array}\hfill \\ \hfill \\ \text{ x = 1}{\text{.07×10}}^{\text{-}}{}^{\text{6}}{\text{M = [H}}^{\text{+}}\text{]}\hfill \\ \hfill \\ \text{pH}\text{=}\text{-log[H+] = -log[1}{\text{.07×10}}^{\text{-}}{}^{\text{6}}\text{]}\text{=}\text{5}\text{.97 }\hfill \end{array}\end{array}$

The concentration of hydrogen ion is calculated from acid dissociation constant (${\text{K}}_{\text{a}}$).  Using the concentration of hydrogen ion, the $\text{pH}$ is determined by taking negative logarithm of concentration of hydrogen ion.

Interpretation Introduction

Interpretation:

The $\text{pH}$ of titration $\text{HCl}\text{Vs}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ has to be calculated.

Concept introduction:

• Titration is used to find out the strength of an unknown solution from a concentration of known solution.  A standard solution whose concentration is known, is dropped slowly to an unknown solution and wait for the chemical reaction finish.
• The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$  in a solution.
• $\text{pH}$ is used to determine the acidity of an aqueous solution.

To calculate: $\text{pH}$ of $\text{25}\text{.0 mL of 0}\text{.100MHCl Vs}35.\text{0 mL of 0}\text{.100 M}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$

Answer to Problem 16.106QP

pH = 10.24

Explanation of Solution

$\begin{array}{l}{\text{HCl}}_{\left(\text{aq}\right)}\text{+}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}{}_{\left(\text{aq}\right)}\to {\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}{\text{Cl}}_{\text{(aq)}}\\ \text{Initial}\text{concentration }\left(\text{M}\right)\text{: }\text{2}\text{.50 ×}{\text{10}}^{\text{-3}}\text{ 3}\text{.50 ×}{\text{10}}^{\text{-3}}\text{0}\\ \text{Change}\text{in}\text{concentration }\left(\text{M}\right)\text{: }\text{-2}\text{.50 ×}{\text{10}}^{\text{-3}}\text{-2}\text{.50 ×}{\text{10}}^{\text{-3}}\text{+2}\text{.50 ×}{\text{10}}^{\text{-3}}\\ \text{Final concentration}\left(\text{M}\right)\text{: }\text{0}\text{1}\text{.00 ×}{\text{10}}^{\text{-3}}\text{2}\text{.50 ×}{\text{10}}^{\text{-3}}\\ \\ \text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}\\ \text{pH}\text{=}\text{-log(2}\text{.3}\text{×}{\text{10}}^{\text{-11}}\text{)}\text{+}\text{log}\frac{\text{(1}\text{.00}\text{×}{\text{10}}^{\text{-3}}\text{)}}{\text{(2}\text{.50}\text{×}{\text{10}}^{\text{-3}}\text{)}}\text{=}\text{10}\text{.24}\end{array}$

The $\text{pH}$ is calculated using Henderson-hasselbalch equation.  By substituting the value of concentration of base and acid and taking logarithm the $\text{pH}$ is determined.