Chapter 17, Problem 17.31QP

(a)

Interpretation Introduction

Interpretation:

The molar mass of acid and ${\text{K}}_{\text{a}}$ of the unknown acid has to be calculated.

Concept introduction:

• Molar mass is the ratio of mass of given substance to that number of moles of compound.
• $\text{Molar}\text{mass}\text{=}\frac{\text{Amount}\text{of}\text{given}\text{substance}}{\text{Number}\text{of}\text{moles}}$
• Acid dissociation constant ${\text{K}}_{\text{a}}$ is defined as measure of acid strength in a given solution
• ${\text{K}}_{\text{a}}\text{=}\frac{{\text{[A}}^{\text{-}}{\text{][H}}^{\text{+}}\text{]}}{\text{[HA]}}$

To calculate: the molar mass of monoprotic acid

Answer to Problem 17.31QP

The molar mass of monoprotic acid is $1.10\times 10^{2}g/mol$.

Explanation of Solution

Amount of monoprotic acid is $\text{0}\text{.1276}\text{g}$

Concentration of $\text{NaOH}$ is $0.0633\text{M}$

Calculate the number of moles of monoprotic acid

${\text{HA}}_{\text{(aq)}}{\text{ + NaOH}}_{\text{(aq)}}\to {\text{ NaA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{monoprotic}\text{acid}\text{=}\text{18}\text{.4}\text{mL}\text{×}\frac{\text{0}\text{.0633}\text{mol}}{\text{1000}\text{mL}}\\ \text{=}\text{0}\text{.00116}\text{mol}\end{array}$

In this reaction one mole of monoprotic acid needs to neutralise one mole of $\text{NaOH}$.  The number of moles of monoprotic acid can be calculated using volume and given concentration of the formic acid.

Calculate the molar mass of monoprotic acid

$\begin{array}{l}\text{Molar}\text{mass}\text{=}\frac{\text{Amount}\text{of}\text{given}\text{substance}}{\text{Number}\text{of}\text{moles}}\\ \text{Molar}\text{mass}\text{=}\frac{\text{0}\text{.1276}\text{g}\text{HA}}{\text{0}\text{.00116}\text{mol HA}}\\ \text{=}\text{1}\text{.10}\text{×}{\text{10}}^{\text{2}}\text{g/mol}\end{array}$

The molar mass of monoprotic acid can be calculated on dividing given weight of monoprotic acid by number of moles of monoprotic acid.

(b)

Interpretation Introduction

Interpretation:

The molar mass of acid and ${\text{K}}_{\text{a}}$ of the unknown acid has to be calculated.

Concept introduction:

• Molar mass is the ratio of mass of given substance to that number of moles of compound.
• $\text{Molar}\text{mass}\text{=}\frac{\text{Amount}\text{of}\text{given}\text{substance}}{\text{Number}\text{of}\text{moles}}$
• Acid dissociation constant ${\text{K}}_{\text{a}}$ is defined as measure of acid strength in a given solution
• ${\text{K}}_{\text{a}}\text{=}\frac{{\text{[A}}^{\text{-}}{\text{][H}}^{\text{+}}\text{]}}{\text{[HA]}}$

To calculate: the number of moles of $\text{NaOH}$ and concentration of hydrogen ion

Answer to Problem 17.31QP

The ${\text{K}}_{\text{a}}$ of the monoprotic acid is $1.6\times 10^{-6}$

Explanation of Solution

Amount of monoprotic acid is $\text{0}\text{.1276}\text{g}$

Concentration of $\text{NaOH}$ is $0.0633\text{M}$

$\text{10}\text{mL}\text{×}\frac{\text{0}\text{.0633}\text{mol}}{\text{1000}\text{mL}}\text{=}\text{6}\text{.33}\text{×}{\text{10}}^{\text{-4}}$

$\begin{array}{l}\text{The}\text{reaction}\text{is}\text{as}\text{follows,}\\ {\text{HA}}_{\text{(aq)}}\text{+}{\text{NaOH}}_{\text{(aq)}}\to {\text{NaA}}_{\text{(aq)}}\text{+}{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\\ \\ \text{Initial concentration(M): }\text{0}\text{.0016 }\text{6}\text{.33}\text{×}{\text{10}}^{\text{-4}}\text{0}\\ \text{Change in concentration (M):}\text{6}\text{.33}\text{×}{\text{10}}^{\text{-4}}\text{6}\text{.33}\text{×}{\text{10}}^{\text{-4}}\text{+6}\text{.33}\text{×}{\text{10}}^{\text{-4}}\\ \text{Final}\text{concentration (M): }\text{5}\text{.2}\text{×}{\text{10}}^{\text{-4}}\text{0}\text{6}\text{.33}\text{×}{\text{10}}^{\text{-4}}\\ \text{The}\text{concentration}\text{of}\text{HA}\text{and}{\text{A}}^{\text{-}}\text{are}\\ \text{[HA]}\text{=}\frac{\text{5}{\text{.2 ×10}}^{\text{-4}}\text{ mol}}{\text{0}\text{.0350}\text{L}}\text{=}\text{0}\text{.015}\text{M}\\ {\text{[A}}^{\text{-}}\text{]}\text{=}\frac{\text{6}{\text{.33 ×10}}^{\text{-4}}\text{ mol}}{\text{0}\text{.0350}\text{L}}\text{=}\text{0}\text{.0181}\text{M}\\ \text{pH}\text{=}{\text{-log[H}}^{\text{+}}\text{]}\\ \text{5}\text{.87}\text{=}{\text{-log[H}}^{\text{+}}\text{]}\\ {\text{[H}}^{\text{+}}\text{]}\text{=}{\text{10}}^{\text{-pH}}\text{=}{\text{10}}^{\text{-5}\text{.87}}\text{=}\text{1}\text{.35}\text{×}{\text{10}}^{\text{-6}}\text{M}\end{array}$

The number of moles of sodium of hydroxide can be calculated using the given concentration divided by thousand.  The weak acid reacts with $\text{NaOH}$ and the concentrations vary from the obtained concentrations, strength of hydrogen ion can be determined.

Calculate the ${\text{K}}_{\text{a}}$ of the monoprotic acid

$\begin{array}{l}\text{The}\text{equilibrium}\text{is}\text{as}\text{follows,}\\ {\text{HA}}_{\text{(aq)}}\to {\text{H}}^{\text{+}}{}_{\text{(aq)}}+{\text{A}}^{\text{-}}{}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{0}\text{.0015 }0\text{0}\text{.0181}\\ \text{Change in concentration (M):}1.35\text{×}{\text{10}}^{\text{-6}}+1.35\text{×}{\text{10}}^{\text{-6}}+1.35\text{×}{\text{10}}^{\text{-6}}\\ \text{Final}\text{concentration (M): }0.0151.35\text{×}{\text{10}}^{\text{-6}}\text{0}\text{.0181}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[A}}^{\text{-}}{\text{][H}}^{\text{+}}\text{]}}{\text{[HA]}}\\ =\frac{\text{(}1.35\text{×}{\text{10}}^{\text{-6}}\text{)(0}\text{.0181)}}{0.015}\\ =1.6\times {10}^{-6}\end{array}$

The ${\text{K}}_{\text{a}}$ can be calculated by using concentrations of reactants and products.  By doing simple mathematical calculations the acid dissociation constant can be determined.