Chapter 2, Problem 2.23P

(a)

Interpretation Introduction

Interpretation:

The mass of Benzoic acid to prepare $\text{250}\text{ml}\text{of 100}\text{.0 mM}$ aqueous solution should be calculated.

Mass of the solute in molar solution:

The preparation of known concentration solution done by the calculated mass of solute is dissolved in required value solution.

The mass of the solute is calculated by using below equation,

$\text{Mass of }\text{solute (g)}\text{=}\text{Molecular weight }\left[\text{g/mol}\right]\text{×Concentrations }\left[\text{mol/L}\right]\text{×Volume }\left[\text{L}\right]$

Answer to Problem 2.23P

The mass of Benzoic acid to prepare $\text{250}\text{mL}\text{of 100}\text{.0 mM}$ aqueous solution is $\text{3}\text{.053}\text{g}$.

Explanation of Solution

To calculate the mass of Benzoic acid to prepare $\text{250}\text{mL}\text{of 100}\text{.0 mM}$ aqueous solution.

Given,

Molar mas of Benzoic acid is $\text{122}\text{. 12g}$

Density of Benzoic acid is $\text{1}\text{.27 g/mL}$

The concentration to be prepare $\text{100}\text{.}0\text{mM}$

$\begin{array}{l}\text{Mass}\text{of}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}\text{=122}\text{.12}\text{g}\text{×0}\text{.1}\text{M×0}\text{.250}\text{L}\\ \text{=}\text{3}\text{.053}\text{g}\end{array}$

The mass of Benzoic acid to $\text{250}\text{ml}\text{of 100}\text{.0 mM}$ aqueous solution is $\text{3}\text{.053}\text{g}$

Hence, $\text{3}\text{.053}\text{g}$ of Benzoic acid is dissolved in $\text{250}\text{mL}$ water to give $\text{250}\text{ml}\text{of 100}\text{.0 mM}$ aqueous solution.

Conclusion

The mass of Benzoic acid to prepare $\text{250}\text{ml}\text{of 100}\text{.0 mM}$ aqueous solution was calculated.

(b)

Interpretation Introduction

Interpretation:

The true mass of Benzoic acid should be calculated.

Concept introduction:

Buoyancy Correction:

The weight of the object in air is less then weight of the object in vacuum because of buoyancy, this weight variation is corrected using Buoyancy equation it is,

$\text{m}\text{=}\frac{\text{m'}\left(\text{1-}\frac{{\text{d}}_{\text{a}}}{{\text{d}}_{\text{w}}}\right)}{\left(\text{1-}\frac{{\text{d}}_{\text{a}}}{\text{d}}\right)}$

Where,

$\text{m}$ is mass of object in vacuum.

$\text{m'}$ is mass of object in air (read on a balance).

${\text{d}}_{\text{a}}$ is the density of air $\left(\text{0}\text{.001 2 g/mL near I bar and 25°C}\right)$.

${\text{d}}_{\text{w}}$ is the density of calibration weights $\text{(8}\text{.0 g/mL)}$.

$\text{d}$ is the density of the object being weighed.

Answer to Problem 2.23P

The true mass of Benzoic acid is $\text{3}\text{.0554}\text{g}$

Explanation of Solution

To calculate the true mass of Benzoic acid.

Given,

The density of Benzoic acid near $\text{25°C}$ is $\text{1}\text{.27 g/mL}$

Benzoic acid mass in air is $\text{3}\text{.053}\text{g}$

Air density is $\text{0}\text{.0012 g/mL}$

$\begin{array}{l}\text{m}\text{=}\frac{\text{(3}\text{.053 g)}\left(\text{1-}\frac{\text{0}\text{.0012}\text{g/mL}}{\text{8}\text{.0}\text{g}\text{/mL}}\right)}{\left(\text{1-}\frac{\text{0}\text{.0012}\text{g/mL}}{\text{1}\text{.27 g/mL}}\right)}\\ \text{=}\text{3}\text{.0554}\text{g}\end{array}$

The give values are plugged in above equation to give a true mass of Benzoic acid.

The true mass of Benzoic acid is $\text{3}\text{.0554}\text{g}$

Conclusion

The true mass of Benzoic acid was calculated.

(c)

Interpretation Introduction

Interpretation:

The preparation of $\text{50}\text{.0 μM}$ Benzoic acid solution by serial dilution method should be explained.

Concept introduction:

Serial dilation:

Serial dilation is the process, which is used to prepare a very low concentration solutions from high concentration solutions by continues step dilution method.

The Serial dilation is carried out by using volumetric principle.

${\text{M}}_{\text{ini}}{\text{V}}_{\text{ini}}\text{=}{\text{M}}_{\text{fnl}}{\text{V}}_{\text{fnl}}$

Where,

${\text{M}}_{\text{ini}}$ is initial concentration.

${\text{V}}_{\text{ini}}$ is initial volume.

${\text{M}}_{\text{fnl}}$ is final concentration.

${\text{V}}_{\text{fnl}}$ is final volume.

Answer to Problem 2.23P

$\text{30}\text{.53}\text{g}$ of benzoic acid is dissolved in $\text{1000}\text{ml}$ volumetric flask with water then pipet 5 mL of this stock solution into 500 mL volumetric flask to give  $\text{2}\text{.5}\text{mM}$ of benzoic acid solution, After pipet 5 mL of $\text{2}\text{.5}\text{mM}$ benzoic acid solution and transfer into 250 mL volumetric flask with water to give a $\text{50}\text{.0 μM}$ solution.

Explanation of Solution

To explain the preparation of $\text{50}\text{.0 μM}$ Benzoic acid solution by serial dilution method.

Stock solution preparation:

Molar mas of Benzoic acid is $\text{122}\text{. 12g}$

The concentration to be prepare $\text{100}\text{.}0\text{mM}$

$\begin{array}{l}\text{Mass}\text{of}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}\text{=122}\text{.12}\text{g}\text{×0}\text{.25}\text{M×1}\text{L}\\ \text{=}30.53\text{g}\end{array}$

The mass of Benzoic acid to $\text{1 liter}\text{of 500}\text{.0 mM}$ aqueous solution is $\text{30}\text{.53}\text{g}$

Hence, $30.53\text{g}$ of Benzoic acid is dissolved in $\text{1000 mL}$ flask with water to give $\text{1000}\text{ml}\text{of 0}\text{.25M}$ aqueous solution.

First dilution:

$\begin{array}{l}\text{=}\frac{\text{0}\text{.25}\text{M×5}}{\text{500}\text{mL}}\\ \text{=}\text{0}\text{.0025}\text{M}\\ \text{=}\text{2}\text{.5}\text{mM}\end{array}$

5 mL of stock solution is diluted in $\text{500 mL}$ volumetric flask with water give a $\text{2}\text{.5}\text{mM}$ solution.

Second dilution:

$\begin{array}{l}\text{=}\frac{\text{0}\text{.0025}\text{M×5}}{\text{250}\text{mL}}\\ \text{=}\text{0}\text{.00005}\text{M}\\ \text{=}\text{5}\text{μM}\end{array}$

5 mL of first dilution solution is diluted in $\text{250 mL}$ volumetric flask with water give a $\text{5μM}$ solution.

Conclusion

The preparation of $\text{50}\text{.0 μM}$ Benzoic acid solution by serial dilution method was explained.