Chapter 26, Problem 26.BE

Interpretation Introduction

Interpretation:

The mass percentages of given compounds should be determined.

Ion-Exchange Chromatography:

Ion-Exchange Chromatography is separation technique, which is work in the principle of exchanging of ions based on attraction to the ion exchanger.

It contains two phases, one is stationary phase and another one is mobile phase.

In generally resins are act as a stationary phase, the positively charged ion exchangers attract solute anions and negatively charged ion exchangers are attract solute cations.

The higher polar eluent is passed through a column the exchangers are releases the solute and they will come out from the column.

In this process, the stationary phase (ion exchangers) is exchange the solute ions into eluent ions therefore it is called as Ion-exchange chromatography.

Volumetric principle:

The volume and concentration of unknown solution is determined by it is titrate with known volume and concentration solution.

The volume and concentration of unknown solution is required equivalent volume and concentration of known solution in the volumetric titration.

${\text{V}}_{\text{1}}{\text{M}}_{\text{1}}\text{=}{\text{V}}_{\text{2}}{\text{M}}_{\text{2}}$

Where,

${\text{V}}_{\text{1}}$ is volume of known solution

${\text{N}}_{\text{1}}$ is concentration of known solution

${\text{V}}_{\text{1}}$ is volume of unknown solution

${\text{V}}_{\text{1}}$ is concentration of unknown solution

Mole:

The product of molarity of solution and volume of solution to give a mole of solute that present in the solution.

$\text{Mole}\text{=}\text{Molarity}\text{×}\text{volume}$

Mass percentage:

$\text{wt\%=Weight}\text{percent=}\frac{\text{mass}\text{of}\text{substance}}{\text{mass}\text{of}\text{total}\text{solution}\text{or}\text{total}\text{sample}}\text{x(100)}$

Answer to Problem 26.BE

The mass percentages of ${\text{VOSO}}_{\text{4}}$, ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ and ${\text{H}}_{\text{2}}\text{O}$ are $\text{80}\text{.9}\text{\%}$, $\text{10}\text{.7}\%$ and $\text{8}\text{.4}\%$ respectively.

Explanation of Solution

To determine the mass percentages of given compounds.

In ion-exchange chromatography, the mole of ions replaced is equal to the mole of ions present in the solute (sample). In cation-exchange column ${\text{H}}^{\text{+}}$ is a Separator ion and the equivalent mole of ${\text{H}}^{\text{+}}$ ion is calculated by titration to find the mole of cation present in the sample.

In the cation exchange chromatography ${\text{OH}}^{\text{-}}$ is a common suppressor ion.

From the above statements, the amount of ${\text{OH}}^{\text{-}}$ is equal to total cation charge that is sum of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ and ${\text{VO}}^{\text{2+}}$.

The total mole of the Mole of $\text{NaOH}$ is,

$\begin{array}{l}\text{Mole}\text{of}\text{NaOH}\text{= 0}\text{.02274}\text{M}\text{×0}\text{.01303}\text{L}\\ \text{=}\text{0}\text{.2963}\text{mmol}\end{array}$

Mole of ${\text{VO}}^{\text{2+}}$ is,

$\begin{array}{l}\text{Mole}\text{of}\text{NaOH}\text{= 0}\text{.024}\text{M}\text{×0}\text{.50}\text{L}\\ \text{=}\text{0}\text{.1}\text{.215}\text{mmol}\end{array}$

Mole of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is,

$\begin{array}{l}\text{Mole}\text{of}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{= }\left(\frac{\left(\text{2}\text{.963-2}\text{.43}\right)}{\text{2}}\right)\\ \text{=}\text{0}{\text{.266}}_{\text{5}}\text{mmol}\end{array}$

From the above calculation, the mass of ${\text{VOSO}}_{\text{4}}$ and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ plugged in above equation.

$\begin{array}{l}\text{1}{\text{.215 mmol VOSO}}_{\text{4}}\text{ = 0}\text{.198 g}\\ \text{0}{\text{.2665 mmol H}}_{\text{2}}{\text{SO}}_{\text{4 }}\text{= 0}\text{.0261 g}\end{array}$

The total mass of the sample is $\text{0}\text{.2447 g}$

Hence the mass percentages of ${\text{VOSO}}_{\text{4}}$, ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ and ${\text{H}}_{\text{2}}\text{O}$ are,

$\begin{array}{l}{\text{VOSO}}_{\text{4}}\text{\%}\text{=}\frac{\text{0}\text{.198}}{\text{0}\text{.2447}}\text{×100}\\ \text{=}\text{80}\text{.9}\text{\%}\\ {\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{\%}\text{=}\frac{\text{0}\text{.026}}{\text{0}\text{.2447}}\text{×100}\\ \text{=}\text{10}\text{.7}\text{\%}\\ {\text{H}}_{\text{2}}\text{O}\text{\%}\text{=}\text{100-}\text{(10}\text{.7+80}\text{.9)}\\ \text{=}\text{8}\text{.4\%}\end{array}$

The calculated moles of the compounds are converted into grams and this gram mass of each compounds are plugged in the above equations to give the mass percentage of each compounds that present in given compound.

Conclusion

The mass percentages of given compounds that was be determined.