Chapter 26, Problem 26.AE

(a)

Interpretation Introduction

Interpretation:

The volume of  $\text{0}\text{.023 1 M NaOH}$ is needed to titrate the eluate form chromatography 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.

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

Answer to Problem 26.AE

The required volume of $\text{NaOH}$ to titrate the $\text{1 0}\text{.00 mL of 0}{\text{.045 8 M KNO}}_{\text{3}}$ is $\text{19}\text{.83}\text{mL}$

Explanation of Solution

To determine the volume of  $\text{0}\text{.023 1 M NaOH}$ is needed to titrate the eluate form chromatography.

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 mole of ${\text{K}}^{\text{+}}$ ions are applied to the column is equal to the mole of ${\text{H}}^{\text{+}}$ realised.

Therefore, the mole of $\text{NaOH}$ required is equal to the mole of ${\text{H}}^{\text{+}}$ realised from the column.

$\begin{array}{l}\text{Mole}\text{of}\text{NaOH}\text{= 0}\text{.0458}\text{M}\times \text{0}\text{.0100}\text{L}\\ \text{=}\text{4}\text{.58}{\text{M10}}^{\text{-4}}\text{mol}\end{array}$

Hence, the volume of $\text{NaOH}$ required is,

$\begin{array}{l}\text{Molarity =}\frac{\text{mole}}{\text{Volume}}\\ \text{Volume}\text{=}\frac{\text{mole}}{\text{Molarity}}\\ \text{=}\text{0}\text{.0458}\text{M×}\text{0}\text{.0100}\text{L}\\ \text{=}\text{19}\text{.83}\text{mL}\end{array}$

The calculated mole and concentration of $\text{NaOH}$ are plugged in above equation to give volume of $\text{NaOH}$ required to titrate the $\text{1 0}\text{.00 mL of 0}{\text{.045 8 M KNO}}_{\text{3}}$

Conclusion

The volume of  $\text{0}\text{.023 1 M NaOH}$ is needed to titrate the eluate form chromatography was determined.

(b)

Interpretation Introduction

Interpretation:

The milli equivalents of cation that present in per gram of sample 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.

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

Answer to Problem 26.AE

The milli equivalents of cation that present in per gram of sample is $\text{4}\text{.27meq}$

Explanation of Solution

To determine the the milli equivalents of cation that present in per gram of sample.

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 mole equivalent of cation equal to the mole of ${\text{OH}}^{\text{-}}$ ion

Therefore, the mole of $\text{NaOH}$ required is equal to the mole of ${\text{H}}^{\text{+}}$ realised from the column.

$\begin{array}{l}\text{Mole}\text{of}\text{NaOH}\text{= 0}\text{.1396}\text{M×}\text{0}\text{.03064}\text{L}\\ \text{=}\text{4}\text{.27mmol}\end{array}$

The volume and concentration of $\text{NaOH}$ are plugged in above equation to give the mole of $\text{NaOH}$ required to titrate.

Hence, the milli equivalents of cation that present in per gram of sample is $\text{4}\text{.277}\text{meq}$

Conclusion

The milli equivalents of cation that present in per gram of sample was determined.

(c)

Interpretation Introduction

Interpretation:

The equivalent mass of $\text{0}\text{.269 2 g}$ sample 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.

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

Answer to Problem 26.AE

The equivalent mass of $\text{0}\text{.269 2 g}$ sample is $\text{62}\text{.94}\text{g /}\text{eq}$

Explanation of Solution

To determine the equivalent mass of $\text{0}\text{.269 2 g}$ sample.

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 mole equivalent of cation equal to the mole of ${\text{OH}}^{\text{-}}$ ion

Therefore, the mole of $\text{NaOH}$ required is equal to the mole of ${\text{H}}^{\text{+}}$ realised from the column.

$\begin{array}{l}\text{Mole}\text{of}\text{NaOH}\text{= 0}\text{.1396}\text{M×}\text{0}\text{.03064}\text{L}\\ \text{=}\text{4}\text{.27mmol}\end{array}$

The volume and concentration of $\text{NaOH}$ are plugged in above equation to give the mole of $\text{NaOH}$ required to titrate.

Hence, the milli equivalents of cation that present in per gram of sample is $\text{4}\text{.277}\text{meq}$

The equivalent mass id given by the mass of substance that contains one equivalent.

The charge of give sample is +1so the equivalent is,

$\begin{array}{l}\text{=}\frac{\text{0}\text{.269}\text{g}}{\text{4}\text{.277}{\text{×10}}^{\text{-3}}\text{eq}}\\ \text{=}\text{62}\text{.94}\text{g /}\text{eq}\end{array}$

The gine mass of sample is divided by calculated milli equivalent to give a equivalent mass of $\text{0}\text{.269 2 g}$ sample.

Conclusion

The equivalent mass of $\text{0}\text{.269 2 g}$ sample was determined.