26-B. Vanadyl sulfate ${\mathbf{VOSO}}_{\mathbf{4}}$, FM 163.00), as supplied commercially, is contaminated with${\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{(}\mathbf{FM}\mathbf{98}\mathbf{.}\mathbf{08}\mathbf{)}$and${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$A solution was prepared by dissolving 0.2447g of impure${\mathbf{VO}}^{\mathbf{2}\mathbf{+}}$in 50.0 mL of water. Spectrophotometric analysis indicated that the concentration of the blueion was 0.0243M. A 5.00-mL sample was passed through a cation-exchange column loaded with${\mathbf{H}}^{\mathbf{+}}$. When${\mathbf{VO}}^{\mathbf{2}\mathbf{+}}$from the 5.00-mL sample became bound to the column, thereleased required 13.03mL of 0.02274M NaOH for titration. Find the weight percent of each component${\mathbf{VOSO}}_{\mathbf{4}}\mathbf{,}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$) in the vanadyl sulfate.

Interpretation:

The mass percentages of given compounds should be determined.

Ion-Exchange Chromatography:

Ion-Exchange Chromatography is separation techniq ue, 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, in the stationary phase (ion exchangers) the solute ions are exchanged into eluent ions; therefore it is called ion-exchange chromatography.

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.

${\mathrm{V}}_1{\mathrm{M}}_1={\mathrm{V}}_2{\mathrm{M}}_2$

where

$V_1$is volume of known solution

$N_1$is concentration of known solution

$V_1$is volume of unknown solution

$V_1$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.

Mole = Molarity *volume

Mass percentage:

$wt\%=Weightpercent=\frac{massofsuhs\tan ce}{massoftotulsolutionortotalsample}x\left(100\right)$

The mass percentages of${\mathrm{VOSO}}_4,{\mathrm{H}}_2{\mathrm{SO}}_4\mathrm{and}{\mathrm{H}}_2\mathrm{O}$ are 80.9%, 10.7% and 8.4% respectively.

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 ${\mathrm{H}}^{+}$is a Separator ion and the equivalent mole of ${\mathrm{H}}^{+}$ion is calculated by titration to find the mole of cation present in the sample

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

From the above statements, the amount of ${\mathrm{OH}}^{-}$is equal to total cation charge that is sum of${\mathrm{H}}_2{\mathrm{SO}}_4$and$V{O}^{2+}$

The total mole of the Mole of NaOH is

$$\begin{gathered} \mathrm{NaOH}=0.02274\mathrm{M}\times 0.01303\mathrm{L} \\ =0.2963\mathrm{mmol} \\ \mathrm{NaOH}=0.024\mathrm{M}\times 0.50\mathrm{L} \\ \mathrm{Mole}\mathrm{of}{\mathrm{H}}_2{\mathrm{SO}}_4=\frac{(2.963-2.43)}{2}) \\ =0.2665\mathrm{mmol} \end{gathered}$$

From the above calculation, the mass${\mathrm{VOSO}}_4,{\mathrm{H}}_2{\mathrm{SO}}_4\mathrm{and}{\mathrm{H}}_2\mathrm{O}$ ofplugged in above equation.

$$\begin{gathered} 1.215{\mathrm{mmolVOSO}}_4=0.198\mathrm{g} \\ 0.2665\mathrm{mmolH}2{\mathrm{SO}}_4=0.0261\mathrm{g} \end{gathered}$$

The total mass of the sample is 0.2447g

Hence the mass percentage of${\mathrm{VOSO}}_4,{\mathrm{H}}_2{\mathrm{SO}}_4\mathrm{and}{\mathrm{H}}_2\mathrm{O}$

$$\begin{gathered} VOS{O}_4\%=\frac{0.198}{0.2447}\times 100 \\ =80.9\% \\ {H}_{2}S{O}_4\%\% \\ =\frac{0.026}{0.2447}\times 100.7\% \\ {H}_{2}O\%=100-(10.7+80.9) \\ =8.4\% \end{gathered}$$

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