Primary-standard-grade arsenic(III) oxide $\left({\mathrm{As}}_4{O}_6\right)$is a useful (but carcinogenic) reagent for standardizing oxidants including${\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{I}}_{\mathbf{3}}^{\mathbf{-}}$. To standardize ${\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{,}{\mathbf{As}}_{\mathbf{4}}{\mathbf{O}}_{\mathbf{6}}$is dissolved in base and then titrated with localid="1655103723885" ${\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}$in acid. A small amount of iodide $\left(I^{-}\right)$or iodate $\left({\mathrm{IO}}_3^{-}\right)$catalyzes the reaction between${\mathbf{H}}_{\mathbf{3}}{\mathbf{AsO}}_{\mathbf{3}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}$.

$$\begin{gathered} {\mathbf{As}}_{\mathbf{4}}{\mathbf{O}}_{\mathbf{6}}\mathbf{+}\mathbf{8}{\mathbf{OH}}^{\mathbf{-}}\mathbf{}\mathbf{֏}\mathbf{}\mathbf{}\mathbf{}\mathbf{4}{\mathbf{HAsO}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \\ {\mathbf{HAsO}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{}\mathbf{֏}\mathbf{}\mathbf{}\mathbf{}{\mathbf{H}}_{\mathbf{3}}{\mathbf{AsO}}_{\mathbf{3}} \\ \mathbf{5}{\mathbf{H}}_{\mathbf{3}}{\mathbf{AsO}}_{\mathbf{3}}\mathbf{+}\mathbf{2}{\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{+}\mathbf{6}{\mathbf{H}}^{\mathbf{+}}\mathbf{\to }\mathbf{5}{\mathbf{H}}_{\mathbf{2}}{\mathbf{AsO}}_{\mathbf{4}}\mathbf{+}\mathbf{2}{\mathbf{Mn}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \end{gathered}$$

(a) A 3.214-g aliquot of${\mathbf{KMnO}}_{\mathbf{4}}$(FM 158.034) was dissolved in 1.000Lof water, heated to cause any reactions with impurities to occur, cooled, and filtered. What is the theoretical molarity of this solution if${\mathbf{KMnO}}_{\mathbf{4}}$were pure and if none was consumed by impurities?

(b) What mass of localid="1655105254654" ${\mathbf{As}}_{\mathbf{4}}{\mathbf{O}}_{\mathbf{6}}$(FM 395.68 ) would be just sufficient to react with$\mathbf{25}\mathbf{.}\mathbf{00}\mathbf{mL}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{KMnO}}_{\mathbf{4}}$solution in part (a)?

(c) It was found that$\mathbf{0}\mathbf{.}\mathbf{1468}\mathbf{g}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{As}}_{\mathbf{4}}{\mathbf{O}}_{\mathbf{6}}$required $\mathbf{29}\mathbf{.}\mathbf{985}\mathbf{}\mathbf{mL}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{KMnO}}_{\mathbf{4}}$ofsolution for the faint color of unreacted ${\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}$to appear. In a blank titration,$\mathbf{0}\mathbf{.}\mathbf{03}\mathbf{}\mathbf{mL}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}$was required to produce enough color to be seen. Calculate the molarity of the permanganate solution.

It is indicated by M and is one of the most extensively used units of concentration. The number of moles of solute present in 1 litre of solution is defined as.

$\mathrm{Molarity}=\frac{\mathrm{No}.\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{solute}}{\mathrm{Volume}\mathrm{of}\mathrm{solutions}\left(\mathrm{in}\mathrm{Litres}\right)}$

a)

To calculate theoretical molarity, we can use these two equations:

$$\begin{gathered} \mathrm{c}\left({\mathrm{KMnO}}_4\right)=\frac{\mathrm{n}\left({\mathrm{KMnO}}_4\right)}{\mathrm{V}\left({\mathrm{KMnO}}_4\right)}.......\left(1\right) \\ \mathrm{n}\left({\mathrm{KMnO}}_4\right)=\frac{\mathrm{m}\left({\mathrm{KMnO}}_4\right)}{\mathrm{M}\left({\mathrm{KMnO}}_4\right)}.......\left(2\right) \end{gathered}$$

By inserting equation (2) into equation (1), we get:

$$\begin{gathered} \mathrm{c}\left({\mathrm{KMnO}}_4\right)=\frac{\mathrm{m}\left({\mathrm{KMnO}}_4\right)}{\mathrm{m}\left({\mathrm{KMnO}}_4\right)\times \mathrm{V}\left({\mathrm{KMnO}}_4\right)} \\ \mathrm{c}\left({\mathrm{KMnO}}_4\right)=\frac{3.214\mathrm{g}}{158.034\mathrm{g}/\mathrm{mol}\times 1\mathrm{L}} \\ \mathrm{c}\left({\mathrm{KMnO}}_4\right)=0.0203\mathrm{M} \end{gathered}$$

Therefore, the molarity of $\mathrm{c}\left({\mathrm{KMnO}}_4\right)\mathrm{is}0.0203\mathrm{M}$.

(b)

To calculate the mass of ${\mathrm{As}}_4{\mathrm{O}}_6$sufficient to react with $25.00{\mathrm{mLKMnO}}_4$we need to find out the ratio of moles they react in.

First we can put${\mathrm{H}}_3{\mathrm{AsO}}_3\mathrm{and}{\mathrm{KMnO}}_4$in ratio:

$\mathrm{n}\left({\mathrm{H}}_3{\mathrm{AsO}}_3\right)=\frac{5}{2}\mathrm{n}\left({\mathrm{KMnO}}_4\right)$

Then we can put ${{\mathrm{HASO}}_3}^{2-}\mathrm{and}{\mathrm{H}}_3{\mathrm{AsO}}_3$in ratio:

$\mathrm{n}\left({{\mathrm{HAsO}}_3}^{2-}\right)=\mathrm{n}\left({\mathrm{H}}_3{\mathrm{AsO}}_3\right)=\frac{5}{2}\mathrm{n}\left({\mathrm{KMnO}}_4\right)$

And lastly, we can put ${{\mathrm{HAsO}}_3}^{2-}\mathrm{and}{\mathrm{As}}_4{\mathrm{O}}_6$in ratio:

$\mathrm{n}\left({\mathrm{As}}_4{\mathrm{O}}_6\right)=\frac{1}{4}\mathrm{n}\left({{\mathrm{HAsO}}_3}^{2-}\right)=\frac{1}{4}\mathrm{n}\left({\mathrm{H}}_3{\mathrm{AsO}}_3\right)=\frac{1}{4}.\frac{5}{2}\mathrm{n}\left({\mathrm{KMnO}}_4\right)=\frac{5}{8}\mathrm{n}\left({\mathrm{KMnO}}_4\right)$

Since we determined that the ratio in which ${\mathrm{As}}_4{\mathrm{O}}_6\mathrm{and}{\mathrm{KMnO}}_4$react is $1:5/8$we can calculate the mass of ${\mathrm{As}}_4{\mathrm{O}}_6$:

$$\begin{gathered} \mathrm{n}\left({\mathrm{As}}_4{\mathrm{O}}_6\right)=\frac{5}{8}\mathrm{n}\left({\mathrm{KMnO}}_4\right) \\ \frac{\mathrm{m}\left({\mathrm{As}}_4{\mathrm{O}}_6\right)}{\mathrm{M}\left({\mathrm{As}}_4{\mathrm{O}}_6\right)}=\frac{5}{8}.\mathrm{c}\left({\mathrm{KMnO}}_4\right).\mathrm{V}\left({\mathrm{KMnO}}_4\right)/.\mathrm{M}\left({\mathrm{As}}_4{\mathrm{O}}_6\right) \\ \mathrm{m}\left({\mathrm{As}}_4{\mathrm{O}}_6\right)=\frac{5}{8}.\mathrm{c}\left({\mathrm{KMnO}}_4\right).\mathrm{V}\left({\mathrm{KMnO}}_4\right).\mathrm{M}\left({\mathrm{As}}_4{\mathrm{O}}_6\right) \\ \mathrm{m}\left({\mathrm{As}}_4{\mathrm{O}}_6\right)=\frac{5}{8}.0.0203\mathrm{mol}/\mathrm{L}.25\times {10}^{-3}\mathrm{H}.395.68\mathrm{g}/\mathrm{mol} \\ \mathrm{m}\left({\mathrm{As}}_4{\mathrm{O}}_6\right)=0.126\mathrm{g} \end{gathered}$$

Therefore, the mass of ${\mathrm{As}}_4{\mathrm{O}}_6\mathrm{is}\mathrm{m}\left({\mathrm{As}}_4{\mathrm{O}}_6\right)=0.126\mathrm{g}$

c)

To calculate the molarity of permanganate solution, we can use the equation from step 4 .

Given data is:

$$\begin{gathered} \mathrm{m}\left({\mathrm{As}}_4{\mathrm{O}}_6\right)=0.1468\mathrm{g} \\ \mathrm{V}\left({\mathrm{KMnO}}_4\right)=29.98\mathrm{mL}-0.03\mathrm{mL}=29.95\mathrm{mL}=29.95\times {10}^{-3}\mathrm{L} \\ \mathrm{M}\left({\mathrm{As}}_4{\mathrm{O}}_6\right)=395.68\mathrm{g}/\mathrm{mol} \end{gathered}$$

By rearraning the equation from step 4 we get:

$\mathrm{m}\left({\mathrm{As}}_4{\mathrm{O}}_6\right)=\frac{5}{8}.\mathrm{c}\left({\mathrm{KMnO}}_4\right).\mathrm{V}\left({\mathrm{KMnO}}_4\right).\mathrm{M}\left({\mathrm{As}}_4{\mathrm{O}}_6\right)/\frac{8}{5.\mathrm{V}\left({\mathrm{KMnO}}_4\right).\mathrm{M}\left({\mathrm{As}}_4{\mathrm{O}}_6\right)}$

The equation for molarity is then:

$$\begin{gathered} \mathrm{c}\left({\mathrm{KMnO}}_4\right)=\frac{8}{5}.\frac{\mathrm{m}\left({\mathrm{As}}_4{\mathrm{O}}_6\right)}{\mathrm{V}\left({\mathrm{KMnO}}_4\right).\mathrm{M}\left({\mathrm{As}}_4{\mathrm{O}}_6\right)} \\ \mathrm{c}\left({\mathrm{KMnO}}_4\right)=\frac{8}{5}.\frac{0.1468\mathrm{g}}{29.95\times {10}^{-3}\mathrm{L}.395.68\mathrm{g}/\mathrm{mol}} \\ \mathrm{c}\left({\mathrm{KMnO}}_4\right)=0.0198\mathrm{M} \end{gathered}$$

Hence, the molarity of the${\mathrm{KMnO}}_4$is$\mathrm{c}\left({\mathrm{KMnO}}_4\right)=0.0198\mathrm{M}$