Volumetric Analysis

Titration of ${\mathrm{Na}}_3{\mathrm{PO}}_4$ with $\mathrm{HCl}$ to phenolphthalein end point produces ${\mathrm{Na}}_2{\mathrm{HPO}}_4$; subsequent titration to methyl orange end point produces ${\mathrm{NaH}}_2{\mathrm{PO}}_4$. Find the $\mathrm{mg}$ of ${\mathrm{Na}}_2{\mathrm{HPO}}_4$ present in a sample which required $25\mathrm{ml},0.12 \mathrm{NHCl}$ solution to reach the phenolophthalein end point and an additional $35\mathrm{ml}$ of the same acid to reach the methyl orange end point.

${\mathrm{H}}_2{\mathrm{O}}_2$ acts as both oxidising as well as reducing agent. As oxidising agent, its product is ${\mathrm{H}}_2\mathrm{O},$ but as reducing agent, its product is ${\mathrm{O}}_2.$ Volume strength has great significance for chemical reactions.

The strength of  $\text{'}10\mathrm{V}\text{'}$ means $1$ volume (or litre) of ${\mathrm{H}}_2{\mathrm{O}}_2$ on decomposition $\left({\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{H}}_2\mathrm{O}+\frac{1}{2}{\mathrm{O}}_2\right)$ gives $10$ volume (or litre) of oxygen at $\mathrm{NTP}.$

$15\mathrm{gmBa}{\left({\mathrm{MnO}}_4\right)}_2$ sample containing inert impurity is completely reacting with $100 \mathrm{ml}$ of $\text{'}11.2 \mathrm{V}\text{'}{\mathrm{H}}_2{\mathrm{O}}_2,$ in acidic medium then what will be the $\%$ purity of $\mathrm{Ba}{\left({\mathrm{MnO}}_4\right)}_2$ in the sample?

(Atomic mas $\mathrm{Ba}=137, \mathrm{Mn}=55$)

Iodine reacts with hypo to give _____.

$12.5 \mathrm{g}$ of a sample of arsenious oxide was dissolved in water containing $7.5 \mathrm{g}$ of sodium bicarbonate and the resulting solution was diluted to $250 \mathrm{mL}. 25 \mathrm{mL}$ of this solution was completely oxidized by $22.4 \mathrm{mL}$ of a solution of iodine, $25 \mathrm{mL}$ of which reacted with $24.5 \mathrm{mL}$ of a solution of ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3·5{\mathrm{H}}_2\mathrm{O}$, containing $27.7 \mathrm{g}$ of this salt in $1 \mathrm{L}$. Calculate the percentage of arsenious oxide in the sample. The reactions involved are:

${\mathrm{As}}_2{\mathrm{O}}_3+6{\mathrm{NaHCO}}_3\to 2{\mathrm{Na}}_3{\mathrm{AsO}}_3+3{\mathrm{H}}_2\mathrm{O}+6{\mathrm{CO}}_2$

${\mathrm{Na}}_3{\mathrm{AsO}}_3+2{\mathrm{I}}_2+2{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{Na}}_3{\mathrm{AsO}}_4+4\mathrm{HI}$

Round off your answer to the nearest integer.