“$\mathrm{A}$” obtained by Ostwald's method involving air oxidation of ${\mathrm{NH}}_3$, upon further air oxidation produces “$\mathrm{B}$”. “$\mathrm{B}$” on hydration forms an oxoacid of Nitrogen along with evolution of “$\mathrm{A}$”. The oxoacid also produces “$\mathrm{A}$” and gives positive brown ring test

Ostwald's method is based upon catalytic oxidation of ${\mathrm{NH}}_3$ by atmospheric oxygen.

$4{\mathrm{NH}}_3 \left(\mathrm{g}\right) +5{\mathrm{O}}_2 \left(\mathrm{g}\right) \underset{500\mathrm{K}, 9 \mathrm{bar}}{\overset{\mathrm{Pt}/\mathrm{Rh} \mathrm{gauge} \mathrm{catalyst}}{\to }} \underset{\left(\mathrm{A}\right)}{4\mathrm{NO} \left(\mathrm{g}\right) }+6{\mathrm{H}}_2\mathrm{O} \left(\mathrm{g}\right)$

Nitric oxide thus formed combines with oxygen giving ${\mathrm{NO}}_2$.

$2\mathrm{NO} \left(\mathrm{g}\right) +{\mathrm{O}}_2 \left(\mathrm{g}\right) \underset{\left(\mathrm{B}\right)}{\rightleftharpoons 2{\mathrm{NO}}_2 \left(\mathrm{g}\right)}$

Nitrogen dioxide so formed, dissolves in water to give ${\mathrm{HNO}}_3$.

$3{\mathrm{NO}}_2 \left(\mathrm{g}\right) + {\mathrm{H}}_2\mathrm{O} \left(\mathrm{l}\right) \to 2{\mathrm{HNO}}_3 \left(\mathrm{aq}\right) + \mathrm{NO} \left(\mathrm{g}\right)$

$\therefore$ A and B are $\mathrm{NO}$ and ${\mathrm{NO}}_2$ respectively.