Question

Among the following acids, if number of acids stronger than $H_{3}P{O}_4$ is $x$ and number of monobasic acid is $y$ then, $\frac{x}{y}$ would be?
$H_{2}C{O}_3,H_{3}P{O}_3,C{H}_{3}COOH,H_{2}S{O}_4,H_{3}P{O}_2,H_4{P}_2{O}_7,HCl{O}_4,H_{3}B{O}_3,HN{O}_3.$

Answer 1

More is the electronegativity of central atom, more is the acidic nature.
So, $H_{2}C{O}_3,C{H}_{3}C{O}_{2}H$ and $H_{3}B{O}_3$ are weaker acids than $H_{3}P{O}_4$.
Here $H_{3}P{O}_3$ and $H_{3}P{O}_2$ are stronger acids than $H_{3}P{O}_4$ due to restrictive resonance.
$\therefore$ The number of stronger acids than $H_{3}P{O}_4$ is $x=6$

Monobasic $\to C{H}_{3}C{O}_{2}H,H_{3}P{O}_2,HCl{O}_4,H_{3}B{O}_3,HN{O}_3$
$H_{3}B{O}_3$ is monobasic as it can accomodate one more $O{H}^{-}$ ion to form $\left[B\right(OH{)}_4{]}^{-}$.
Dibasic $\to H_{2}C{O}_3,H_{3}P{O}_3,H_{2}S{O}_4$
Tetrabasic $\to H_4{P}_2{O}_7$
$\therefore y=5$
So, $\frac{x}{y}=\frac{6}{5}=1.2$