Question

In the following reaction:
$P_4\left(s\right)+O{H}^{-}\left(aq\right)\to P{H}_3\left(g\right)+H_{2}P{O}_2^{-}\left(aq\right)$
Calculate the stoichiometric coefficient of $O{H}^{-}$ and $H_{2}O$ respectively present in a balanced equation in a basic medium?

• A. $3H_{2}O,3O{H}^{-}$
• B. $6H_{2}O,6O{H}^{-}$
• C. $3H_{2}O,6O{H}^{-}$
• D. $6H_{2}O,3O{H}^{-}$

Answer 1

The correct option is A $3H_{2}O,3O{H}^{-}$

${\stackrel{0}{P}}_4\left(s\right)+O{H}^{-}\left(aq\right)\to \stackrel{-3}{P}{\stackrel{+1}{H}}_3\left(g\right)+{\stackrel{+1}{H}}_2\stackrel{+1}{P}\stackrel{-2}{O_2^{-}}\left(aq\right)$

Total decrease in O.N of $P_4$ in $P{H}_3=12$.
Total increase in O.N of $P_4$ in $H_{2}P{O}_2^{-}=1\times 4=4$
To balance increase and decrease in O.N, multiply $P{H}_3$ by 1 and $H_{2}P{O}_2^{-}$ by 3.
$P_4\left(s\right)+O{H}^{-}\left(aq\right)\to P{H}_3\left(g\right)+3H_{2}P{O}_2^{-}\left(aq\right)$
To balance $O$ atoms, we have to multiply $O{H}^{-}$ by 6, we get
$P_4\left(s\right)+6O{H}^{-}\left(aq\right)\to P{H}_3\left(g\right)+3H_{2}P{O}_2^{-}\left(aq\right)$
To balance $H$ atoms
$P_4\left(s\right)+6O{H}^{-}+3H_{2}O\to P{H}_3\left(g\right)+3H_{2}P{O}_2^{-}\left(aq\right)+3O{H}^{-}$
$P_4+3O{H}^{-}+3H_{2}O\to P{H}_3\left(g\right)+3H_{2}P{O}_2^{-}\left(aq\right)$