Question

The correct formula to calculate the hydroxyl ion concentration of an aqueous solution of $N{{H}_{4}}N{{O}_{3}}$ is,
A) $\sqrt{\dfrac{C\times {{K}_{w}}}{{{K}_{b}}}}$
B) $\sqrt{\dfrac{{{K}_{w}}\times {{K}_{b}}}{C}}$
C) $\sqrt{\dfrac{C\times {{K}_{w}}}{{{K}_{a}}}}$
D) $\sqrt{\dfrac{{{K}{a}}\times {{K}{w}}}{C}}$$$

Answer 1

$N{{H}_{4}}N{{O}_{3}}$ is the salt of a weak base and a strong acid. It is the salt of $N{{H}_{4}}OH$ and $HN{{O}_{3}}$ $\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]={{K}_{W}}$, ${{K}_{W}}$ is the ionization constant of water

Complete step by step solution:
The requirement of the question is to find an equation that is used for the calculation of the concentration of aqueous $N{{H}_{4}}N{{O}_{3}}$ solution from the above options given.
So we know that the$N{{H}_{4}}N{{O}_{3}}$ is the salt of $N{{H}_{4}}OH$ and $HN{{O}_{3}}$
$N{{H}_{4}}N{{O}_{3}}\to N{{H}_{4}}OH+HN{{O}_{3}}$
(Weak base) (Strong acid)
As $N{{H}_{4}}OH$is a weak base, it undergoes hydrolysis in aqueous solution,
$N{{H}_{4}}^{+}+{{H}_{2}}O\to N{{H}_{4}}OH+{{H}^{+}}$
When time changes the concentration of the reactants and products are also changing.

$N{{H}_{4}}^{+}+{{H}_{2}}O\to N{{H}_{4}}OH+{{H}^{+}}$

When time, t =o | C 0 0
When time, t=t | C(1-h) Ch Ch

C is the concentration
So we write the hydrolysis constant,${{K}_{h}}$ as,
${{K}_{h}}=\dfrac{Ch\times Ch}{C(1-h)}$
Here$(1-h)\sim 1$, applying this condition on the above equation.
Hence the equation becomes,${{K}_{h}}=\dfrac{{{(Ch)}^{2}}}{C}=C{{h}^{2}}$
$h=\sqrt{\dfrac{{{K}_{h}}}{C}}$
Here as the $N{{H}_{4}}^{+}$ ion is acting as the acid hence ${{K}_{h}}$ can be written as,
${{K}_{h}}={{K}_{a}}=\dfrac{{{K}_{w}}}{{{K}_{b}}}$ ,
${{K}_{a}}$=Dissociation or ionization constant of acid
${{K}_{b}}$= Ionization constant of base
${{K}_{w}}$= ionization of water

Concentration of $\left[ {{H}^{+}} \right]$ can be written as, $$\left[ {{H}^{+}} \right]=Ch=\sqrt{\dfrac{{{K}_{w}}\times C}{{{K}_{b}}}}$$ We know the relation, $\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]={{K}_{w}}$ So writing the equation for $\left[ O{{H}^{-}} \right]$ Hence, $\left[ O{{H}^{-}} \right]=\dfrac{{{K}_{w}}}{\left[ {{H}^{+}} \right]}$ $$\left[ O{{H}^{-}} \right]=\sqrt{\dfrac{{{K}_{w}}\times {{K}_{b}}}{C}}$$$$$$$$$$$$ **So the correct option for the answer is option (B)** **Note:** Derivation should be done with utmost care while substituting one equation with the other. And from the standard equation we could find the concentration term, dissociation constant of acid or dissociation constant of base. And so the extent to which ions have dissociated. ${{K}_{h}}={{K}_{a}}=\dfrac{{{K}_{w}}}{{{K}_{b}}}$ ${{K}_{a}}$is related to ${{K}_{b}}$,by the equation ,$$$$ We should always write the ionization equations of the ions for the better results. As the chemical equation the dissociation of compounds to respective ions and it is easy for us to interpret.