Question

How do you calculate the pH of 0.15 M aqueous solution of hydrazine?

Answer 1

Hint: Hydrazine is a compound of nitrogen and hydrogen atoms, and its formula is ${{H}_{2}}N-N{{H}_{2}}$, when it is dissolved in water the reaction will be:
${{H}_{2}}N-N{{H}_{2}}+{{H}_{2}}O\rightleftharpoons {{H}_{3}}{{N}^{+}}-N{{H}_{2}}+O{{H}^{-}}$
The formulas used for solving the question will be: ${{K}_{eq}}=\dfrac{[{{N}_{2}}H_{5}^{+}][O{{H}^{-}}]}{[{{N}_{2}}{{H}_{4}}]}$, 14 = pH + pOH, $pOH=-{{\log }_{10}}[O{{H}^{-}}]$.

Complete answer:
Hydrazine is a compound of nitrogen and hydrogen atoms, and its formula is ${{H}_{2}}N-N{{H}_{2}}$, when it is dissolved in water the reaction will be:
${{H}_{2}}N-N{{H}_{2}}+{{H}_{2}}O\rightleftharpoons {{H}_{3}}{{N}^{+}}-N{{H}_{2}}+O{{H}^{-}}$
The equilibrium constant for this reaction will be:
${{K}_{eq}}=\dfrac{[{{N}_{2}}H_{5}^{+}][O{{H}^{-}}]}{[{{N}_{2}}{{H}_{4}}]}$
The value of equilibrium constant of this strong base is $1.0\text{ x 1}{{\text{0}}^{-6}}$
This can be written as:
${{K}_{eq}}=\dfrac{[{{N}_{2}}H_{5}^{+}][O{{H}^{-}}]}{[{{N}_{2}}{{H}_{4}}]}=1.0\text{ x 1}{{\text{0}}^{-6}}$
Given the molarity of the compound is 0.15 M, so after equilibrium, x moles will be formed in the products, and in the reactants, the concentration will be 0.15 – x.
The equilibrium constant equation will be:
${{K}_{eq}}=\dfrac{(x)\text{ x }(x)}{0.15-x}=1.0\text{ x 1}{{\text{0}}^{-6}}$
$\dfrac{{{x}^{2}}}{0.15-x}=1.0\text{ x 1}{{\text{0}}^{-6}}$
Since, the value of equilibrium constant is very small than the value of concentration, then the concentration of reactant after equilibrium can be taken as 0.15
$\dfrac{{{x}^{2}}}{0.15}=1.0\text{ x 1}{{\text{0}}^{-6}}$
$x=3.87\text{ x 1}{{\text{0}}^{-4}}$ mol /L
So, the concentration of ions in the reaction will be:
$[{{N}_{2}}H_{5}^{+}]=[O{{H}^{-}}]=3.87\text{ x 1}{{\text{0}}^{-4}}\text{ mol /L}$
As we know that:
pH + pOH = 14
and we have the concentration hydroxyl ions, then we can calculate the pOH by using the formula:
$pOH=-{{\log }_{10}}[O{{H}^{-}}]$
Putting the values, we can write:
$pOH=-{{\log }_{10}}(3.87\text{ x 1}{{\text{0}}^{-4}})=3.41$
Now, we can calculate the pH, as:
pH = 14 – 3.41 = 10.6
Therefore, the pH of 0.15 M solution of hydrazine is 10.6.

Note:
If the value of the concentration of hydrogen ions is known instead of hydroxyl ions then we can calculate the pH of the solution by using the formula:
$pH=-{{\log }_{10}}[{{H}^{+}}]$