Solveeit Logo
Login

Question

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

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

Explanation

Solution

Hydrazine is a compound of nitrogen and hydrogen atoms, and its formula is H2N−NH2{{H}_{2}}N-N{{H}_{2}}, when it is dissolved in water the reaction will be:
H2N−NH2+H2O⇌H3N+−NH2+OH−{{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: Keq=[N2H5+][OH−][N2H4]{{K}_{eq}}=\dfrac{[{{N}_{2}}H_{5}^{+}][O{{H}^{-}}]}{[{{N}_{2}}{{H}_{4}}]}, 14 = pH + pOH, pOH=−log⁡10[OH−]pOH=-{{\log }_{10}}[O{{H}^{-}}].

Complete answer:
Hydrazine is a compound of nitrogen and hydrogen atoms, and its formula is H2N−NH2{{H}_{2}}N-N{{H}_{2}}, when it is dissolved in water the reaction will be:
H2N−NH2+H2O⇌H3N+−NH2+OH−{{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:
Keq=[N2H5+][OH−][N2H4]{{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 x 10−61.0\text{ x 1}{{\text{0}}^{-6}}
This can be written as:
Keq=[N2H5+][OH−][N2H4]=1.0 x 10−6{{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:
Keq=(x) x (x)0.15−x=1.0 x 10−6{{K}_{eq}}=\dfrac{(x)\text{ x }(x)}{0.15-x}=1.0\text{ x 1}{{\text{0}}^{-6}}
x20.15−x=1.0 x 10−6\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
x20.15=1.0 x 10−6\dfrac{{{x}^{2}}}{0.15}=1.0\text{ x 1}{{\text{0}}^{-6}}
x=3.87 x 10−4x=3.87\text{ x 1}{{\text{0}}^{-4}} mol /L
So, the concentration of ions in the reaction will be:
[N2H5+]=[OH−]=3.87 x 10−4 mol /L[{{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[OH−]pOH=-{{\log }_{10}}[O{{H}^{-}}]
Putting the values, we can write:
pOH=−log⁡10(3.87 x 10−4)=3.41pOH=-{{\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+]pH=-{{\log }_{10}}[{{H}^{+}}]