Chapter 16, Problem 73PS

Hydrazine, N₂H₄, can interact with water in two steps.

N₂H₄(aq) + H₂O(ℓ) ⇄ N₂H₅⁺(aq) + OH⁻(aq) K_b1 = 8.5 × 10⁻⁷

N₂H₅⁺(aq) + H₂O(ℓ) ⇄ N₂H₆²⁺(aq) + OH⁻(aq) K_b2 = 8.9 × 10⁻¹⁶

1. (a) What is the concentration of OH⁻, N₂H₅⁺ and N₂H₆²⁺ in a 0.010M aqueous solution of hydrazine?
2. (b) What is the pH of the 0.010M solution hydrazine?

Interpretation Introduction

Interpretation:

The concentration of ${\text{OH}}^{\text{-}}{\text{, N}}_{\text{2}}{\text{H}}_{\text{5}}^{\text{+}}{\text{ and N}}_{\text{2}}{\text{H}}_{\text{6}}^{\text{2+}}$ in a 0.010M aqueous solution of hydrazine has to be determined.

Concept Introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

${\text{K}}_{\text{a}}$ is acid constant for equilibrium reactions.

$\begin{array}{l}{\text{HA + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+ A}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{\text{[HA]}}\end{array}$

${\text{K}}_{\text{b}}$: It is a base constant for equilibrium reaction.

$\begin{array}{l}{\text{BOH + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{B}}^{\text{+}}{\text{+ OH}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[B}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}{\text{[BOH]}}\end{array}$

Answer to Problem 73PS

The concentration of ${\text{OH}}^{\text{-}}{\text{, N}}_{\text{2}}{\text{H}}_{\text{5}}^{\text{+}}{\text{ and N}}_{\text{2}}{\text{H}}_{\text{6}}^{\text{2+}}$ in a 0.010M aqueous solution of hydrazine

  $\begin{array}{l}{\text{[OH}}^{\text{-}}\text{] = 9}{\text{.2×10}}^{\text{-5}}\text{M}\\ {\text{[N}}_{\text{2}}{\text{H}}_{\text{5}}^{\text{+}}\text{] = 9}{\text{.2×10}}^{\text{-5}}\text{M}\\ {\text{[N}}_{\text{2}}{\text{H}}_{\text{6}}^{\text{2+}}\text{] = 8}{\text{.9×10}}^{\text{-16}}\text{M}\end{array}$

Explanation of Solution

Hydrazine, ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ can interact with water in two steps.

First ionization:

${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(aq) + H}{\text{2}}_{}\text{O(l) }\rightleftharpoons {\text{N}}_{\text{2}}{\text{H}}_5^{\text{+}}{\text{(aq) + OH}}^{\text{-}}{\text{(aq) K}}_{\text{b1}}\text{=8}{\text{.5×10}}^{\text{-7}}$

$\begin{array}{l}\text{Equilibrium expression:}\\ {\text{K}}_{\text{b1}}\text{= }\frac{{\text{[N}}_{\text{2}}{\text{H}}_5^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}{{\text{[N}}_{\text{2}}{\text{H}}_{\text{4}}\text{]}}\end{array}$

Second ionization:

${\text{N}}_{\text{2}}{\text{H}}_{\text{5}}^{\text{+}}{\text{(aq) + H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{N}}_{\text{2}}{\text{H}}_{\text{6}}^{\text{2+}}{\text{(aq) + OH}}^{\text{-}}{\text{(aq) K}}_{\text{b2}}\text{=8}{\text{.9×10}}^{\text{-16}}$

$\begin{array}{l}\text{Equilibrium expression:}\\ {\text{K}}_{\text{b2}}\text{= }\frac{{\text{[N}}_{\text{2}}{\text{H}}_6^{\text{2+}}{\text{][OH}}^{\text{-}}\text{]}}{{\text{[N}}_{\text{2}}{\text{H}}_5^{+}\text{]}}\end{array}$

From the ${\text{K}}_{\text{b1}}{\text{ and K}}_{\text{b2}}$ values, ${\text{ K}}_{\text{b2}}$ is smaller than the ${\text{ K}}_{\text{b1}}$.

Therefore, ${\text{OH}}^{\text{-}}$ is almost produced entirely from.

Let’s calculate the ${\text{OH}}^{\text{-}}$ from ${\text{ K}}_{\text{b1}}$.

Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.

$\begin{array}{l}{\text{ N}}_2{\text{H}}_4{\text{(aq) + H}}_{\text{2}}\text{O (aq)}\rightleftharpoons {\text{ N}}_2{\text{H}}_5^{+}{\text{(aq) + OH}}^{-}\text{(aq)}\\ \text{I 0}\text{.010 -- -- --}\\ \text{C -x -- +x +x}\\ \text{E (0}\text{.010-x) -- x x}\end{array}$

$\begin{array}{l}\text{Equilibrium expression:}\\ {\text{K}}_{\text{b1}}\text{= }\frac{{\text{[N}}_{\text{2}}{\text{H}}_5^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}{{\text{[N}}_{\text{2}}{\text{H}}_{\text{4}}\text{]}}\end{array}$

$\begin{array}{l}\text{8}{\text{.5×10}}^{\text{-7}}\text{ = }\frac{\text{(x)(x)}}{\text{0}\text{.010- x}}\\ \text{8}{\text{.5×10}}^{\text{-7}}\text{ = }\frac{{\text{(x)}}^{\text{2}}}{\text{0}\text{.010- x}}\\ \text{(0}\text{.010-x) approximately equals to 0}\text{.010}\\ \text{8}\text{.5 }\times {\text{10}}^{-7}\text{ = }\frac{{\text{(x)}}^{\text{2}}}{\text{0}\text{.010}}\\ {\text{ x}}^{\text{2}}\text{= (0}\text{.010)(8}{\text{.5×10}}^{\text{-7}}\text{)}\\ \text{ x = }\sqrt{\text{(0}\text{.010)(8}{\text{.5×10}}^{\text{-7}}\text{)}}\\ \text{ = 9}{\text{.2×10}}^{-5}\\ \text{Therefore,}\\ {\text{[OH}}^{-}\text{] =9}{\text{.2×10}}^{\text{-5}}\text{M}\\ {\text{[N}}_2{\text{H}}_5^{+}\text{] =9}{\text{.2×10}}^{\text{-5}}\text{M}\end{array}$

Let’s calculate the ${\text{N}}_{\text{2}}{\text{H}}_{\text{5}}^{\text{+}}{\text{ and N}}_{\text{2}}{\text{H}}_{\text{6}}^{\text{2+}}$ from second ionization.

${\text{N}}_{\text{2}}{\text{H}}_{\text{5}}^{\text{+}}{\text{(aq) + H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{N}}_{\text{2}}{\text{H}}_{\text{6}}^{\text{2+}}{\text{(aq) + OH}}^{\text{-}}{\text{(aq) K}}_{\text{b2}}\text{=8}{\text{.9×10}}^{\text{-16}}$

$\begin{array}{l}\text{Equilibrium expression:}\\ {\text{K}}_{\text{b2}}\text{= }\frac{{\text{[N}}_{\text{2}}{\text{H}}_6^{\text{2+}}{\text{][OH}}^{\text{-}}\text{]}}{{\text{[N}}_{\text{2}}{\text{H}}_5^{+}\text{]}}\end{array}$

Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.

$\begin{array}{l}{\text{ N}}_2{\text{H}}_5^{+}{\text{(aq) + H}}_{\text{2}}\text{O (aq)}\rightleftharpoons {\text{ N}}_2{\text{H}}_6^{+}{\text{(aq) + OH}}^{-}\text{(aq)}\\ \text{I 9}\text{.2}\times {\text{10}}^{-5}\text{ -- -- 9}\text{.2}\times {\text{10}}^{-5}\\ \text{C -x -- +y +y}\\ \text{E (9}\text{.2}\times {\text{10}}^{-5}\text{- x) -- y (9}\text{.2}\times {\text{10}}^{-5})\end{array}$

$\begin{array}{l}\text{Equilibrium expression:}\\ {\text{K}}_{\text{b2}}\text{= }\frac{{\text{[N}}_{\text{2}}{\text{H}}_6^{\text{2+}}{\text{][OH}}^{\text{-}}\text{]}}{{\text{[N}}_{\text{2}}{\text{H}}_5^{+}\text{]}}\end{array}$

$\begin{array}{l}\text{8}{\text{.9×10}}^{\text{-16}}\text{= }\frac{\text{y(9}{\text{.2×10}}^{\text{-5}}\text{ + y)}}{\text{(9}{\text{.2×10}}^{\text{-5}}\text{- y)}}\\ \text{y is very small compared to 9}{\text{.2×10}}^{\text{-5}}\\ \text{Therefore,}\\ \text{8}{\text{.9×10}}^{\text{-16}}\text{= }\frac{\text{y(9}{\text{.2×10}}^{\text{-5}}\text{)}}{\text{(9}{\text{.2×10}}^{\text{-5}}\text{)}}\\ \text{y = 8}{\text{.9×10}}^{\text{-16}}\\ {\text{[N}}_{\text{2}}{\text{H}}_{\text{6}}^{\text{2+}}\text{]= 8}{\text{.9×10}}^{\text{-16}}\text{M}\end{array}$

Interpretation Introduction

Interpretation:

$\text{pH}$ of the 0.010M solution hydrazine has to be calculated

Concept Introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

${\text{K}}_{\text{a}}$ is acid constant for equilibrium reactions.

$\begin{array}{l}{\text{HA + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+ A}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{\text{[HA]}}\end{array}$

${\text{K}}_{\text{b}}$: It is a base constant for equilibrium reaction.

$\begin{array}{l}{\text{BOH + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{B}}^{\text{+}}{\text{+ OH}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[B}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}{\text{[BOH]}}\end{array}$

Answer to Problem 73PS

The pH of the 0.010M solution hydrazine is $9.96$.

Explanation of Solution

$\begin{array}{c}{\text{K}}_{\text{w}}{\text{= [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ {\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{{\text{[OH}}^{\text{-}}\text{]}}\\ {\text{[OH}}^{\text{-}}\text{]=9}{\text{.2×10}}^{\text{-5}}\\ \text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{9}{\text{.2×10}}^{\text{-5}}}\\ \text{=1}{\text{.09×10}}^{\text{-10}}\text{M}\\ {\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] =1}{\text{.09×10}}^{\text{-10}}\text{M}\end{array}$

$pH$ can be calculated using the concentration of $H_{3}O$

$\begin{array}{c}pH=-\log \left[H_3{O}^{+}\right]\\ =-\log \left(\text{1}{\text{.09×10}}^{\text{-10}}\right)\\ =9.96\end{array}$