Respuesta :
Kb is called the equilibrium constant of basicity. These are actually constants for weak acids. The value for ammonia, NH₃, is actually 1.8×10⁻⁵. This is used to know how much moles of ammonia could react in an equilibrium reaction.
But I'm still gonna show you how it's solved. When we use Kb, we hydrate ammonia. The equilibrium reaction is
NH₃ + H₂O ↔ NH₄⁺ + OH⁻
Now, we use the ICE(Initial-Change-Equilibrium) analysisL
NH₃ + H₂O ↔ NH₄⁺ + OH⁻
Initial 0.55 - 0 0
Change -x - +x +x
--------------------------------------------------
Equilibrium 0.55 - x x x
The variable x here denotes the moles of the substances that is involved in the reaction. They are balanced out by their stoichiometric ratio which is equal to 1:1. Then, we apply the equation for equilibrium constants
Kb = [NH₄⁺][OH⁻]/[NH₃]
the concentration of water is not included because the solution is very dilute. Substituting the equilibrium amounts:
Kb = [x][x]/[0.55-x]
Since we are given the pH, we can use the relationship
pH = 14 - pOH ; pOH = -log [OH⁻]
11.5 = 14 - pOH
pOH = 2.5
2,5 = -log [OH⁻]
[OH⁻]=[x] = 3.162×10⁻³
Thus,
Kb = [3.162×10⁻³][3.162×10⁻³]/]0.55- 3.162×10⁻³]
Kb = 1.82 × 10⁻⁵
But I'm still gonna show you how it's solved. When we use Kb, we hydrate ammonia. The equilibrium reaction is
NH₃ + H₂O ↔ NH₄⁺ + OH⁻
Now, we use the ICE(Initial-Change-Equilibrium) analysisL
NH₃ + H₂O ↔ NH₄⁺ + OH⁻
Initial 0.55 - 0 0
Change -x - +x +x
--------------------------------------------------
Equilibrium 0.55 - x x x
The variable x here denotes the moles of the substances that is involved in the reaction. They are balanced out by their stoichiometric ratio which is equal to 1:1. Then, we apply the equation for equilibrium constants
Kb = [NH₄⁺][OH⁻]/[NH₃]
the concentration of water is not included because the solution is very dilute. Substituting the equilibrium amounts:
Kb = [x][x]/[0.55-x]
Since we are given the pH, we can use the relationship
pH = 14 - pOH ; pOH = -log [OH⁻]
11.5 = 14 - pOH
pOH = 2.5
2,5 = -log [OH⁻]
[OH⁻]=[x] = 3.162×10⁻³
Thus,
Kb = [3.162×10⁻³][3.162×10⁻³]/]0.55- 3.162×10⁻³]
Kb = 1.82 × 10⁻⁵
The equilibrium constant for ammonia[tex]\left({{{\text{K}}_{\text{b}}}}\right)[/tex] is[tex]\boxed{1.82\times{{10}^{-5}}}[/tex].
Further explanation:
Chemical equilibrium is the state in which the concentration of reactants and products become constant and do not change with time. This is because the rate of forward and backward direction becomes equal. The general equilibrium reaction is as follows:
[tex]{\text{A(g)}}+{\text{B(g)}}\rightleftharpoons{\text{C(g)}}+{\text{D(g)}}[/tex]
Equilibrium constant is the constant that relates the concentration of product and reactant at equilibrium. The formula to calculate the equilibrium constant for general reaction is as follows:
[tex]{\text{K}}=\frac{{\left[{\text{D}}\right]\left[{\text{C}}\right]}}{{\left[{\text{A}}\right]\left[{\text{B}}\right]}}[/tex]
Here, K is the equilibrium constant.
The equilibrium constant for the dissociation of acid is known as [tex]{{\text{K}}_{\text{a}}}[/tex]and equilibrium constant for the dissociation of base is known as[tex]{{\text{K}}_{\text{b}}}[/tex].
The expression that relates pH and pOH is given as follows:
[tex]{\text{pH}}+{\text{pOH}}=14[/tex] …… (1)
Rearrange equation (1) to calculate pOH.
[tex]{\text{pOH}}=14-{\text{pH}}[/tex] …… (2)
Substitute 11.50 for the value of pH in equation (2).
[tex]\begin{aligned}{\text{pOH}}&=14-{\text{11}}{\text{.50}}\\&={\text{2}}{\text{.5}}\\\end{aligned}[/tex]
pOH is the measure of hydroxide ion concentration. The formula to calculate pOH is as follows:
[tex]{\text{pOH}}=-\log\left[{{\text{O}}{{\text{H}}^-}}\right][/tex] …… (3)
Here,
[tex]\left[{{\text{O}}{{\text{H}}^-}}\right][/tex] is the concentration of hydroxide ion.
Rearrange equation (3) to calculate [tex]\left[{{\text{O}}{{\text{H}}^-}}\right][/tex].
[tex]\left[{{\text{O}}{{\text{H}}^-}}\right]={10^{-{\text{pOH}}}}[/tex] …… (4)
Substitute 2.5 for pOH in equation (4).
[tex]\begin{aligned}\left[{{\text{O}}{{\text{H}}^-}}\right]&={10^{-2.5}}\\&=0.0031622\\\end{aligned}[/tex]
The given equilibrium reaction is,
[tex]{\text{N}}{{\text{H}}_{\text{3}}}+{{\text{H}}_2}{\text{O}}\rightleftharpoons{\text{NH}}_4^++{\text{O}}{{\text{H}}^-}[/tex]
The expression of [tex]{{\text{K}}_{\text{b}}}[/tex]for the above reaction is as follows:
[tex]{{\text{K}}_{\text{b}}}=\frac{{\left[{{\text{NH}}_4^+}\right]\left[{{\text{O}}{{\text{H}}^-}}\right]}}{{\left[{{\text{N}}{{\text{H}}_3}}\right]}}[/tex] …... (5)
The equilibrium concentration of both [tex]{\text{NH}}_4^+[/tex] and [tex]{\text{O}}{{\text{H}}^-}[/tex] is the same.
0.0031622 M of [tex]{\text{O}}{{\text{H}}^-}[/tex]is present at equilibrium so 0.0031622 M out of 0.55 M of has reacted.
The initial concentration of the aqueous solution is 0.55 M. So the concentration of [tex]{\text{N}}{{\text{H}}_3}[/tex] left at equilibrium is calculated as follows:
[tex]\begin{aligned}\left[{{\text{N}}{{\text{H}}_3}}\right]&={\text{Initial concentration of N}}{{\text{H}}_{\text{3}}}-{\text{Reacted concentration of N}}{{\text{H}}_{\text{3}}}\\&={\text{0}}{\text{.55 M}}-{\text{0}}{\text{.0031622 M}}\\&={\text{0}}{\text{.5468378 M}}\\\end{aligned}[/tex]
The value of[tex]\left[{{\text{O}}{{\text{H}}^-}}\right][/tex]is 0.0031622 M.
The value of [tex]\left[{{\text{N}}{{\text{H}}_3}}\right][/tex] is 0.5468378 M.
The value of [tex]\left[{{\text{NH}}_4^+}\right][/tex] is 0.0031622 M.
Substitute these values in equation (5).
[tex]\begin{aligned}{{\text{K}}_{\text{b}}}&=\frac{{\left({0.0031622\;{\text{M}}}\right)\left({0.0031622\;{\text{M}}}\right)}}{{\left({0.546837{\text{8 M}}}\right)}}\\&=1.82861\times{10^{-5}}\\&\approx1.82\times{10^{-5}}\\\end{aligned}[/tex]
Therefore, equilibrium constant for ammonia is[tex]{\mathbf{1}}{\mathbf{.82\times1}}{{\mathbf{0}}^{{\mathbf{-5}}}}[/tex].
Learn more:
1. Calculation of equilibrium constant of pure water at 25°c: https://brainly.com/question/3467841
2. Complete equation for the dissociation of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex](aq): https://brainly.com/question/5425813
Answer details:
Grade: Senior school
Subject: Chemistry
Chapter: Equilibrium
Keywords: NH3, OH-, NH4+, H2O, equilibrium, kb, pH, pOH, 14, 11.5, 2.5, aqueous solution, 0.0031622 M, 0.5468378 M.
