Chemistry, 27.09.2019 15:00 scastillo8

The ph of a 0.55 m aqueous solution ammonia, nh3, at 25.0°c is 11.50. what is the value of kb for nh3?

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Answer from: mdkester11

The equilibrium constant for ammonia $\left({{{\text{K}}_{\text{b}}}}\right)$ is $\boxed{1.82\times{{10}^{-5}}}$ .

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:

${\text{A(g)}}+{\text{B(g)}}\rightleftharpoons{\text{C(g)}}+{\text{D(g)}}$

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:

${\text{K}}=\frac{{\left[{\text{D}}\right]\left[{\text{C}}\right]}}{{\left[{\text{A}}\right]\left[{\text{B}}\right]}}$

Here, K is the equilibrium constant.

The equilibrium constant for the dissociation of acid is known as ${{\text{K}}_{\text{a}}}$ and equilibrium constant for the dissociation of base is known as ${{\text{K}}_{\text{b}}}$ .

The expression that relates pH and pOH is given as follows:

${\text{pH}}+{\text{pOH}}=14$ …… (1)

Rearrange equation (1) to calculate pOH.

${\text{pOH}}=14-{\text{pH}}$ …… (2)

Substitute 11.50 for the value of pH in equation (2).

$\begin{aligned}{\text{pOH}}&=14-{\text{11}}{\text{.50}}\\&={\text{2}}{\text{.5}}\\\end{aligned}$

pOH is the measure of hydroxide ion concentration. The formula to calculate pOH is as follows:

${\text{pOH}}=-\log\left[{{\text{O}}{{\text{H}}^-}}\right]$ …… (3)

Here,

$\left[{{\text{O}}{{\text{H}}^-}}\right]$ is the concentration of hydroxide ion.

Rearrange equation (3) to calculate $\left[{{\text{O}}{{\text{H}}^-}}\right]$ .

$\left[{{\text{O}}{{\text{H}}^-}}\right]={10^{-{\text{pOH}}}}$ …… (4)

Substitute 2.5 for pOH in equation (4).

$\begin{aligned}\left[{{\text{O}}{{\text{H}}^-}}\right]&={10^{-2.5}}\\&=0.0031622\\\end{aligned}$

The given equilibrium reaction is,

${\text{N}}{{\text{H}}_{\text{3}}}+{{\text{H}}_2}{\text{O}}\rightleftharpoons{\text{NH}}_4^++{\text{O}}{{\text{H}}^-}$

The expression of ${{\text{K}}_{\text{b}}}$ for the above reaction is as follows:

${{\text{K}}_{\text{b}}}=\frac{{\left[{{\text{NH}}_4^+}\right]\left[{{\text{O}}{{\text{H}}^-}}\right]}}{{\left[{{\text{N}}{{\text{H}}_3}}\right]}}$ …... (5)

The equilibrium concentration of both ${\text{NH}}_4^+$ and ${\text{O}}{{\text{H}}^-}$ is the same.

0.0031622 M of ${\text{O}}{{\text{H}}^-}$ 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 ${\text{N}}{{\text{H}}_3}$ left at equilibrium is calculated as follows:

$\begin{aligned}\left[{{\text{N}}{{\text{H}}_3}}\right]&={\text{Initial concentration of N}}{{\text{H}}_{\text{3}}}-{\text{Reacted concentration of N}}{{\text{H}}_{\text{3}}}\\&={\text{0}}{\text{.55 M}}-{\text{0}}{\text{.0031622 M}}\\&={\text{0}}{\text{.5468378 M}}\\\end{aligned}$

The value of $\left[{{\text{O}}{{\text{H}}^-}}\right]$ is 0.0031622 M.

The value of $\left[{{\text{N}}{{\text{H}}_3}}\right]$ is 0.5468378 M.

The value of $\left[{{\text{NH}}_4^+}\right]$ is 0.0031622 M.

Substitute these values in equation (5).

$\begin{aligned}{{\text{K}}_{\text{b}}}&=\frac{{\left({0.0031622\;{\text{M}}}\right)\left({0.0031622\;{\text{M}}}\right)}}{{\left({0.546837{\text{8 M}}}\right)}}\\&=1.82861\times{10^{-5}}\\&\approx1.82\times{10^{-5}}\\\end{aligned}$

Therefore, equilibrium constant for ammonia is ${\mathbf{1}}{\mathbf{.82\times1}}{{\mathbf{0}}^{{\mathbf{-5}}}}$ .

Learn more:

1. Calculation of equilibrium constant of pure water at 25°c:

2. Complete equation for the dissociation of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ (aq):

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.

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The ph of a 0.55 m aqueous solution ammonia, nh3, at 25.0°c is 11.50. what is the value of kb for nh...

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