Question

Ephedrine $\mathbf{\left(}{\mathbf{\text{C}}}_{\mathbf{10}}{\mathbf{\text{H}}}_{\mathbf{15}}\mathbf{\text{ON}}\mathbf{\right)}$is a base that is used in nasal sprays as a decongestant.

(a) Write an equation for its equilibrium reaction with water.

(b) The${\mathit{K}}_{\mathbf{\text{b}}}$for ephedrine is$\mathbf{1.4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$. Calculate the${\mathit{K}}_{\mathbf{a}}$for its conjugate acid.

(c) Is ephedrine a weaker or a stronger base than ammonia?

Short answer

(a) The equation for ephedrine for its reaction with water is written as:

${\mathbf{C}}_{10}{\mathbf{H}}_{15}{\mathbf{ON}}_{\left(aq\right)}+{\mathbf{H}}_2{\mathbf{O}}_{\left(l\right)}\rightleftharpoons {\mathbf{C}}_{10}{\mathbf{H}}_{15}{\mathbf{ONH}}_{\left(aq\right)}^{+}+{{\mathbf{OH}}^{-}}_{\left(aq\right)}$

(b) The value of${\mathrm{K}}_{\mathrm{a}}$ for its conjugate acid is${\text{7.1×10}}^{\text{-11}}$.

(c) Ephidrine base is stronger than ammonia,

Step-by-step solution

Understanding the concecp of equilibrium constant.

Equilibrium constant is the ratio between the concentration of products and the concetrations of reactant at equilibrium.

For example,

$\mathbf{j}\mathbf{}\mathbf{A}\mathbf{+}\mathbf{k}\mathbf{}\mathbf{B}\mathbf{=}\mathbf{lC}\mathbf{+}\mathbf{mD}$

Here,

A,B,C,D are chemical species and j,k,l and m are coefficients of the balanced chemical equation.

The equilibrium constant will be;

$\mathit{K}\mathbf{=}\frac{\mathbf{\left[}\mathbf{C}\mathbf{\right]}{\mathbf{\left[}\mathbf{D}\mathbf{\right]}}^{\mathbf{m}}}{{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}^{\mathbf{j}}{\mathbf{\left[}\mathbf{B}\mathbf{\right]}}^{\mathbf{k}}}$

Part (a) Step 2: Writing an equation.

Write an equation for its equilibrium reaction with water.

Given compound:${\text{C}}_{10}{\text{H}}_{15}\text{ON}$

_{${\mathbf{C}}_{10}{\mathbf{H}}_{15}{\mathbf{ON}}_{\left(\mathrm{aq}\right)}+{\mathbf{H}}_2{\mathbf{O}}_{\left(\mathrm{l}\right)}\rightleftharpoons {\mathbf{C}}_{10}{\mathbf{H}}_{15}{\mathbf{ONH}}_{\left(\mathrm{aq}\right)}^{+}+{{\mathbf{OH}}^{-}}_{\left(\mathrm{aq}\right)}$}

Part (b) Step 3: Calculate the Ka for its conjugate acid.

Formulas we will need to get a solution:

${\mathbf{\text{K}}}_{\mathbf{\text{w}}}{\mathbf{\text{=K}}}_{\mathbf{\text{a}}}{\mathbf{\text{×K}}}_{\mathbf{\text{b}}}$

where:

data-custom-editor="chemistry" ${\mathbf{\text{K}}}_{\mathbf{\text{w}}}$is ionic product constant for water which value isdata-custom-editor="chemistry" $\mathbf{1.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{14}}\mathbf{}{\mathit{K}}_{\mathbf{b}}$ is base ionization constantdata-custom-editor="chemistry" ${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$ is acid ionization constant.

As we are given the data for base ionization constant we will plug that data in formula from above:

data-custom-editor="chemistry" ${\text{K}}_{\text{w}}{\text{=K}}_{\text{a}}{\text{×K}}_{\text{b}}$

Here, we are given:

data-custom-editor="chemistry" ${\text{K}}_{\text{b}}{\text{=1.4×10}}^{\text{-4}}$

Plug this into formula:

data-custom-editor="chemistry" ${\text{1.0×10}}^{\text{-14}}{\text{=K}}_{\text{a}}{\text{×1.4×10}}^{\text{-4}}$

Substituting the values in this equation, we get:

data-custom-editor="chemistry" ${\text{K}}_{\text{a}}\text{=}\frac{{\text{1.0×10}}^{\text{-14}}}{{\text{1.4×10}}^{\text{-4}}}$

Divide these numbers and we get a solution fordata-custom-editor="chemistry" ${\mathrm{K}}_{\mathrm{a}}$ :

data-custom-editor="chemistry" ${\text{K}}_{\text{a}}{\text{=7.1×10}}^{\text{-11}}$

Part (c) Step 4 : Compare the base ionization constant of ephedrine and ammonia.

In general we know that when base has a larger${\text{K}}_{\text{b}}$ then the base is stronger.

To solve this question, we will have to compare base ionization constant of ephadrine and base ionization constant of ammonia.

From the textbook we know that${\text{K}}_{\text{b}}$ for ammonia is${\text{1.5×10}}^{\text{-5}}$ and the${\text{K}}_{\text{b}}$ for ephedrine is${\text{1.4×10}}^{\text{-4}}$.

We see that:

${\text{K}}_{\text{b}}{\text{(ephedrine)>K}}_{\text{b}}\text{(ammonia)}$

Therefore, the ephedrine is a stronger base than ammonia.