Chapter 16, Problem 111IL

Interpretation Introduction

Interpretation:

The bases has to be arranged in order of their increasing basicity.

Concept introduction:

The $\text{pH}$ is a logarithmic scale used to specify the acidity or basicity of an aqueous solution.

The expression for $\text{pH}$ is,

$\text{pH}=-\text{log}\left[{\text{H}}^{+}\right]$                                                                                                         (1)

The $\text{pH}$ varies from $0$ to $14$ for an aqueous solution.

$\begin{array}{l}\text{pH}=7\left(\text{Neutral}\right)\\ \text{pH}>7\left(\text{Basic}\right)\\ \text{pH}<7\left(\text{Acidic}\right)\end{array}$

The value of $\text{pOH}$ is a measure of hydroxide ion concentration while $\text{pH}$ is a measure of hydrogen ion concentration. The expression for $\text{pOH}$ is,

$\text{pOH}=-\text{log}\left[{\text{OH}}^{-}\right]$                                                                                                    (2)

The sum of $\text{pH}$ and $\text{pOH}$ is equal to $14$ at $25{}^{°}\text{C}$.

$\text{pH}+\text{pOH}=14$                                                                                                          (3)

Answer to Problem 111IL

The increasing order of basicity among three given bases is,

$C{O}_3{}^{-2}<C{N}^{-}<C{H}_{3}N{H}_2$

Explanation of Solution

An equilibrium constant $\left(K\right)$ is the ratio of the concentration of products and reactants raised to appropriate stoichiometric coefficient at equilibrium.

For base ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$,

  ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{CH}}_{\text{3}}{\text{NH}}_3{}^{+}\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right)$

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows,

$K_{\text{b}}=\frac{\left[{\text{CH}}_{\text{3}}{\text{NH}}_3{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\right]}$

An equilibrium constant $\left(K\right)$ with subscript $b$ indicates that it is an equilibrium constant of the base in water.

For base ${\text{CN}}^{-}$,

    ${\text{CN}}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons \text{HCN}\left(\text{l}\right)+{\text{OH}}^{-}\left(\text{aq}\right)$

The equilibrium constant, $K_{\text{b}}$ for ${\text{CN}}^{-}$ in water is expressed as,

$K_{\text{b}}=\frac{\left[{\text{CN}}^{-}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{HCN}\right]}$

For base ${\text{CO}}_{\text{3}}{}^{-2}$,

    ${\text{CO}}_{\text{3}}{}^{-2}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{HCO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right)$

The equilibrium constant, $K_{\text{b}}$ for ${\text{CO}}_{\text{3}}{}^{-2}$ in water is expressed as,

$K_{\text{b}}=\frac{\left[{\text{HCO}}_{\text{3}}{}^{-}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{H}}_2{\text{CO}}_{\text{3}}\right]}$

Polyprotic acid is capable of donating more than one proton. The ionization constant for each successive loss of a proton is about ${10}^4$ to ${10}^6$ which is smaller than the previous ionization step. Because of these reason, the first ionization constant $\left(K_{\text{b1}}\right)$, is much larger than the second ionization constant $\left(K_{\text{b2}}\right)$, so the hydroxide ion concentration in the solution results almost entirely from the first step. The hydroxide ion produced in the second step can be neglected. Therefore, hydroxide ion concentration in the solution of ${\text{CO}}_{\text{3}}{}^{-2}$ is calculated entirely from the first ionization step.

Given:

Refer Appendix H and I for $K_b$ values of given bases.

The value $K_{\text{b}}$ for ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ is $5\times {10}^{-4}$.

The value $K_{\text{b}}$ for ${\text{CN}}^{-}$ is $2.5\times {10}^{-5}$.

The value $K_{\text{b}}$ for ${\text{CO}}_3{}^{-2}$ is $2.4\times {10}^{-8}$.

The initial concentration of each solution is $0.10\text{M}$.

Set up an ICE table for the reaction of any base B with water.

$\begin{array}{cccccc}\text{Equilibrium}& \text{B}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)& \rightleftharpoons & {\text{BH}}^{+}\left(\text{aq}\right)& \text{+}& {\text{OH}}^{-}\left(\text{aq}\right)\\ \text{Initial}\left(\text{M}\right)& \text{0}\text{.10}& & \text{0}& & \text{0}\\ \text{Change}& -x& & \text{+}x& & \text{+}x\\ \text{Equilibrium}\left(\text{M}\right)& \text{0}\text{.10 }-x& & x& & x\end{array}$

The $K_{\text{b}}$ is written in terms of $x$ at equilibrium,

$\begin{array}{c}{K}_{\text{b}}=\frac{x^2}{0.1-x}\\ x=\sqrt{K_{\text{b}}\times 0.1}\end{array}$ (4)

Substitute the value of $K_{\text{b}}$ for ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ in equation (4) to calculate the value of $x$,

$\begin{array}{c}x=\sqrt{5\times {10}^{-4}\times 0.1}\\ =0.7\times {10}^{-2}\end{array}$

The value of $x$ is equal to the concentration of hydroxide ion, substitute the value of hydroxide ion in equation (2) to calculate the value of $\text{pOH}$,

$\begin{array}{c}\text{pOH}=-\log \left(0.7\times {10}^{-2}\right)\\ =2.15\end{array}$

Substitute the value of $\text{pOH}$ in equation (3) to calculate $\text{pH}$ of the solution,

$\begin{array}{c}14=2.15+\text{pH}\\ \text{pH}=\text{11}\text{.85}\end{array}$

Substitute the value of $K_{\text{b}}$ for ${\text{CN}}^{-}$ in equation (4) to calculate the value of $x$,

$\begin{array}{c}x=\sqrt{2.5\times {10}^{-5}\times 0.1}\\ =0.16\times {10}^{-2}\end{array}$

The value of $x$ is equal to the concentration of hydroxide ion.

Substitute the value of hydroxide ion in equation (2) to calculate the value of $\text{pOH}$,

$\begin{array}{c}\text{pOH}=-\log \left(0.16\times {10}^{-2}\right)\\ =2.8\end{array}$

Substitute the value of $\text{pOH}$ in equation (3) to calculate $\text{pH}$ of the solution,

$\begin{array}{c}14=2.8+\text{pH}\\ \text{pH}=\text{11}\text{.20}\end{array}$

Substitute the value of $K_{\text{b}}$ for ${\text{CO}}_3{}^{-2}$ in equation (4) to calculate the value of $x$,

$\begin{array}{c}x=\sqrt{2.4\times {10}^{-8}\times 0.1}\\ =0.48\times {10}^{-4}\end{array}$

The value of $x$ is equal to the concentration of hydroxide ion.

Substitute the value of hydroxide ion in equation (2) to calculate the value of $\text{pOH}$,

$\begin{array}{c}\text{pOH}=-\log \left(0.48\times {10}^{-4}\right)\\ =4.31\end{array}$

Substitute the value of $\text{pOH}$ in equation (3) to calculate $\text{pH}$ of the solution,

$\begin{array}{c}14=4.31+\text{pH}\\ \text{pH}=\text{9}\text{.69}\end{array}$

From the $\text{pH}$ values, calculated for given bases, one can conclude that higher the value of $\text{pH}$, more basic the solution and among three given bases ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ has the highest value of $\text{pH}$ and ${\text{CO}}_{\text{3}}{}^{-2}$ has the lowest value of $\text{pH}$.

The increasing order of basicity among three given bases is,

$C{O}_3{}^{-2}<C{N}^{-}<C{H}_{3}N{H}_2$

Conclusion

The increasing order of their basicity is $C{O}_3{}^{-2}<C{N}^{-}<C{H}_{3}N{H}_2$.