Question

Calculate the degree of ionization and $\text{ pH }$ of $\text{ 0}\text{.05 M }$ the solution of a weak base having the ionization constant $\text{ }\left( {{\text{K}}_{\text{C}}} \right)\text{ }$ is$\text{ }1.77\text{ }\times \text{ }{{10}^{-5}}\text{ }$ . Also, calculate the ionization constant of the conjugate acid of this base.

Answer 1

According to the Arrhenius concept, the base is the substance that dissociates into hydroxide ions. Here, the reaction of the dissociation of the base is given as:
$\text{ BOH }\rightleftharpoons \text{ }{{\text{B}}^{\text{+}}}\text{ + O}{{\text{H}}^{-}}$
The equilibrium constant or the $\text{ }{{\text{K}}_{\text{b}}}\text{ }$is related to the concentration of the solution and the degree of dissociation. It is the extent to which the base dissociates into the solution. The sum of $\text{pOH}$ and $\text{pH}$ is equal to the 14.

Complete step by step answer:
Let’s consider a weak monobasic base $\text{ BOH }$, its dissociation by the Arrhenius acid-base concept may be represented by the following equation:
$\text{ BOH }\rightleftharpoons \text{ }{{\text{B}}^{\text{+}}}\text{ + O}{{\text{H}}^{-}}$
The equilibrium constant or the dissociation constant $\text{ }{{\text{K}}_{\text{b}}}\text{ }$of the weak base is represented as:
$\text{ }{{\text{K}}_{\text{b}}}\text{=}\dfrac{\left[ {{\text{B}}^{\text{+}}} \right]\left[ \text{O}{{\text{H}}^{-}} \right]}{\left[ \text{BOH} \right]}\text{ }$
If the initial concentration of the weak base is ‘c’ moles per litre and if $\text{ }\alpha$ is the degree of dissociation, then the weak monobasic base dissociates as follows:

{} & \text{BOH} & \rightleftharpoons & {{\text{B}}^{+}} & \+ & \text{O}{{\text{H}}^{-}} \\\ \text{Initial conc}\text{.} & c & {} & 0 & {} & 0 \\\ \text{Final conc}\text{.} & c(1-\alpha ) & {} & c\alpha & {} & c\alpha \\\ \end{matrix}$$ The dissociation constant $\text{ }{{\text{K}}_{\text{b}}}\text{ }$of the weak base is given as: $\text{ }{{\text{K}}_{\text{b}}}\text{ = }\dfrac{c\alpha \text{ }\times \text{ }c\alpha \text{ }}{c(1-\alpha )}=\text{ }\dfrac{c{{\alpha }^{2}}}{(1-\alpha )}$ Since, $\alpha \ll 1$ , the dissociation constant of a weak base is rewritten as : $\text{ }{{\text{K}}_{\text{b}}}\text{ = }c{{\alpha }^{2}}$ Let’s rearrange the equation concerning the dissociation constant we get, $$\text{ }\alpha =\sqrt{\dfrac{{{\text{K}}_{\text{b}}}}{c}}$$ We have given the following data: The dissociation constant of a weak base,${{\text{K}}_{\text{b}}}\text{ = }1.77\text{ }\times \text{ }{{10}^{-5}}\text{ }$ The concentration of weak base solution,$\text{ 0}\text{.05 M }$ To find: a) Degree of ionisation b) $$\text{ pH }$$Of solution c) Ionisation constant of the conjugate acid a) Lets, calculate the value of the degree of dissociation, $$\text{ }\alpha =\sqrt{\dfrac{{{\text{K}}_{\text{b}}}}{c}}\text{ = }\sqrt{\dfrac{1.77\times {{10}^{-5}}}{0.05}}\text{ = 0}\text{.0188}$$ Therefore, the degree of dissociation $(\alpha )$ is $$\text{0}\text{.0188}$$ or $$\text{ 1}\text{.88}\times \text{1}{{\text{0}}^{\text{-2}}}$$ . b) Now, we are interested to find out $$\text{ pH }$$about the solution. The concentration of hydroxide ions is given as: $\left[ \text{O}{{\text{H}}^{-1}} \right]=\text{ c}\alpha \text{ }$ Let’s calculate the concentration of hydroxide ions. $\left[ \text{O}{{\text{H}}^{-1}} \right]=\text{ c}\alpha \text{ = }\left( 0.05\text{ }\times \text{ 1}\text{.8}\times \text{1}{{\text{0}}^{\text{-2}}} \right)\text{ = 9}\text{.0}\times \text{1}{{\text{0}}^{\text{-4}}}$ We know that, $\text{pOH = }-\log \left[ \text{O}{{\text{H}}^{-}} \right]$ Therefore, the $\text{pOH}$ is $-\log \left[ 9.0\times {{10}^{-4}} \right]=\text{ 3}\text{.04}$ The sum of $\text{pOH}$ and $\text{pH}$ is equal to the 14. Therefore, the $\text{pH}$ of the solution is, $\begin{aligned} & 14=\text{pH + pOH} \\\ & 14=\text{pH + 3}\text{.04} \\\ & \text{pH}=\text{14 }-\text{ 3}\text{.04} \\\ & \therefore \text{pH = 10}\text{.95} \\\ \end{aligned}$ So, $\text{pH}$ of the weak base is$\text{10}\text{.95}$. c) Let's calculate the ionisation constant of the conjugate acid of a weak base. The product of the dissociation constant of a base and its conjugate acid is always equal to the dissociation constant of water ${{\text{K}}_{\text{w}}}$ . Thus , on applying this we can calculate the ionisation constant of conjugate acid ${{\text{K}}_{\text{a}}}$. $$\begin{aligned} & {{\text{K}}_{\text{w}}}\text{ = }{{\text{K}}_{\text{base}}}\text{ }\\!\\!\times\\!\\!\text{ }{{\text{K}}_{\text{conjugate acid}}} \\\ & \Rightarrow {{\text{K}}_{\text{conjugate acid}}}\text{ = }\dfrac{{{\text{K}}_{\text{w}}}\text{ }}{{{\text{K}}_{\text{base}}}} \\\ \end{aligned}$$ Let’s substitute the values, $${{\text{K}}_{\text{conjugate acid}}}\text{ = }\dfrac{{{\text{K}}_{\text{w}}}\text{ }}{{{\text{K}}_{\text{base}}}}\text{ = }\dfrac{{{10}^{-14}}}{1.77\times {{10}^{-5}}}\text{ = 5}\text{.65 }\times {{10}^{-10}}$$ Thus, the ionisation constant of the conjugate acid is equal to the$$\text{5}\text{.65 }\times {{10}^{-10}}$$. **Hence, a) The degree of dissociation $\text{ }(\alpha)\text{ }$ of a weak base is$$\text{ 1}\text{.88}\times \text{1}{{\text{0}}^{\text{-2}}}$$. b) The $$\text{ pH }$$ of the weak base solution is equal to$\text{10}\text{.95}$. c) The ionisation constant of conjugate acid $${{\text{K}}_{\text{conjugate acid}}}$$ or weak base is equal to$$\text{5}\text{.65 }\times {{10}^{-10}}$$.** **Note:** According to the Lowry-Bronsted acid-base theory, the base dissociates into the solution and produces its corresponding conjugate acid. The base loses the hydroxide ion, the reaction is as shown below: $\begin{matrix} \text{BOH} & \text{+} & {{\text{H}}^{\text{+}}} & \rightleftharpoons & {{\text{B}}^{\text{+}}} & \text{+} & {{\text{H}}_{\text{2}}}\text{O} \\\ \text{(Base)} & {} & \text{(Acid)} & {} & \text{(Conjugate acid)} & {} & \text{(Conjugate base)} \\\ \end{matrix}$ Thus, the conjugate acid concentration can be obtained from the equilibrium constant.