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Question: Calculate the degree of ionization and pH of 0.05 M solution of a weak base having the ionization co...

Calculate the degree of ionization and pH of 0.05 M solution of a weak base having the ionization constant Kc{K_c} is 1.77×10−51.77 \times {10^{ - 5}}. Also calculate the ionization constant of the conjugate acid of this base.

Explanation

Solution

The degree of ionization is denoted by α\alpha , it is calculated by first assigning the concentration of the base and its constituent ion in terms of CαC\alpha where C is concentration. The ionization constant is calculated by dividing the ionization constant of water by the ionization constant of the base.

Complete step by step answer:
Given,
Concentration of solution is 0.05 M.
The ionization constant Kc{K_c} is 1.77×10−51.77 \times {10^{ - 5}}.
A weak base on dissolving in water partially dissociates into its constituent ion.
The dissociation reaction of the weak base is shown below.
BOH→B++OH−BOH \to {B^ + } + O{H^ - }
The base dissociates to give a cation and hydroxide anion.
Let the concentration of BOH be C(1−α)C(1 - \alpha ).
The concentration of B+{B^ + } be CαC\alpha .
The concentration of OH−O{H^ - } be CαC\alpha .
The dissociation is calculated by dividing the concentration of the individual constituent ions by the total concentration of the base solution.
The dissociation constant of base BOH is given as shown below.
Kc=[B+][OH−][BOH]{K_c} = \dfrac{{[{B^ + }][O{H^ - }]}}{{[BOH]}}
The dissociation constant Kc{K_c} for the above reaction is given as shown below.
KC=C2α2C(1−α){K_C} = \dfrac{{{C^2}{\alpha ^2}}}{{C(1 - \alpha )}}
∴α<<1\therefore \alpha < < 1
So,
Kc=Cα2{K_c} = C{\alpha ^2}
α=KcC\alpha = \sqrt {\dfrac{{{K_c}}}{C}}
To find the value of degree of ionization, substitute the value of dissociation constant and concentration in the above equation.
⇒α=1.77×10−55×10−2\Rightarrow \alpha = \sqrt {\dfrac{{1.77 \times {{10}^{ - 5}}}}{{5 \times {{10}^{ - 2}}}}}
⇒α=0.0188\Rightarrow \alpha = 0.0188
To calculate the pH, first we need to find the value of pOH, the concentration of OH is given as CαC\alpha .
Substitute the value of concentration and degree of ionization to find the value.
[OH−]=0.05×0.0188[O{H^ - }] = 0.05 \times 0.0188
[OH−]=9.4×10−4[O{H^ - }] = 9.4 \times {10^{ - 4}}
[H+][OH−]=Kw[{H^ + }][O{H^ - }] = {K_w}
Where,Kw{K_w} is an ionization constant of water.
[H+]=Kw[OH−][{H^ + }] = \dfrac{{{K_w}}}{{[O{H^ - }]}}
Kw{K_w}is 10−14{10^{ - 14}}
Substitute the values in the above equation.
⇒[H+]=10−149.4×10−4\Rightarrow [{H^ + }] = \dfrac{{{{10}^{ - 14}}}}{{9.4 \times {{10}^{ - 4}}}}
⇒[H+]=1.06×10−11\Rightarrow [{H^ + }] = 1.06 \times {10^{ - 11}}
The pH is defined as the negative logarithm of hydrogen ion concentration.
The formula to calculate pH is given below.
pH=−log⁡[H+]pH = - \log [{H^ + }]
Substitute the value of hydrogen ion concentration in the above equation.
⇒pH=−log⁡[1.06×10−11]\Rightarrow pH = - \log [1.06 \times {10^{ - 11}}]
⇒pH=11−log⁡(1.06)\Rightarrow pH = 11 - \log (1.06)
⇒pH=10.97\Rightarrow pH = 10.97
The formula to calculate the ionization constant is given as shown below.
Ka=KwKc{K_a} = \dfrac{{{K_w}}}{{{K_c}}}
Where,
Ka{K_a} is ionization constant.
Substitute the values in the above equation.
⇒Ka=10−141.77×10−5\Rightarrow {K_a} = \dfrac{{{{10}^{ - 14}}}}{{1.77 \times {{10}^{ - 5}}}}
⇒Ka=5.65×10−10\Rightarrow {K_a} = 5.65 \times {10^{ - 10}}

Note:
In this question Kc{K_c} and Ka{K_a} are both ionization constant but Kc{K_c} is ionization constant of weak base and Ka{K_a} is the ionization constant of its conjugate acid. The ionization constant is also known as dissociation constant.