Question

The pH of a 0.1 M solution of $N{{H}_{4}}OH$ (having ${{K}_{b}}=1\times {{10}^{-5}}$) is equal to:
A. 10
B. 6
C. 11
D. 12

Answer 1

$N{{H}_{4}}OH$ is a strong base. As the value of ${{K}_{b}}$ is also given, try out to find the pOH of the solution. From that, we can easily find out the pH.

Complete step by step solution:
A strong base is a substance that dissociates, or splits itself into ions, very rapidly, when they are in their aqueous state. The term ${{K}_{b}}$ is called the base dissociation constant, and is used to determine how completely the dissociation of a particular basic substance takes place in water, or a similar aqueous solution. $N{{H}_{4}}OH$ being a strong base, dissociates into the $N{{H}_{4}}^{+}$ and $O{{H}^{-}}$ ions. There is another term called degree of dissociation, or α that tells us how much amount has been broken down at a particular time.
Now,
$\begin{aligned} & N{{H}_{4}}OH\rightleftharpoons N{{H}_{4}}^{+}+O{{H}^{-}} \\\ & \,\,\,\,\,\,\,c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\\ \end{aligned}$ , where c is the concentration
This is when the time t=0, as no products are formed, so they have 0 concentration.
After some time t, we have
$\begin{aligned} & N{{H}_{4}}OH\rightleftharpoons N{{H}_{4}}^{+}+O{{H}^{-}} \\\ & \,\,\,\,c-c\alpha \,\,\,\,\,\,\,\,\,\,\,c\alpha \,\,\,\,\,\,\,\,\,\,\,\,c\alpha \\\ \end{aligned}$ ……………( 1 ) , where $\alpha$ is degree of dissociation
So, assuming after time t, product formation is $c\alpha$ (one mole for each product), we get the equation of the base dissociation constant, ${{K}_{b}}$ as:

${{K}_{b}}=\dfrac{{{c}^{2}}{{\alpha }^{2}}}{c(1-\alpha )}$ , by definition
So, we get
${{K}_{b}}=\dfrac{c{{\alpha }^{2}}}{1-\alpha }$
Assuming the value of $\alpha$ to be very small i.e negligible, we get $1-\alpha$ almost equal to $\alpha$. So,
$\begin{aligned} & {{K}_{b}}=c{{\alpha }^{2}} \\\ & or,\,\,\alpha =\sqrt{\dfrac{{{K}_{b}}}{c}} \\\ \end{aligned}$

From (1), we get the concentration of OH:
$\begin{aligned} & [O{{H}^{-}}]=c\alpha \\\ & =c\times \sqrt{\dfrac{{{K}_{b}}}{c}} \\\ & =\sqrt{{{K}_{b}}\times c} \\\ \end{aligned}$
Now by putting values of ${{K}_{b}}$ and c from the question, we have
$[O{{H}^{-}}]=\sqrt{1\times {{10}^{-5}}\times 0.1}={{10}^{-3}}$
Also, the dissociation constant for water is ${{K}_{W}}$, which is
$\begin{aligned} & {{K}_{w}}=[{{H}^{+}}][O{{H}^{-}}] \\\ & ={{10}^{-14}}=[{{H}^{+}}][{{10}^{-3}}] \\\ & =[{{H}^{+}}]={{10}^{-11}} \\\ \end{aligned}$ , as value of ${{K}_{w}}$ is ${{10}^{-14}}$ (remember)
Also, we know,
$\begin{aligned} & pH=-\log [{{H}^{+}}] \\\ & As[{{H}^{+}}]={{10}^{-11}}, \\\ & pH=-\log [{{10}^{-11}}] \\\ & pH=11 \\\ \end{aligned}$

So, the pH of the solution is 11, which gives option c as the answer.

Note: The dissociation constants can be changed by application of temperature and also in the presence of a catalyst. The smaller the dissociation constant, the less likely it is to be split into ions, which means there is a strong intermolecular force that holds the atoms together.