pH of CH3CH2NH2 buffer solution
Part A
-   Calculating  the initial $\pu{pH}$(before adding $\pu{0.02 mol}$ of $\ce{HCl})$

The equilibrium here is 
$$\ce{CH3CH2NH2 + H2O &lt;=&gt; CH3CH2NH3+  + OH−}$$
For  $\ce{ CH3CH2NH2} , \mathrm{K_b} = 4.6\times{10^{−4} }$

$$\mathrm{K_b} = 4.6\times{10^{−4} }=\frac{[\ce{CH3CH2NH3+}][\ce{OH−}]}{ [\ce{CH3CH2NH2}]}$$
The nominal concentrations of the base and conjugate acid  are respectively $$\mathrm{C_b} =\pu{0.170 M}~\text{ and } \mathrm{ C_a } =\pu{0.150 M}$$.

The mass balance expressions are:
$$ [\ce{CH3CH2NH3+} ]+ [\ce{CH3CH2NH2}] = \mathrm{C_a} + \mathrm{C_b} = \pu{0.32M}~\text{ and}~ [\ce{Cl−}] = \mathrm{C_a}=\pu{0.150M}$$
and the charge balance is given by:
$$[\ce{Cl−}] + [\ce{OH−}] = [\ce{H+}] + [\ce{CH3CH2NH3+}] \approx{[\ce{ CH3CH2NH3+}]}$$
in which the approximation shown above is justified  that the solution will be weakly alkaline.

The equilibrium concentrations of the conjugate species are then:
$$[\ce{CH3CH2NH3+}] = 0.150 + [\ce{OH−}]$$ 

\begin{array}{  } 
[\ce{CH3CH2NH2}] &amp;= &amp;(\mathrm{C_a} + \mathrm{C_b}) &amp;-&amp;([\ce{ CH3CH2NH3+} ])\\ &amp;= &amp;(0.32) &amp;–&amp;([\ce{ CH3CH2NH3+} ])\\ &amp;= &amp;0.320 &amp;- &amp;(0.150 + [\ce{OH−}])\\&amp;=&amp;0.170 &amp;- &amp;[\ce{OH−}]\end{array}
Because $\mathrm{C_a}$ and $\mathrm{C_b}$ are large compared to $[\ce{OH−}]$ (the solution is not expected to be strongly alkaline),
the $[\ce{OH−}]$ terms in the above expressions can be dropped and the equilibrium expression becomes
$$\mathrm{K_b}=5.6\times{10^{-4}} =\frac{ (0.150)[\ce{OH−}]}{(0.170)}$$
from which we find $[\ce{OH−}] = \pu{6.347\times{10^{-4}} M}$ , $\pu{pOH} = 3.2$ and $\pu{pH} = 14- 3.2 = 10.8$

Part B : Calculating  the final $ \pu{pH}$(after adding $\pu{0.020 mol}$ of $\ce{ HCl}$)


Calculate the initial amount of $\ce{ CH3CH2NH3+}$ and the initial amount of $\ce{CH3CH2NH2}$:
$$\begin{align}n_\ce{ CH3CH2NH3+} &amp;=n_\ce{ CH3CH2NH3Cl}\\ &amp;=V_\text{(solution)}\times{[\ce{ CH3CH2NH3Cl}]_\text{I}}\\ &amp;=
\frac{\pu{550.0 ml}}{\pu{1000 ml/L}}\times{\pu{0.150 mol/L}}\\
&amp;=\pu{0.083 mol}
 \end{align}$$
$$\begin{align}n_\ce{ CH3CH2NH2} &amp;=V_\text{(solution)}\times{[\ce{ CH3CH2NH2}]_\mathrm{I}}\\ &amp;=
\frac{\pu{550 ml}}{ \pu{1000 ml/L}}\times{\pu{0.170 mol/L}}\\
&amp;=\pu{0.094 mol}
 \end{align}$$
Addition of the strong acid $\pu{ 0.020 mol}$ of $\ce{ HCl}$ , consumes this same amount of $\ce{ CH3CH2NH2}$ and produces $\pu{ 0.020 mol}$ of $\ce{ CH3CH2NH3+}$ ion as explained in the following table:


\begin{array}{c | c c c c c c c}
\text{rxn}&amp;\ce{ CH3CH2NH2} &amp;+ &amp;\ce{HCl} &amp;-&gt; &amp;\ce{CH3CH2NH3+} &amp;+ &amp;\ce{Cl-}\\\hline
 \text{I}&amp;(0.094)&amp;  &amp;(0.020)&amp;  &amp;(0.083)&amp; &amp;(0.083)\\ 
\text{C} &amp;(0.094-0.020)&amp; &amp;(0.020-0.020)&amp;   &amp;(0.083+ 0.020)&amp; 
   &amp;(0.083+ 0.020)  \\
\text{F} &amp;(\pu{0.074 mol})&amp;  &amp;(\pu{0.000 mol})&amp;  &amp;(\pu{0.103 mol})&amp;  &amp;(\pu{0.103 mol})    
 \end{array}


We can continue using the approximations for the concentrations of the conjugate species
that we developed in the preceding part A.
The values of C$_\mathrm{b }$ and C$_\mathrm{a }$ are now


$$\mathrm{C}_\mathrm{b }= [\ce{CH3CH2NH2}]_\mathrm{I} =\frac{n_\ce{CH3CH2NH2}}{V}=\frac{\pu{0.074mol}}{\pu{0.550 L}}=\pu{0.133 M}\approx{[\ce{CH3CH2NH2}]_\mathrm{e}}$$
and
$$\mathrm{C}_\mathrm{a }= [\ce{CH3CH2NH3+}]_\text{I} =\frac{n_\ce{CH3CH2NH3+}}{V}=\frac{\pu{0.103 mol}}{\pu{0.55 L}}=\pu{0.187 M}\approx{[\ce{CH3CH2NH3+}]_\mathrm{e}}$$


Substituting these into the equilibrium constant expression yields
$$\begin{align}[\ce{OH−}] &amp;=(\mathrm{K_b})\frac{[\ce{CH3CH2NH2}]} {[\ce{CH3CH2NH3+} ]}\\
&amp;=(5.6\times{10^{-4}})\times{\frac{\pu{0.133 M}} {\pu{0.187M}}}\\&amp;=\pu{3.983\times{10^{-4}} M }
\end{align}$$
the new $\pu{pOH}= 3.40$ ,and the new $\pu{pH}=14- 3.40=10.60$, so addition of the acid has changed the $\pu{pH}$, a decrease of only $0.20~~\pu{pH}$ unit.