pH of CH3CH2NH2 buffer solution [closed]








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> 
> For $\pu{550.0 mL}$ of a buffer solution that is $\pu{0.170 M}$ in $\ce{CH3CH2NH2}$ and $\pu{0.150 M}$ in $\ce{CH3CH2NH3Cl}$, calculate the initial pH and the final pH after adding $\pu{0.020 mol}$ of $\ce{HCl}$.
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The following is my attempt on solving this problem:


Since Kb for ethylalamine = $5.6 \times 10^{-4}$.


pKb = $-\log( 5.6 \times 10^{-4}) = 3.25$


pOH = $3.25 + \log (0.150/0.170)$


pOH = $3.196$


pH = $14 - 3.196 = 10.80$ for Initial


Final pH:


$\pu{0.170 M} \* \pu{0.55 L}= 0.0935 mol$ $\ce{CH3CH2NH2}$


$\pu{0.150 M} \* \pu{0.55 L} = \pu{0.0825 mol} \ce{CH3CH2NH3Cl}$,


$0.0825 - 0.02 = \pu{0.0625 mol} \ce{CH3CH2NH3Cl}$,


$0.0935 + 0.02 = \pu{0.1135 mol} \ce{CH3CH2NH2}$


pOH = $3.25 + \log(0.0625/0.1135)$


pOH = $2.991$


pH = $14 - 2.991 = 11.01$ for final


However, these values turn out to be wrong. Please help, thank you!




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


The equilibrium here is 
$$\ce{CH3CH2NH2 + H2O <=> CH3CH2NH3+ + OH−}$$
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}] &= &(\mathrm{C\_a} + \mathrm{C\_b}) &-&([\ce{ CH3CH2NH3+} ])\\ &= &(0.32) &–&([\ce{ CH3CH2NH3+} ])\\ &= &0.320 &- &(0.150 + [\ce{OH−}])\\&=&0.170 &- &[\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 $\ce{CH3CH2NH2}$:
$$\begin{align}n\_\ce{ CH3CH2NH3+} &=n\_\ce{ CH3CH2NH3Cl}\\ &=V\_\text{(solution)}\times{[\ce{ CH3CH2NH3Cl}]\_\text{I}}\\ &=
\frac{\pu{550.0 ml}}{\pu{1000 ml/L}}\times{\pu{0.150 mol/L}}\\
&=\pu{0.083 mol}
 \end{align}$$
$$\begin{align}n\_\ce{ CH3CH2NH2} &=V\_\text{(solution)}\times{[\ce{ CH3CH2NH2}]\_\mathrm{I}}\\ &=
\frac{\pu{550 ml}}{ \pu{1000 ml/L}}\times{\pu{0.170 mol/L}}\\
&=\pu{0.094 mol}
 \end{align}$$
* 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}&\ce{ CH3CH2NH2} &+ &\ce{HCl} &-> &\ce{CH3CH2NH3+} &+ &\ce{Cl-}\\\hline
 \text{I}&(0.094)& &(0.020)& &(0.083)& &(0.083)\\ 
\text{C} &(0.094-0.020)& &(0.020-0.020)& &(0.083+ 0.020)& 
 &(0.083+ 0.020) \\
\text{F} &(\pu{0.074 mol})& &(\pu{0.000 mol})& &(\pu{0.103 mol})& &(\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
$$\begin{align}[\ce{OH−}] &=(\mathrm{K\_b})\frac{[\ce{CH3CH2NH2}]} {[\ce{CH3CH2NH3+} ]}\\
&=(5.6\times{10^{-4}})\times{\frac{\pu{0.133 M}} {\pu{0.187M}}}\\&=\pu{3.983\times{10^{-4}} M }
\end{align}$$
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.


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