# pH of CH3CH2NH2 buffer solution [closed]

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 $$\ce{HCl}$$.

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

## closed as off-topic by Tyberius, A.K., MaxW, Todd Minehardt, Jon CusterMar 31 at 17:51

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• Hi! This site has a policy that homework-like questions have to show some effort to solve them. Hence, could you please edit your post and add, how you would solve it and what exactly is your difficulty here? Thank you! – Arsak Mar 30 at 9:38
• Hi! Thank you for that info. I edited my question and added my attempt, but the values that I calculated turns out to be wrong. – toffee Mar 30 at 17:38
• "$0.0825 - 0.02 = \pu{0.0625 mol} \ce{CH3CH2NH3Cl}$",$$0.0825 + 0.02 = \pu{0.1025 mol} \ce{CH3CH2NH3Cl}$$ – Adnan AL-Amleh Mar 31 at 3:03
• "$0.0935 + 0.02 = \pu{0.1135 mol} \ce{CH3CH2NH2}$" $$0.0935 - 0.02 = \pu{0.0735 mol} \ce{CH3CH2NH2}$$ – Adnan AL-Amleh Mar 31 at 3:06

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.