Chapter 19, Problem 19.77QA

Interpretation Introduction

To find:

a) Sketch the titration curve for the titration of $125 mL$ of a $0.015 M$ solution of methylamine with $0.100 M HCl$.

b) Label the curve with the $pH$ of the analyte solution, the $pKa$ of the analyte, and with the $pH$ and titrant volumes halfway to the equivalence point and at the equivalence point.

c) Draw the structures of the species present in the solution at the equivalence point.

Answer to Problem 19.77QA

Solution:

a) and b) Labeled titration curve for the titration of $125 mL$ of a $0.015 M$ solution of methylamine with $0.100 M HCl$ is as follows:

c) At the equivalence point, the structure of the species present in the solution is

Explanation of Solution

1) Concept:

We can calculate moles of weak base present and the moles of $H^{+}$ ions added from their respective volumes and their molarities. As the $HCl$ is added to the methylamine solution, the following reaction occurs:

$C{H}_{3}N{H}_2\left(aq\right)+H^{+}\left(aq\right)\to C{H}_{3}N{H}_3^{+}\left(aq\right)$

We know how much of the weak base remains and how much of its conjugate acid produced because one mole of base reacts with one mole of $H^{+}$ ions to produce one mole of methylammonium ions. We can calculate $\left[C{H}_{3}N{H}_3^{+}\right]$ and then use the Henderson-Hasselbalch equation to determine the pH of the solution. The $p{K}_b$ for methylamine can be found in Appendix 5; it equals 3.36.

2) Formula:

i) $K_b= \frac{\left[H{B}^{+}\right]\left[O{H}^{-}\right]}{\left[B\right]}$

ii) $pH= -\mathrm{l}\mathrm{o}\mathrm{g}\left[H_3{O}^{+}\right]$

iii) $pOH= -\mathrm{l}\mathrm{o}\mathrm{g}\left[{OH}^{-}\right]$

iv) $pH+pOH=14$

v) $p{K}_a+p{K}_b=14$

vi) $pH=p{K}_a+\mathrm{l}\mathrm{o}\mathrm{g}\frac{\left[base\right]}{\left[acid\right]}$

vii) $x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

3) Given:

i) $\left[C{H}_{3}N{H}_2\right]=0.015 M$

ii) Volume of methylamine = $125 mL$

iii) $\left[HCl\right]=0.100 M$

iv) For methylamine, $p{K}_b=3.36$

v) For methylamine, $K_b=4.4 \times {10}^{-4}$

4) Calculation:

a) Calculation of initial $pH$ of methylamine solution.

Reaction	$C{H}_{3}N{H}_2\left(aq\right)+H_{2}O\left(aq\right)\to C{H}_{3}N{H}_3^{+}\left(aq\right)+O{H}^{-}\left(aq\right)^ale nion of initial pH of methylamine solution. determine the pH of the solution. The p mole of methylammonium ions. We can$
	$C{H}_{3}N{H}_2 \left(M\right)$	$C{H}_{3}N{H}_3^{+} \left(M\right)$	$O{H}^{-}\left(M\right)$
Initial	$0.015$	$0$	$0$
Change	$-x$	$+x$	$+x$
Final	$0.015-x$	$x$	$x$

Applying law of mass action,

$K_b= \frac{\left[C{H}_{3}N{H}_3^{+}\right]\left[O{H}^{-}\right]}{\left[C{H}_{3}N{H}_2\right]}$

$4.4 \times {10}^{-4}= \frac{\left(x\right)\left(x\right)}{(0.015-x)}$

$6.60\times {10}^{-6}-\left(4.4\times {10}^{-4}\right)x=x^2$

$x^2+ \left(4.4\times {10}^{-4}\right)x-\left(6.60\times {10}^{-6}\right)=0$

Solving this equation using quadratic equation formula:

$x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x= \frac{-\left(4.4\times {10}^{-4}\right) \pm \sqrt{{\left(4.4\times {10}^{-4}\right)}^2-4\left(1\right)(6.60\times {10}^{-6})}}{2\left(1\right)}$

$x= \frac{-\left(4.4\times {10}^{-4}\right)\pm \sqrt{2.66\times {10}^{-5}}}{2}$

$x= -0.002798$ and $x=0.002358$

The negative value of concentration has no significance. Therefore, we will proceed with $x=0.002358$.

So, $\left[O{H}^{-}\right]=x=0.002358$

$pOH= -\log \left[{OH}^{-}\right]= -\log \left(0.002358\right)=2.6$

$pH+pOH=14$

$pH=14-2.6=11.4$

Therefore, initial $pH$ of methylamine solution is $11.4$.

At half way to the equivalence point,$pH=p{K}_a$.

$p{K}_a+p{K}_b=14$

$p{K}_a=14-3.36=10.64$

Therefore, at half way to the equivalence point, $pH=p{K}_a=10.64$

At equivalence point, the moles of base is the same as moles of acid.

Calculation of moles of weak base methylamine:

$125 mL\times \frac{1L}{1000 mL}\times \frac{0.015 mol}{L}=0.001875 mol$

Calculation of moles of $HCl$:

$0.001875 mol C{H}_{3}N{H}_2\times \frac{1 mol HCl}{1 mol C{H}_{3}N{H}_2}=0.001875 mol HCl$

Calculation of volume of $HCl$ needed:

$0.001875 mol HCl \times \frac{1 L}{0.100 mol}=0.01875 L$

Reaction	$C{H}_{3}N{H}_2\left(aq\right)+H^{+}\left(aq\right)\to C{H}_{3}N{H}_3^{+}\left(aq\right)^ale nion of initial pH of methylamine solution. determine the pH of the solution. The p mole of methylammonium ions. We can$
	$C{H}_{3}N{H}_2 \left(moles\right)$	$H^{+} \left(moles\right)$	$C{H}_{3}N{H}_3^{+} \left(moles\right)$
Initial	$0.001875$	$0.001875$	$0$
Change	$-0.001875$	$-0.001875$	$+0.001875$
Final	$0$	$0$	$0.001875$

Therefore, at equivalence point, only $C{H}_{3}N{H}_3^{+}$ and $C{l}^{-}$ ions will be in the solution.

Calculation of total volume of solution at equivalence point:

Total volume = $0.125 L+0.01875L=0.14375 L$

Calculation of concentration of $C{H}_{3}N{H}_3^{+}$:

$\left[C{H}_{3}N{H}_3^{+}\right]=\frac{0.001875 mol}{0.14375 L}=0.0130 M$

Reaction	$C{H}_{3}N{H}_3^{+}\left(aq\right)+H_{2}O\left(aq\right)\to C{H}_{3}N{H}_2\left(aq\right)+H_3{O}^{+}\left(aq\right)^ale nion of initial pH of methylamine solution. determine the pH of the solution. The p mole of methylammonium ions. We can$
	$C{H}_{3}N{H}_3^{+} \left(M\right)$	$C{H}_{3}N{H}_2 \left(M\right)$	$H_3{O}^{+}\left(M\right)$
Initial	$0.013$	$0$	$0$
Change	$-x$	$+x$	$+x$
Final	$0.013-x$	$x$	$x$

Calculation of $K_a$ from $K_b$:

$p{K}_a=14-p{K}_b=14-3.36=10.64$

$K_a={10}^{-p{K}_a}={10}^{-10.64}$

$K_a=2.291\times {10}^{-11}$

Applying the law of mass action:

$K_a= \frac{\left[C{H}_{3}N{H}_2\right]\left[H_3{O}^{+}\right]}{\left[C{H}_{3}N{H}_3^{+}\right]}$

$2.291\times {10}^{-11}= \frac{\left(x\right)\left(x\right)}{(0.013-x)}$

We assume that the value of $x$ will be very small as compared to $0.013$. Therefore, ignoring $x$ from denominator and solving for $x$:

$x^2=2.978\times {10}^{-13}$

$x=5.457 \times {10}^{-7} M$

Therefore, $\left[H_3{O}^{+}\right]=x=5.457 \times {10}^{-7} M$

$pH=-\log \left[H_3{O}^{+}\right]= -\log \left(5.457 \times {10}^{-7}\right)$

$pH=6.26$

At equivalence point, the $pH$ of solution is $6.26$.

b) Label the curve with the $pH$ of the analyte solution:

c) At the equivalence point, the structure of the species present in the solution is

Conclusion:

The initial $pH$ of methylamine, at half way to equivalence point, and equivalence point are calculated, and graph is plotted.