Q19.57P - Chemistry: Molecular Nature Of Matter And Change

Question

Find the pH during the titration of $\mathbf{\text{20.00 mL}}$ of $\mathbf{\text{0.1000M}}$triethylamine,  ${\mathbf{\left(}{\mathbf{\text{CH}}}_{\mathbf{\text{3}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}\mathbf{\right)}}_{\mathbf{\text{3}}}\mathbf{\text{ N}}\mathbf{\left(}{\mathbf{\text{K}}}_{\mathbf{\text{b}}}{\mathbf{\text{=5.2×10}}}^{\mathbf{\text{-4}}}\mathbf{\right)}$with  $\mathbf{\text{1000 MNaOH}}$ solution after the following additions of titrant:

(a)  $\mathbf{\text{0 mL}}$

(b)  $\mathbf{\text{10.00 mL}}$

(c)  $\mathbf{\text{15.00 mL}}$

(d)  $\mathbf{\text{19.00 mL}}$

(c)  $\mathbf{\text{19.95 mL}}$

(f)  $\mathbf{\text{20.00 mL}}$

(g)  $\mathbf{\text{20.05 mL}}$

(h)  $\mathbf{\text{25.00 mL}}$

Step-by-Step Solution

Answer

The required pH values during the tritration are:

$\begin{array}{l}\text{a) }11.84\\ \text{b) }10.72\\ \text{c) }10.24\\ \text{d) }9.44\\ \text{e) }8.12\\ \text{f) }6.01\\ \text{g) }3.9\\ \text{h) }1.95\end{array}$

Step 1: Concept introduction

We have a weak base that is titrated by a strong acid in this situation. To answer this problem, we must consider a weak base calculation at the outset, a calculation of the buffer solution using the Henderson-Hasselbalch equation, and a weak base calculation at the conclusion. $\mathbf{\text{HClKb}}\mathbf{=}\mathbf{5}\mathbf{.2}\mathbf{\backslash }\mathbf{times}\mathbf{10}\mathbf{\{}\mathbf{\wedge }\mathbf{\{}\mathbf{-}\mathbf{4}\mathbf{\left\}}\mathbf{\right\}}$  We only have triethylamine in the solution because no strong acid has been added, thus we must use the following equation to calculate the pH of a weak base:

Step 2: Calculation of the buffer solution

a.  Because there is only triethylamine in this solution and no strong acid, we must use the equation for a weak base for the pH calculation:

 $\begin{array}{l}\text{pOH = -log}\left[{\text{OH}}^{\text{-}}\right]\\ \text{ = -log}\sqrt{\text{Kb×[A]}}\\ \text{ = -log}\sqrt{{\text{5.2×10}}^{\text{-4}}\text{×0.1M}}\\ \text{ = 2.14.}\\ \\ \text{pH = 14-pOH}\\ \text{ = 11.86.}\end{array}$

 

Therefore, the pH value is  $\mathrm{11.84.}$

Step 3: Calculation of the solution

b) In this situation, we added $10\text{\hspace{0.17em}ml}$  of a strong acid, therefore, the weak base is greater than , and we can use the Henderson-Hasselbalch equation to compute the pH:

 

$\begin{array}{c}\text{pH = pKa+log}\frac{\text{nT rieth }}{\text{nHCl}}\\ \\ \text{pKa = -log}\frac{\text{Kw}}{\text{Kb}}\\ \text{ = 10.72.}\\ \\ \text{pH = 10.72+log}\frac{\text{0.02l×0.1M-0.01l×0.1M}}{\text{0.01l×0.1M}}\\ \text{ = 10.72.}\end{array}$

Therefore, the pH value is  $\text{10.72.}$

Step 4: Calculation of part (c)

c) We added $15\text{\hspace{0.17em}ml}$  of a strong acid, therefore, the weak base is greater than $5\text{\hspace{0.17em}ml,}$  and we can compute the pH using the Henderson-Hasselbalch equation:

$\begin{array}{c}\text{pH = pKa+log}\frac{\text{nTrieth}}{\text{nHCl}}\\ \\ \text{pKa = -log}\frac{\text{Kw}}{\text{Kb}}\\ \text{ =10.72.}\\ \\ \text{pH = 10.72+log}\frac{\text{0.02l×0.1M-0.015l×0.1M}}{\text{0.015l×0.1M}}\\ \text{ =10.24.}\end{array}$

Therefore, the pH value is  $\mathrm{10.24.}$

Step 5: Calculation of the part (d)

(d) We added  $19\text{\hspace{0.17em}ml}$ of a strong acid in this case, thus the weak base is greater than  and we can compute the pH using the Henderson-Hasselbalch equation:

 $\begin{array}{c}\mathrm{pH}=\mathrm{pKa}+\log \frac{\mathrm{n}\text{ Trieth }}{\mathrm{nHCl}}\\ \\ \mathrm{pKa}=-\log \frac{\mathrm{Kw}}{\mathrm{Kb}}\\ =\mathrm{10.72.}\\ \\ \mathrm{pH}=10.72+\log \frac{0.02\mathrm{l}\times 0.1\mathrm{M}-0.019\mathrm{l}\times 0.1\mathrm{M}}{0.019\mathrm{l}\times 0.1\mathrm{M}}\\ =\mathrm{9.44.}\end{array}$

Therefore, the pH value is  $\mathrm{9.44.}$

Step 6: Calculation of the part (e)

e) We added   $19.95\text{\hspace{0.17em}ml}$ of a strong acid in this case, thus the weak base is greater than  $0.05\text{\hspace{0.17em}ml,}$ and we can compute the pH using the Henderson-Hasselbalch equation:

$\begin{array}{c}\mathrm{pH}=\mathrm{pKa}+\log \frac{\mathrm{n}\text{ Trieth }}{\mathrm{nHCl}}\\ \\ \mathrm{pKa}=-\log \frac{\mathrm{Kw}}{\mathrm{Kb}}\\ =\mathrm{10.72.}\\ \\ \mathrm{pH}=10.72+\log \frac{0.02\mathrm{l}\times 0.1\mathrm{M}-0.01995\mathrm{l}\times 0.1\mathrm{M}}{0.01995\mathrm{l}\times 0.1\mathrm{M}}\\ =\mathrm{8.12.}\end{array}$

Therefore, the pH value is  $\mathrm{8.12.}$

Step 7: Calculation of the part (f)

f) In this case, we added  $20\text{\hspace{0.17em}ml}$ of a strong acid, and the equivalence point was attained because the molarities and the volumes were the same. At this time, all of the weak bases have been transformed to weak acids, and the solution contains only weak acids. We can solve this problem by computing the  

$\begin{array}{l}\text{TotalVvalue = }0.02\mathrm{l}+0.02\mathrm{l}\\ =0.04\mathrm{l}.\\ \\ \left[{\text{H}}_3{\text{O}}^{+}\right]=\sqrt{\mathrm{Ka}\times \frac{0.02\mathrm{l}\times 0.1\mathrm{M}}{0.04\mathrm{l}}}\\ =\sqrt{\frac{\mathrm{Kw}}{\mathrm{Kb}}\times \frac{0.02\mathrm{l}\times 0.1\mathrm{M}}{0.04\mathrm{l}}}\\ =9.797\times {10}^{-7}.\\ \\ \mathrm{pH}=-\log \left[{\text{H}}_3{\text{O}}^{+}\right]\\ =\mathrm{6.01.}\end{array}$
 

Therefore, the pH value is  $\mathrm{6.01.}$

Step 8: Calculation of the part (g)

g) We are past the equivalence point in this case, and we added $20.05\text{\hspace{0.17em}ml}$  of a strong acid, therefore, the strong acid is in excess, and we solve the problem by calculating the  $\left[{\text{H}}_3{\text{O}}^{+}\right]:$

$\begin{array}{c}\text{ Total\hspace{0.17em}value = 0.02l+0.02005l}\\ \text{ = 0.04005l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \text{ = }\frac{\text{0.02005l×0.1M-0.02l×0.1M}}{\text{0.04005l}}\\ \text{ = 0.000124\hspace{0.17em}M}\\ \\ \text{pH = -log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \text{ = 3.9.}\end{array}$

 

Therefore, the pH value is  $\mathrm{3.9.}$

Step 9: Calculation of the part (h)

h) We are past the equivalence point in this example, and we added  $25\text{\hspace{0.17em}ml}$ of a strong acid, therefore, the strong acid is in excess, and we solve the problem by calculating the  $\left[{\text{H}}_3{\text{O}}^{+}\right]:$

$\begin{array}{c}\text{Total\hspace{0.17em}value = 0.02l+0.025l}\\ \text{ = 0.045l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \text{ = }\frac{\text{0.025l×0.1M-0.02l×0.1M}}{\text{0.045l}}\\ \text{ = 0.0111\hspace{0.17em}\hspace{0.17em}M.}\\ \\ \text{pH = -log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \text{ =1.95.}\end{array}$

Therefore, the pH value is  $\mathrm{1.95.}$