91Ó°ÊÓ

Solution'spHvalue, which measures the concentration of hydrogen ions, reveals whether it is acidic or alkaline.

a)

First, we can calculate mols of the ${\text{C}}_5{\text{H}}_5\text{N}$ :

$2.65\text{l}×0.0750\text{M}=0.19875\text{mol}$

$$\begin{gathered} 1:1=0.19875\text{mol}:x \\ x=0.19875\text{molHNO} \end{gathered}$$

Now, we can calculate the number of role="math" localid="1663666602420" $\text{ml}$of${\text{HNO}}_3$ required to reach the equivalence point:

$$\begin{gathered} V=\frac{n}{c} \\ V=\frac{0.19875\text{mol}}{0.447\text{M}} \\ =444.6{\text{mlHNO}}_3 \end{gathered}$$

Now we have to calculate the pH value at the equivalence point:

$\mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{†}\right]$

At the equivalence point all of the ${\text{C}}_5{\text{H}}_5\text{N}$has transformed into ${\text{C}}_5{\text{H}}_5\text{N}$, its weak conjugate acid, so we can apply calculation for acid base and Ka:

$$\begin{gathered} \mathrm{Kb}\left({\mathrm{C}}_5{\mathrm{H}}_5\mathrm{N}\right)=1.7×{10}^9 \\ \mathrm{Kw}=1×{10}^{-14} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\sqrt{\frac{Kw}{Kb}×\frac{2.65\text{l}×0.0750\text{M}}{2.65\text{l}+0.4446\text{l}}} \\ \mathrm{pH}=3.21 \end{gathered}$$

Therefore, the required pH is 3.21.

b)

First, we can calculate mols of the ${\text{NH}}_2{\text{CH}}_2{\text{CH}}_2{\text{NH}}_2$:

$0.188\text{l}×0.250\text{M}=0.047\text{mol}$

Because these two substances react in molar ratio 1: 1, we can use the number of mols of ${\text{NH}}_2{\text{CH}}_2{\text{CH}}_2\text{N}{{\text{H}}_2}^2\text{Co}$

Calculate number of mols of ${\text{HNO}}_3$

$$\begin{gathered} 1:1=0.047\text{mol}:x \\ x=0.047\text{molHNO} \end{gathered}$$

Now, we can calculate number of ml of${\text{HNO}}_3$required to reach equivalence point:

$$\begin{gathered} V=\frac{n}{c} \\ V=\frac{0.047\text{mol}}{0.447\text{M}} \\ =105.1{\text{mlHNO}}_3 \end{gathered}$$

Now we have to calculate pH value of the first equivalence point:

$$\begin{gathered} \mathrm{Kb}1=8.5×{10}^5 \\ \mathrm{Kb}2=7.1×{10}^8 \\ \mathrm{Ka}1=\frac{\mathrm{Kw}}{\mathrm{Kb}1} \\ =1.17×{10}^{10} \\ \mathrm{Ka}2=\frac{\mathrm{Kw}}{\mathrm{Kb}2} \\ =1.4×{10}^7 \end{gathered}$$

pH calculation for amphoteric substances$\left({\text{NH}}_2{\text{CH}}_2{\text{CH}}_2{\text{NH}}_3\right)$:

$$\begin{gathered} \mathrm{pH}=\frac{1}{2}(\mathrm{pKa}1+\mathrm{pKa}2) \\ \mathrm{pH}=8.39 \end{gathered}$$

Second equivalence point:

In the second equivalence point the number of ml of ${\text{HNO}}_3$required is double the value of ml in the first equivalence point because number of mols of ethylenediamine at the first equivalence point is the same as the number of mols of ${\text{NH}}_2{\text{CH}}_2{\text{CH}}_2{\text{NH}}_3$at the second equivalence point and we need exact same amount of ${\text{HNO}}_3$.

$$\begin{gathered} \text{V}=105.1\text{ml}+105.1\text{ml} \\ =210.2{\text{mlHNO}}_3 \end{gathered}$$

Now we have to calculate pH value at the equivalence point:

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}_3{\text{O}}^{+}\right] \\ \mathrm{pH}=-\log \frac{\mathrm{Kw}}{\left[\text{OH}\right]} \end{gathered}$$

At the equivalence point all of the ${\text{NH}}_2{\text{CH}}_2{\text{CH}}_2{\text{NH}}_3$ has transformed into ${\text{NH}}_3{\text{CH}}_2{\text{CH}}_2{\text{NH}}_3$, its weak conjugate acid, so we can apply calculation for weak acid and Ka:

$$\begin{gathered} \mathrm{Kb}2=7.1×{10}^8 \\ \mathrm{Kw}=1×{10}^{14}\left[{\mathrm{H}}_3{\mathrm{O}}^{⊤}\right] \\ =\sqrt{\frac{\mathrm{Kw}}{\mathrm{Kb}2}×\frac{0.188\mathrm{l}×0.250\mathrm{M}}{0.0188\mathrm{l}+0.2102\mathrm{l}}} \\ \mathrm{pH}=3.89 \end{gathered}$$

Therefore, the required values are:

$$\begin{gathered} {\text{V}}_1=105.1\text{ml.} \\ {\text{pH}}_1=8.39. \\ {\text{V}}_2=210.2\text{ml.} \\ {\text{pH}}_2=3.89. \end{gathered}$$