Q19.60P - Chemistry: Molecular Nature Of Matter And Change

Question

Find the pH of the equivalence point(s) and the volume (mL) of  needed to reach it in titrations of

(a) $\mathbf{65}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{\text{mL}}$ of $\mathbf{0}\mathbf{.}\mathbf{234}\mathbf{ }\mathbf{\text{M}}\mathbf{ }\mathbf{ }{\mathbf{\text{NH}}}_{\mathbf{3}}$

(b) $\mathbf{21}\mathbf{.}\mathbf{8}\mathbf{ }\mathbf{\text{mL}}$ of $\mathbf{1}\mathbf{.}\mathbf{11}\mathbf{ }\mathbf{\text{M}}\mathbf{ }\mathbf{ }{\mathbf{\text{CH}}}_{\mathbf{3}}{\mathbf{\text{NH}}}_{\mathbf{2}}$.

Step-by-Step Solution

Answer

a) The equivalence point(s) and volume of the given problem are:

 $$\begin{gathered} \text{ V}=122.6\text{ml} \text{HCl} \\ \text{pH}=5.17 \end{gathered}$$

b) The equivalence point(s) and volume of the given problem are:

 $$\begin{gathered} \text{V}=193.6\text{ml} \text{HCl} \\ \text{pH}=5.8 \end{gathered}$$

Step 1: Definition of pH

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

Step 2: Apply calculation for acids base and Ka

a)

We have two weak base-strong acid titrations in this case.

$0.125\text{M} \text{HCl}$ and $65.5\text{ml} 0.234 \text{M} {\text{NH}}_3$.

First, we can calculate mols of the ${\text{NH}}_3$:

  $0.0655 \mathrm{l}\times 0.234 \mathrm{M}=0.0153 \text{mol}$.

Because these two chemicals react in a 1:1 molar ratio, we can compute the number of mols of HCl by multiplying the number of mols of ${\text{NH}}_3$ by the number of mols of HCl:

$$\begin{gathered} 1:1=0.0153mol:x \\ x=0.0153 \text{mol HCl} \end{gathered}$$

 

Now we can figure out how many ml of HCI we'll need to accomplish equivalence:

 $$\begin{gathered} V=\frac{n}{c} \\ V=\frac{0.0153 \text{mol}}{0.125 \mathrm{M}} \\ =122.4 \text{ml } \text{HCl.} \end{gathered}$$

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

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

At the equivalence point all of the  has transformed into ${\text{NH}}_4$, its weak conjugate acid, so we can apply calculation for acid base and Ka:

 $$\begin{gathered} \mathrm{Kb}\left({\text{NH}}_3\right)=1.8\times {10}^{-5} \\ \mathrm{Kw}=1\times {10}^{-14}\left[H_3{O}^{+}\right] \\ =\sqrt{\frac{\mathrm{Kw}}{\mathrm{Kb}}\times \frac{0.0655 \mathrm{l}\times 0.234 \mathrm{M}}{0.0655 \mathrm{l}+0.1224 \mathrm{l}}} \\ \mathrm{pH}=5.17. \end{gathered}$$

Therefore, the required  $\mathrm{pH}=5.17$.

Step 2: Apply calculation for acids base and Ka

b) $21.8 \text{ml} 1.11\text{M} {\text{CHCH}}_3{\text{NH}}_2$ .

First, we can calculate mols of the ${\text{CH}}_3{\text{NH}}_2$:

$0.0218\text{ l}\times 1.11\text{M}=0.0242 \text{mol}$ .

Because these two chemicals react in a 1:1 molar ratio, we can determine the amount of mols of HCl using the number of mols of  ${\text{CH}}_3{\text{NH}}_2$

 $$\begin{gathered} 1:1=0.0242 \text{mol}:x \\ x=0.0242 \text{mol HCl} \end{gathered}$$

Now we can figure out how many ml of HCL  we'll need to accomplish equivalence:

$$\begin{gathered} \mathrm{V}=\frac{n}{c} \\ \mathrm{V}=\frac{0.0242 \text{mol}}{0.125\text{ M}} \\ =193.6\text{ml HCl} \end{gathered}$$

 

We must now determine the pH value at the equivalence point:

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

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

$$\begin{gathered} \mathrm{Kb}\left({\text{CH}}_3{\text{NH}}_2\right)=4.38\times {10}^4 \\ \mathrm{Kw}=1\times {10}^{14} \\ \left[{\text{H}}_3\text{O}\right]=\sqrt{\frac{\mathrm{Kw}}{\mathrm{Kb}}\times \frac{0.0218 \mathrm{l}\times 1.11 \mathrm{M}}{0.0218 \mathrm{l}+0.1936 \mathrm{l}}} \\ \mathrm{pH}=5.8. \end{gathered}$$

 

Therefore, the pH is $5.8$.