91Ó°ÊÓ

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

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 ${\text{NH}}_3$:

role="math" localid="1663382310147" $0.0655\mathrm{l}×0.234 \mathrm{M}=0.0153\text{mol}$.

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:

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

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:

role="math" localid="1663382523820" $\text{pH=-log}\left[{\mathrm{H}}_3{\mathrm{O}}^{\mathrm{H}}\right]$

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:

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

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

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

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

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

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 ${\text{CH}}_3{\text{NH}}_2$

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

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

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

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

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

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

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

Therefore, the pH is$5.8$.