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The negative logarithm of such H+ ion concentration is referred to as pH. In conclusion, hydrogen power is recognized as the name's meaning.

$\mathbf{pH}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{log}\mathbf{}\mathbf{}\mathbf{[}\mathbf{H}\mathbf{+}\mathbf{]}$

The value of a chemical reaction's reaction quotient at chemical equilibrium, which is the condition that a dynamic chemical system reaches after enough time has passed and in which its composition has no detectable tendency to change further, is known as the equilibrium constant.

${\mathbf{K}}_{\mathbf{c}}\mathbf{=}{\mathbf{k}}^{\mathbf{!}}\mathbf{/}\mathbf{Γ}\mathbf{=}\frac{{\left[R\right]}^{\mathbf{ÒÏ}}{\left[S\right]}^{\mathbf{ÒÏ}}\mathbf{.}\mathbf{.}\mathbf{.}}{{\left[A\right]}^{\mathbf{Î\pm }}{\left[B\right]}^{\mathbf{β}}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}}$

In this task we have a titration of a weak base (benzylamine) with a strong acid (HCl).

To make the calculations easier we can find the equivalent point.

That is the volume of acid needed to neutralize the base:

$$\begin{gathered} \mathrm{V}\left(\mathrm{HCl}\right)=\frac{\mathrm{c}(\mathrm{B})â‹\dots \mathrm{V}(\mathrm{B})}{\mathrm{c}(\mathrm{HCl})} \\ \mathrm{V}\left(\mathrm{HCl}\right)=\frac{50.0\mathrm{mL}â‹\dots 0.0319\mathrm{M}}{0.050\mathrm{M}} \\ \mathrm{V}\left(\mathrm{HCl}\right)=31.9\mathrm{mL} \end{gathered}$$

Now we can determine four regions of the titration:

1. Before any acid is added: 0mL

2. Before the equivalence point: $12\mathrm{mL},15.95\mathrm{mL}(1/2{\mathrm{V}}_{\mathrm{e}})$and 30mL

3. At the equivalence point: 31.9mL

4. After the equivalence point: 35.0mL

Before any acid is added, at 0mL we have a weak base solution and the equilibrium equation is:

$$\begin{gathered} \mathrm{B}+{\mathrm{H}}_2\mathrm{O}⇌{\mathrm{BH}}^{+}+{\mathrm{OH}}^{-} \\ \left[\mathrm{B}\right]=0.0319-\mathrm{x} \\ \left[\mathrm{BH}\right]=\mathrm{x} \\ \left[{\mathrm{OH}}^{-}\right]=\mathrm{x} \end{gathered}$$

Now we can insert the unknowns in the equilibrium equation of the weak base:

${\mathrm{K}}_{\mathrm{b}}=\frac{\left[\mathrm{BH}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[\mathrm{B}\right]}=\frac{{\mathrm{x}}^2}{0.0319-\mathrm{x}}$

The value can be calculated by using the appendix G

${\mathrm{K}}_{\mathrm{b}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}}=\frac{{10}^{-14}}{4.5×{10}^{-10}}=2.22×{10}^{-5}$

By rearranging the equation we get a quadratic equation:

${\mathrm{x}}^2+2.22×{10}^{-5}-\mathrm{x}-7.08×{10}^{-7}=0$

By solving the quadratic equation we get the pH at:0 ml

$$\begin{gathered} {\mathrm{x}}^2+2.22×{10}^{-5}\mathrm{x}-7.08×{10}^{-7}=0 \\ \mathrm{pOH}=3.08 \\ \mathrm{pH}=14-3.0 \\ =10.92 \end{gathered}$$

Since the volume of needed to reach the equivalence point is 31.9mL, when we add 12.0mL of it, the remaining volume of B is 19.9 mL and the volume of ${\mathrm{BH}}^{+}$12 mL since it's equivalent to the volume of the added acid.

Now we calculate the pH before the equivalence point by using the Henderson-Hasselbalch equation:

At 12 mL:

$$\begin{gathered} \mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[{\mathrm{BH}}^{+}\right]}{\left[\mathrm{B}\right]} \\ =4.65+\log \frac{12/31.9)}{19.9/31.9} \\ =4.43 \\ \mathrm{pH}=14-4.43 \\ =9.57 \end{gathered}$$

At 15.95mL the volume of HCl added and volume of B consumed are equal so the pOH in that point is equal to the$p{K}_b$ value of the B:

$$\begin{gathered} \mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}=4.65 \\ \mathrm{pH}=14-5.00=9.35 \end{gathered}$$

At 30 mL the equations are equal to the ones in step 4 . We can calculate pOH by using the weak base equilibrium:

$$\begin{gathered} \mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[{\mathrm{BH}}^{+}\right]}{\left[\mathrm{B}\right]} \\ =4.65+\log \frac{30/31.9}{1.9/31.9} \\ =5.85 \\ \mathrm{pH}=14-5.85=8.15 \end{gathered}$$

Now we have to calculate pH at the equivalence point (31.9mL). First we need to write the reaction of weak acid with water

${\mathrm{BH}}^{+}+{\mathrm{H}}_2\mathrm{O}⇌\mathrm{B}+{\mathrm{H}}^{+}$

The concentrations are

$$\begin{gathered} \left[B{H}^{+}\right]=F^{\text{'}}-x \\ \left[B\right]=x \\ \left[H^{+}\right]=x \end{gathered}$$

The only thing we need to do before putting the concentrations in the weak acid equation is to calculate the formal concentration,$F\text{'}$

$$\begin{gathered} {\mathrm{F}}^{\text{'}}=\mathrm{concentrationofinitialB}\frac{\mathrm{initialvolumeofB}}{\mathrm{total}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{solution}} \\ {\mathrm{F}}^{\text{'}}=0.0319\mathrm{M}â‹\dots \frac{50.0\mathrm{mL}}{81.9\mathrm{mL}} \\ {\mathrm{F}}^{\text{'}}=0.0195\mathrm{M} \end{gathered}$$

Also, we can calculate ${\mathrm{K}}_{\mathrm{a}}$from the following equation:

${\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{b}}}=\frac{{10}^{-14}}{2.22×{10}^{-5}}=4.5×{10}^{-10}$

Now we can insert it in the equation for the equilibrium of the weak acid:

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[\mathrm{B}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{BH}}^{+}\right]}=\frac{{\mathrm{x}}^2}{0.0195-\mathrm{x}}$

By rearranging the equation we get a quadratic equation:

${\mathrm{x}}^2+4.5×{10}^{-10}\mathrm{x}-8.775×{10}^{-12}=0$

By solving the quadratic equation we get the ph at 31.9mL:

$$\begin{gathered} \mathrm{x}=\left[{\mathrm{H}}^{+}\right]=2.96×{10}^{-6}\mathrm{M} \\ \mathrm{pH}=5.53 \end{gathered}$$

Therefore the pH value at $\mathrm{o}\mathrm{mL},12\mathrm{mL}15.95\mathrm{mL},30\mathrm{mL},31.9\mathrm{mL}\mathrm{and}35\mathrm{mL}$are $10.92,9.57,9.35,8.15,5.53\mathrm{and}2.74$respectively.