Question

Use activity coefficients to calculate thepHafter10.0mLof0.100Mtrimethylammonium bromide were titrated with4.0mLof0.100MNaOH.

Short answer

The pH after titrating 10.0mL of 0.100M trimethylammonium bromide with of 0.100MNaOH is 9.72

Step-by-step solution

Calculate the PH value using the Henderson-Hasselbalch equation.

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

$\mathrm{pH}=-\log \left[\mathrm{H}+\right]$

• A simple buffer solution consists of an acid and a conjugate base salt.
• A buffer solution's greatest distinguishing characteristic is its capacity to withstand pH fluctuations even when a little quantity of acid or base is added to it.

$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[\mathrm{B}\right]{\mathrm{\gamma }}_{\mathrm{B}}}{\left[{\mathrm{BH}}^{+}\right]{\mathrm{\gamma }}_{\mathrm{BH}}}$

Calculate the initial number of moles of reactants.

In this task, we have a titration of a weak acid with a strong base. First, we are going to write the chemical equation for the reaction :

${\left({\mathrm{CH}}_3\right)}_3{\mathrm{NH}}^{+}+\mathrm{NaOH}\rightleftharpoons {\left({\mathrm{CH}}_3\right)}_3\mathrm{N}+{\mathrm{H}}_2\mathrm{O}$

Now we have to calculate the initial number of moles of reactants:

$$\begin{gathered} \mathrm{n}\left({\left({\mathrm{CH}}_3\right)}_3{\mathrm{NH}}^{+}\right)=10.0\mathrm{mL}·0.100\mathrm{M}=1\mathrm{mmol} \\ \mathrm{n}\left(\mathrm{NaOH}\right)=4.0\mathrm{mL}·0.1\mathrm{M}=0.4\mathrm{mmol} \end{gathered}$$

After the base was added we have the weak acid that didn't react with the base and the conjugate base of the acid so the final number of moles are:

$$\begin{gathered} \mathrm{n}\left({\left({\mathrm{CH}}_3\right)}_3{\mathrm{NH}}^{+}\right)=1\mathrm{mmol}-0.4\mathrm{mmol}=0.6\mathrm{mmol} \\ \mathrm{n}\left({\mathrm{CH}}_3\mathrm{N}\right)=4.0\mathrm{mL}·0.1\mathrm{M}=0.4\mathrm{mmol} \end{gathered}$$

Calculating the concentrations of (CH3)3NH+, NaOHandBr- 

Now we need to calculate the concentrations of the${\left({\mathrm{CH}}_3\right)}_3{\mathrm{NH}}^{+},{\mathrm{NaOHandBr}}^{-}$ by using the total volume of the solution

$$\begin{gathered} \left[{\left({\mathrm{CH}}_3\right)}_3{\mathrm{NH}}^{+}\right]=\frac{0.6\mathrm{mmol}}{14\mathrm{mL}}=0.0429\mathrm{M} \\ \left[\mathrm{NaOH}\right]=\frac{0.4\mathrm{mmol}}{14\mathrm{mL}}=0.0286\mathrm{M} \\ {\left[\mathrm{Br}\right]}^{-}=\frac{1\mathrm{mmol}}{14\mathrm{mL}}=0.0714\mathrm{M} \end{gathered}$$

Calculating the ionic strength.

The ionic strength is given as

$\mathrm{\mu }=\frac{1}{2}\underset{}{\sum {\mathrm{c}}_{\mathrm{i}}{\mathrm{z}}_{\mathrm{i}}=\frac{1}{2}\left(0.0429+0.0714\right)}=0.071\mathrm{M}$

Using the Henderson-Hasselbalch equation.

To calculate pH, we need value of trimethylamine that can be found in appendix G.

We use Henderson-Hasselbalch equation

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[\mathrm{B}\right]{\mathrm{\gamma }}_{\mathrm{B}}}{\left[{\mathrm{BH}}^{+}\right]{\mathrm{\gamma }}_{\mathrm{BH}}} \\ \mathrm{pH}=9.799+\log \frac{0.0286.1.00}{0.0429.0.80} \\ \mathrm{PH}=9.72 \end{gathered}$$

Therefore the pH after titrating $10.0\mathrm{mL}$of $0.100\mathrm{M}$trimethylammonium bromide with $4.0\mathrm{mL}$of $0.100\mathrm{MNaOH}$is$9.72$