When${}^{22.63\mathrm{ml}}$of aqueous${}^{\mathrm{NaOH}}$were added to${}^{1.214\mathrm{g}}$of cyclohexylaminoethanesulfonic acid${}^{(\mathrm{FM}207.29}$ structure in Table ${}^{9-2)}$dissolved in${}^{41.37\mathrm{ml}}$ of water, the pH was ${}^{9.24}$. Calculate the molarity of the${}^{\mathrm{NaOH}}$

The concentration of a chemical solution is measured by both molarity and molality. The major distinction between the two is one of mass vs. volume. The molality refers to the number of moles of a solute in relation to the mass of the solvent. The molarity refers to the number of moles of a solute in relation to the volume of a solution. It is given by the formula,

$\mathbf{M}\mathbf{}\mathbf{=}\mathbf{}\mathbf{moles}\mathbf{}\mathbf{solute}\mathbf{}\mathbf{/}\mathbf{}\mathbf{liters}\mathbf{}\mathbf{solution}\mathbf{}$.

Here we have the titration of a weak acid with strong base.

First we have to calculate the moles of acid added to the solution.

$\mathbf{n}\left(\mathrm{HA}\right)\mathbf{=}\frac{\mathbf{m}\left(\mathrm{HA}\right)}{\mathbf{M}\left(\mathrm{HA}\right)}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{214}\mathbf{g}}{\mathbf{207}\mathbf{.}\mathbf{29}\mathbf{mol}\mathbf{}\mathbf{/}\mathbf{}\mathbf{g}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{857}\mathbf{}\mathbf{mmol}$

Now write the equilibrium equation.

$\mathbf{HA}\mathbf{+}{\mathbf{OH}}^{\mathbf{-}}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}{\mathbf{A}}^{\mathbf{-}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

Initial number of moles are:

$$\begin{gathered} \left[\mathrm{HA}\right]\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{857}\mathbf{-}\mathbf{x} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left[{\mathrm{OH}}^{-}\right]\mathbf{=}\mathbf{x} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left[A^{-}\right]\mathbf{}\mathbf{=}\mathbf{x} \end{gathered}$$

Hence the number of moles at the end of the reaction is$$\begin{gathered} \mathbf{\left[}\mathbf{HA}\mathbf{\right]}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{857}\mathbf{-}\mathbf{x} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}\mathbf{=}\mathbf{x} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left[}{\mathbf{A}}^{\mathbf{-}}\mathbf{\right]}\mathbf{}\mathbf{=}\mathbf{x} \end{gathered}$$ .

Use Henderson-Hasselbalch equation to calculate the molarity of ${}^{\mathbf{NaOH}}$.

$\mathbf{pH}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{pk}}_{\mathbf{a}}\mathbf{+}\mathbf{log}\frac{\left[A^{-}\right]}{\left[\mathrm{HA}\right]}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{39}\mathbf{+}\mathbf{log}\frac{\mathbf{x}}{\mathbf{5}\mathbf{.}\mathbf{857}\mathbf{-}\mathbf{x}}$

Rearrange the equation and solve:

$$\begin{gathered} {\mathbf{10}}^{\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{15}}\mathbf{=}\frac{\mathbf{x}}{\mathbf{5}\mathbf{.}\mathbf{857}\mathbf{-}\mathbf{7}} \\ \mathbf{x}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{708}\mathbf{x}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{147} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{708}\mathbf{x}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{147} \\ \mathbf{x}\mathbf{=}\left[{\mathrm{OH}}^{-}\right]\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{428}\mathbf{}\mathbf{mmol} \end{gathered}$$

Use molarity equation:

$\mathbf{c}\left(\mathrm{NaOH}\right)\mathbf{}\mathbf{=}\frac{\mathbf{n}\left(\mathrm{nAOH}\right)}{\mathbf{V}\left(\mathrm{solution}\right)}\mathbf{}\mathbf{=}\frac{\mathbf{2}\mathbf{.}\mathbf{428}\mathbf{mmol}}{\mathbf{22}\mathbf{.}\mathbf{63}\mathbf{mL}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{107}\mathbf{M}$

Hence the molarity of${}^{\mathbf{NaOH}}$ is ${}^{\mathbf{0}\mathbf{.}\mathbf{107}\mathbf{M}}$ .