Q4.55P - Chemistry: Molecular Nature Of Matter And Change

Question

An unknown amount of acid can often be determined by adding an excess of base and then “back-titrating” the excess. A 0.3471-g sample of a mixture of oxalic acid, which has two ionizable protons, and benzoic acid, which has one, is treated with 100.0 mL of 0.1000 M NaOH. The excess NaOH is titrated with 20.00 mL of 0.2000 M HCl. Find the mass % of benzoic acid.

Step-by-Step Solution

Answer

The mass % of benzoic acid is 70.38.

Step 1: first write down the balanced molecular equation and find out the amount of NaOH used in the reaction:

Molecular equation:

 $\mathit{N}\mathit{a}\mathit{O}\mathit{H}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathit{H}\mathit{C}\mathit{l}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\to }\mathit{N}\mathit{a}\mathit{C}\mathit{l}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{l}\mathbf{\right)}$

Moles of excess NaOH are:

 $\mathbf{moles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{excess}\mathbf{ }\mathbf{NaOH}\mathbf{=}\mathbf{\left(}\mathbf{20}\mathbf{mlHCl}\mathbf{\right)}\mathbf{ }\left(\frac{1\mathrm{lit}}{1000\mathrm{ml}}\right)\mathbf{ }\left(\frac{0.2000\mathrm{mol} \mathrm{HCl}}{1\mathrm{lit} \mathrm{HCl} \mathrm{solution}}\right)\mathbf{ }\left(\frac{1\mathrm{mol} \mathrm{NaOH}}{1\mathrm{mol} \mathrm{HCl}}\right)$ 

$\mathbf{=}\mathbf{4}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{m}\mathit{o}\mathit{l}\mathit{N}\mathit{a}\mathit{O}\mathit{H}$

Now, the total moles of NaOH are:

$\mathbf{=}\mathbf{(}\mathbf{100}\mathit{m}\mathit{l}\mathbf{ }\mathit{N}\mathit{a}\mathit{O}\mathit{H}\mathbf{)}\mathbf{ }\left(\frac{1\mathrm{lit}}{1000\mathrm{ml}}\right)\mathbf{ }\left(\frac{0.1000\mathrm{mol} \mathrm{NaOH}}{1\mathrm{lit} \mathrm{NaOH}}\right)$

Total moles of NaOH 

$\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0100}\mathit{m}\mathit{o}\mathit{l}\mathit{N}\mathit{a}\mathit{O}\mathit{H}$

Used moles of NaOH are:

 $\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathit{o}\mathit{f}\mathit{N}\mathit{a}\mathit{O}\mathit{H}\mathit{u}\mathit{s}\mathit{e}\mathit{d}\mathbf{=}\mathit{t}\mathit{o}\mathit{t}\mathit{a}\mathit{l}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathit{o}\mathit{f}\mathit{N}\mathit{a}\mathit{O}\mathit{H}\mathbf{-}\mathit{e}\mathit{x}\mathit{c}\mathit{e}\mathit{s}\mathit{s}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathit{o}\mathit{f}\mathit{N}\mathit{a}\mathit{O}\mathit{H}$

$$\begin{gathered} \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}\mathit{m}\mathit{o}\mathit{l}\mathit{N}\mathit{a}\mathit{O}\mathit{H}\mathbf{-}\mathbf{4}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{m}\mathit{o}\mathit{l}\mathit{N}\mathit{a}\mathit{O}\mathit{H} \\ \mathbf{=}\mathbf{6}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{m}\mathit{o}\mathit{l}\mathit{N}\mathit{a}\mathit{O}\mathit{H} \end{gathered}$$

Step 2: Now write down the molecular equation for back titration and calculate the mass% of benzoic acid by calculating its mass:

Back – titration equation:

 ${\mathit{C}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}{\mathit{C}}_{\mathbf{7}}{\mathit{H}}_{\mathbf{6}}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{3}\mathit{N}\mathit{a}\mathit{O}\mathit{H}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\to }{\mathit{C}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{4}}\mathit{N}{\mathit{a}}_{\mathbf{2}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}{\mathit{C}}_{\mathbf{7}}{\mathit{H}}_{\mathbf{5}}{\mathit{O}}_{\mathbf{2}}\mathit{N}\mathit{a}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$

Mass of benzoic acid is:

 $\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{benzoic}\mathbf{ }\mathbf{acid}\mathbf{=}\mathbf{\left(}\mathbf{6}\mathbf{x}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{molNaOH}\mathbf{\right)}\left(\frac{1{\mathrm{molC}}_7{H}_6{O}_2}{3\mathrm{molNaOH}}\right)\mathbf{ }\left(\frac{122.13{\mathrm{gmC}}_7{H}_6{O}_2}{1{\mathrm{molC}}_7{H}_6{O}_2}\right)$

The mass percent of benzoic acid is:

 

$\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{2443}\mathbf{g}\mathbf{m}{\mathbf{C}}_{\mathbf{7}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{2}}}{\mathbf{0}\mathbf{.}\mathbf{3471}\mathbf{g}\mathbf{m}{\mathbf{C}}_{\mathbf{7}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{2}}}\mathit{x}\mathbf{100}\mathbf{\%}$

$\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathit{o}\mathit{f}\mathit{b}\mathit{e}\mathit{n}\mathit{z}\mathit{o}\mathit{i}\mathit{c}\mathit{a}\mathit{c}\mathit{i}\mathit{d}\mathbf{=}\mathbf{70}\mathbf{.}\mathbf{38}\mathbf{\%}$