91Ó°ÊÓ

Titration of sulfuric acid by sodium hydroxide produces sodium sulfate and water. The balanced equation is like

$2\mathrm{NaOH}\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)→{\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

According to the question,

Volume of H₂SO₄ = 50mL = 0.05L

Molarity of H₂SO₄ = 0.0782M

Again you know,

$\mathrm{moles}=\mathrm{molarity}×\mathrm{volume}$

Now, moles of H₂SO₄

$\begin{array}{l}=(0.0782×0.05)\left(\mathrm{mol}\right)\\ =0.00391\left(\mathrm{mol}\right)\end{array}$

Hence, moles of H₂SO₄ are 0.00391mol.

According to the balanced equation

$2\mathrm{NaOH}\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)→{\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

2 moles of NaOH react with 1 mole of H₂SO₄

Now, moles of NaOH

$\begin{array}{l}\mathbf{=}\mathbf{0}\mathbf{.00391}\mathbf{×}\frac{\mathbf{2}}{\mathbf{1}}\mathbf{\left(}\mathbf{mol}\mathbf{\right)}\\ \mathbf{=}\mathbf{0}\mathbf{.00782}\mathbf{\left(}\mathbf{mol}\mathbf{\right)}\end{array}$

Hence, moles of H₂SO₄ are 0.00782mol.

According to the question,

Volume of NaOH = 18.4mL = 0.0184L

Again you know,

_{$\mathrm{Molarity}=\frac{\mathrm{moles}}{\mathrm{volume}}$}

Now, molarity of NaOH

$\begin{array}{l}=\frac{0.00782}{0.0184}\left(\mathrm{M}\right)\\ =0.425\left(\mathrm{M}\right)\end{array}$

Hence, molarity of NaOH is 0.425M.

The balanced equation is

$\mathrm{NaOH}\left(\mathrm{aq}\right)+\mathrm{HCl}\left(\mathrm{aq}\right)→\mathrm{NaCl}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

According to the question,

Volume of NaOH required = 27.5mL = 0.0275L

Molarity of NaOH calculated = 0.425M

Again you know,

$\mathrm{Moles}=\mathrm{molarity}×\mathrm{volume}$

Now, moles of NaOH

$\begin{array}{l}=(0.425×0.0275)\left(\mathrm{mol}\right)\\ =0.01169\left(\mathrm{mol}\right)\end{array}$

Hence, moles of NaOH required to titrate HCl are 0.01169mol.

According to the balanced equation

$\mathrm{NaOH}\left(\mathrm{aq}\right)+\mathrm{HCl}\left(\mathrm{aq}\right)→\mathrm{NaCl}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

2 moles of NaOH reacts with 1 mole of HCl

Now, moles of HCl

^{$\begin{array}{l}=0.01169×\frac{1}{1}\left(\mathrm{mol}\right)\\ =0.01169\left(\mathrm{mol}\right)\end{array}$}

Hence, moles of HCl are 0.01169mol.

According to the question,

Volume of HCl = 100mL = 0.1L

Again you know,

_{$\mathrm{Molarity}=\frac{\mathrm{moles}}{\mathrm{volume}}$}

Now, molarity of HCl

$\begin{array}{l}=\frac{0.01169}{0.1}\left(\mathrm{M}\right)\\ =0.1169\left(\mathrm{M}\right)\end{array}$

Hence, molarity of HCl is 0.1169M.

Hence, molarity of NaOH is 0.425M and molarity of HCl is 0.1169M.