91Ó°ÊÓ

The reaction for the given process is,

${\mathbf{HNO}}_{\mathbf{3}}\left(\mathrm{aq}\right)\mathbf{+}\mathbf{NaOH}\left(\mathrm{aq}\right)\mathbf{→}{\mathbf{NaNO}}_{\mathbf{3}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(I\right)$

From the reaction it can be concluded that 1mol of HNO₃ reacts with 1 mol of NaOH to produce 1 mol of NaNO₃ and 1 mol of H₂O.

Volume of NaOH to reach the end point is,

$\mathbf{20}\mathbf{.}\mathbf{00}\mathbf{mL}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{22}\mathbf{mL}\mathbf{=}\mathbf{23}\mathbf{.}\mathbf{22}\mathbf{mL}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{02322}\mathbf{L}$

Molarity of NaOH = 0.0502 M

Number of moles of NaOH are,

$$\begin{gathered} \mathbf{Moles}\mathbf{=}\mathbf{molarity}\mathbf{×}\mathbf{volume} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0502}\mathbf{M}\mathbf{×}\mathbf{0}\mathbf{.}\mathbf{02322}\mathbf{L} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{17}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{mol} \end{gathered}$$

From the reaction, 1 mol HNO₃= 1 mol NaOH.

So, moles of HNO₃ are 1.17x10^-3 mol

Total volume of HNO₃ is,

$\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{mL}\mathbf{+}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{mL}\mathbf{=}\mathbf{80}\mathbf{.}\mathbf{0}\mathbf{mL}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{08}\mathbf{L}$

Now, the concentration of HNO₃ is,

$$\begin{gathered} \mathbf{Concentration}\mathbf{=}\frac{\mathbf{moles}}{\mathbf{volume}\mathbf{}\mathbf{inL}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{17}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}{\mathbf{0}\mathbf{.}\mathbf{08}\mathbf{L}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0146}\mathbf{M} \end{gathered}$$

Thus, the concentration of HNO₃ solution is 0.0146 M.

Calculate the moles of acid and base used in the titration.

Initially, 20 mL of NaOH is used. So, the moles of NaOH are,

$$\begin{gathered} \mathbf{Moles}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{02}\mathbf{L}\mathbf{×}\mathbf{0}\mathbf{.}\mathbf{0502}\mathbf{M} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{004}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{mol} \end{gathered}$$

Initially 50 mL of HNO₃ is used. SO, the moles of HNO₃ are,

$$\begin{gathered} \mathbf{Moles}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{L}\mathbf{×}\mathbf{0}\mathbf{.}\mathbf{0146}\mathbf{M} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{3}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{mol} \end{gathered}$$

The difference between the HNO₃ and NaOH moles is the excess moles of NaOH used.

Now, the excess of NaOH are,

$$\begin{gathered} \mathbf{ExcessNaOH}\mathbf{=}\left(1.004×{10}^{-3}\mathrm{molNaOH}\right)\mathbf{-}\left(7.3×{10}^{-4}{\mathrm{molHNO}}_3\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{8}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{mol} \end{gathered}$$

Thus, the excess moles of NaOH are 2.8x10^-4 mol.