Question

In a titration of HNO₃, you add a few drops of phenolphthalein indicator to 50.00 mL of acid in a flask. You quickly add 20.00 mL of 0.0502 MNaOH but overshoot the end point, and the solution turns deep pink. Instead of starting over, you add 30.00 mL of the acid, and the solution turns colorless. Then, it takes 3.22 mL of the NaOH to reach the end point. (a) What is the concentration of the HNO₃solution? (b) How many moles of NaOH were in excess after the first addition?

Short answer

a. The concentration of HNO₃ solution is 0.0146 M.

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

Step-by-step solution

Determination of concentration of HNO3

The reaction for the given process is,

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

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,

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

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,

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

Thus, the concentration of HNO₃ solution is 0.0146 M.

Determination of moles of NaOH in excess

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

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

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

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

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

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

Now, the excess of NaOH are,

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

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