Chapter 16, Problem 60A

Interpretation Introduction

Interpretation:

  $\mathrm{p}\mathrm{H}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{p}\mathrm{O}\mathrm{H}$ of a resulting solution is to be calculated.

Concept introduction:

Molarity is number of moles of solute present per liter of solution.

When acid and base are mixed, they neutralize each other. So their molar concentrations can be calculated by applying mole concept.

Answer to Problem 60A

The resulting solution is acidic.

  $\left[{\mathrm{H}}^{+}\right]=0.627\mathrm{M}$

  $\mathrm{p}\mathrm{H}=0.203$

  $\mathrm{p}\mathrm{O}\mathrm{H}=13.797$

Solution turns pink

Explanation of Solution

Given data:

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}}=90\mathrm{m}\mathrm{L}\end{array}$

  $\begin{array}{c}\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{o}\mathrm{f}\mathrm{H}\mathrm{N}{\mathrm{O}}_3=1.75\text{}\mathrm{M}\end{array}$

  $\begin{array}{c}\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{o}\mathrm{f}\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}=0.4\text{}\mathrm{M}\end{array}$

Calculation:

  $\begin{array}{c}{\mathrm{n}}_{\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}}=\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{t}\mathrm{y}\times \mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{i}\mathrm{n}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}\end{array}$

  $\begin{array}{c}=0.4\times \frac{90}{1000}\end{array}$

  $\begin{array}{c}=3.6\times {10}^{-2}\mathrm{m}\mathrm{o}\mathrm{l}\end{array}$

When nitric acid reacts with sodium hydroxide reaction is given by:

  $\mathrm{H}\mathrm{N}{\mathrm{O}}_3+\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}\to \mathrm{N}\mathrm{a}\mathrm{N}{\mathrm{O}}_3+{\mathrm{H}}_2\mathrm{O}$

Thus one mol acid reacts with one mol base.

This impies,

  ${\mathrm{n}}_{\mathrm{H}\mathrm{N}{\mathrm{O}}_3}=3.6\times {10}^{-2}\mathrm{m}\mathrm{o}\mathrm{l}$

  $\begin{array}{c}\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{H}\mathrm{N}{\mathrm{O}}_3\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}=\frac{3.6\times {10}^{-2}\mathrm{m}\mathrm{o}\mathrm{l}}{1.75\mathrm{M}}\end{array}$

              $\begin{array}{c}=0.02057\mathrm{L}\end{array}$

              $\begin{array}{c}=20.57\mathrm{m}\mathrm{L}\end{array}$

Thus, $20.57\mathrm{m}\mathrm{L}$ nitric acid is neutralised by given amount of sodium hydroxide. So, the resulting solution is acidic.

  $\begin{array}{c}\mathrm{A}\mathrm{m}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{t}\mathrm{o}\mathrm{f}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{c}\mathrm{i}\mathrm{d}\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}=30-20.57\end{array}$

              $\begin{array}{c}=9.43\mathrm{m}\mathrm{L}\end{array}$

Density of nitric acid=1.51 g/ml

  $\begin{array}{c}\mathrm{A}\mathrm{m}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{t}\mathrm{o}\mathrm{f}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{c}\mathrm{i}\mathrm{d}=\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\times \mathrm{d}\mathrm{e}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{y}\end{array}$

            $\begin{array}{c}\begin{array}{l}=9.43\times 1.51\hfill \\ =14.24\mathrm{g}\hfill \end{array}\end{array}$

And,

  $\begin{array}{c}\text{Molar mass of nitric acid=1+14+48}\end{array}$

              $\begin{array}{c}=63\text{ g/mol}\end{array}$

Thus,

  $\begin{array}{c}\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}=\frac{14.24\times 1000}{63\times 360}\end{array}$

               $\begin{array}{c}=0.627\text{ M}\end{array}$

Therefore, $\left[{\mathrm{H}}^{+}\right]=0.627\mathrm{M}$

Now,

  $\begin{array}{c}\mathrm{p}\mathrm{H}=-\log \left[{\mathrm{H}}^{+}\right]\end{array}$

    $\begin{array}{c}=-\log (0.627)\end{array}$

    $\begin{array}{c}=0.203\end{array}$

  $\begin{array}{c}\mathrm{p}\mathrm{O}\mathrm{H}=14-\mathrm{p}\mathrm{H}\end{array}$

     $\begin{array}{c}=14-0.203\end{array}$

     $\begin{array}{c}=13.797\end{array}$

As overall solution is acidic, it will turn litmus pink.

Conclusion

Thus, the resulting solution is acidic.

  $\left[{\mathrm{H}}^{+}\right]=0.627\mathrm{M}$

  $\mathrm{p}\mathrm{H}=0.203$

  $\mathrm{p}\mathrm{O}\mathrm{H}=13.797$

Solution turns pink.