Chapter 17, Problem 17.72QP

Interpretation Introduction

Interpretation:

The concentration of sodium hydroxide has to be determined using the given data.

Concept introduction:

Molarity is defined as strength of substance in terms of amount of moles of solute dissolved in one litre of solution.

The molarity of ${\text{HNO}}_{\text{2}}$ and $\text{NaOH}$

Answer to Problem 17.72QP

The concentration of  sodium hydroxide is $\text{1}\text{.28M}$

Explanation of Solution

Given,

  $\text{Strength}\text{of}{\text{HNO}}_{\text{2}}\text{is}\text{2}\text{.00M}$

$\begin{array}{l}\begin{array}{l}{\text{HNO}}_{\text{2}}{}_{\left(\text{aq}\right)}\to {\text{H}}^{\text{+}}{}_{\left(\text{aq}\right)}\text{ + }{\text{NO}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\hfill \\ \text{Initial concentration}\left(\text{M}\right)\text{: 2}\text{.00 }\text{0 0}\hfill \end{array}\\ \text{Change in concentration}\left(\text{M}\right)\text{: -x }\text{+x +x }\\ \text{Equilibrium concentration}\left(\text{M}\right)\text{: }\text{2}\text{.00-x }\text{x x}\\ \\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{][NO}}_{\text{2}}^{\text{-}}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}\\ \text{4}\text{.5}\text{×}{\text{10}}^{\text{-4}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{2}\text{.00}\text{-}\text{x}}\approx \frac{{\text{x}}^{\text{2}}}{\text{2}\text{.00}}\\ {\text{x = [H}}^{\text{+}}\text{] = 0}\text{.030M}\\ \text{pH = -log}\left(\text{0}\text{.030}\right)\text{ = 1}\text{.52}\end{array}$

The $\text{pH}$ after addition of sodium hydroxide is $1.5$ times greater

  $\begin{array}{l}\text{pH=1}\text{.52+1}\text{.50=3}\text{.02}\\ {\text{[H}}^{\text{+}}{\text{]= 10}}^{\text{-}}{}^{\text{pH}}{\text{ = 10}}^{\text{-}}{}^{\text{3}\text{.02}}\text{ = 9}\text{.55}\text{×}{\text{10}}^{\text{-}}{}^{\text{4}}\text{M}\\ \\ \text{After}\text{addition of}\text{NaOH,}\text{2}\text{.00M}{\text{HNO}}_{\text{2}}\text{get}\text{diluted}\\ \\ \begin{array}{l}{\text{M}}_{\text{i}}{\text{V}}_{\text{i}}{\text{ = M}}_{\text{f}}{\text{V}}_{\text{f}}\hfill \\ \left(\text{2}\text{.00M}\right)\left(\text{400 mL}\right){\text{ = M}}_{\text{f}}\left(\text{600 mL}\right)\hfill \\ {\text{ M}}_{\text{f}}\text{ = 1}\text{.33M}\hfill \end{array}\end{array}$

The $\text{pH}$ of nitrous acid is calculated using acid dissociation constant (${\text{K}}_{\text{a}}$).  From the $\text{pH}$, the molarity of nitrous acid after addition of sodium hydroxide is calculated.

Now we have buffer solution and not reached the equivalence point.

${\text{HNO}}_{\text{2}}{}_{\left(\text{aq}\right)}{\text{+NaOH}}_{\left(\text{aq}\right)}\to {\text{ NaNO}}_{\text{2}}{}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Because the nitrous acid and sodium hydroxide are in same ratio, the reduction in concentration of nitrous acid is same as sodium hydroxide.

  $\begin{array}{l}\begin{array}{l}{\text{HNO}}_{\text{2}}{}_{\left(\text{aq}\right)}\to {\text{H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+ NO}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\hfill \\ \text{Initial concentration}\left(\text{M}\right)\text{: 1}\text{.33 0 0}\hfill \end{array}\\ \text{Change in concentration}\left(\text{M}\right)\text{: -x }\text{+x }\\ \text{Equilibrium concentration}\left(\text{M}\right)\text{: 1}\text{.33-x }\text{9}{\text{.55×10}}^{\text{-}}{}^{\text{4}}\text{x}\\ \\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{][NO}}_{\text{2}}^{\text{-}}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}\\ \text{4}\text{.5}\text{×}{\text{10}}^{\text{-4}}\text{=}\frac{\text{(9}\text{.55}\text{×}{\text{10}}^{\text{-4}}\text{)(x)}}{\text{1}\text{.33}\text{-}\text{x}}\\ \text{x = 0}\text{.426M}\end{array}$

After dilution to $\text{600mL}$ the concentration of sodium hydroxide is same as decrease in concentration of nitrous acid.  The concentration of $\text{NaOH}$ has to calculate for $\text{600mL}$

  $\begin{array}{l}{\text{M}}_{\text{i}}{\text{V}}_{\text{i}}{\text{ = M}}_{\text{f}}{\text{V}}_{\text{f}}\\ {\text{M}}_{\text{i}}\left(\text{200 mL}\right)\text{ = }\left(\text{0}\text{.426M}\right)\left(\text{600 mL}\right)\\ \left[\text{NaOH}\right]{\text{ = M}}_{\text{i}}\text{ = 1}\text{.28M}\end{array}$

The concentration of nitrous acid is calculated using acid dissociation constant expression.  After dilution to $\text{600mL}$ the concentration of sodium hydroxide is same as decrease in concentration of nitrous acid.  The concentration of $\text{NaOH}$ is calculated for $\text{600mL}$

Conclusion

The concentration of sodium hydroxide was determined using the given data.