Chapter 6, Problem 6.26P

Interpretation Introduction

Interpretation:

• Table for initial and final concentration of ${\text{Na}}^{\text{+}}{\text{,OH}}^{\text{-}}\text{and NaOH}$ in 1F Sodium hydroxide has to be prepared.
• At equilibrium the fraction of $\text{Na}$ in the form of ${\text{NaOH}}_{\left(\text{aq}\right)}$ has to be calculated.

Answer to Problem 6.26P

Table for initial and final concentration of ${\text{Na}}^{\text{+}}{\text{,OH}}^{\text{-}}\text{and NaOH}$ is

$\begin{array}{l}{\text{Na}}^{\text{+}}{\text{+OH}}^{\text{-}}\stackrel{}{\rightleftharpoons }\text{NaOH}\\ \overline{\underset{\_}{\begin{array}{l}Initialconcentration:\text{1}\text{1}\text{0}\\ Finalconcentration:1-x1-xx\end{array}}}\end{array}$

The fraction of $\text{Na}$ in the form of ${\text{NaOH}}_{\left(\text{aq}\right)}$ is $6.85$ or 0.15

Explanation of Solution

Given

$\begin{array}{l}{\text{Li}}^{\text{+}}{\text{+OH}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{LiOH}}_{\left(\text{aq}\right)}{\text{K}}_{\text{1}}\text{=0}\text{.83}\\ \\ {\text{Na}}^{\text{+}}{\text{+OH}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{NaOH}}_{\left(\text{aq}\right)}{\text{K}}_{\text{1}}\text{=0}\text{.20}\end{array}$

To prepare: table for initial and final concentration of ${\text{Na}}^{\text{+}}{\text{,OH}}^{\text{-}}\text{and NaOH}$

$\begin{array}{l}{\text{Na}}^{\text{+}}{\text{+OH}}^{\text{-}}\stackrel{}{\rightleftharpoons }\text{NaOH}\\ \overline{\underset{\_}{\begin{array}{l}Initialconcentration:\text{1}\text{1}\text{0}\\ Finalconcentration:1-x1-xx\end{array}}}\end{array}$

To calculate: fraction of $\text{Na}$ in the form of ${\text{NaOH}}_{\left(\text{aq}\right)}$

Concentration = $\begin{array}{l}\frac{\left[\text{NaOH}\right]}{\left[{\text{Na}}^{\text{+}}\right]\left[{\text{OH}}^{\text{\_}}\right]}\text{=}\frac{\text{x}}{{\left(\text{1-x}\right)}^{\text{2}}}\\ \\ \text{x = 0}\text{.15M}\\ \\ \text{x = 0}\text{.2 - 0}\text{.4x + 0}{\text{.2x}}^{\text{2}}\\ \\ \text{0}{\text{.2x}}^{\text{2}}\text{-1}\text{.4x+0}\text{.2=0}\\ \\ \text{x=}\frac{\text{-b±}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2a}}\text{=}\frac{\text{1}\text{.4±}\sqrt{\text{1}{\text{.4}}^{\text{2}}\text{-4}\left(\text{0}\text{.2}\right)\left(\text{0}\text{.2}\right)}}{\text{2}\left(\text{0}\text{.2}\right)}\\ \\ \text{= 6}\text{.85or0}\text{.15}\end{array}$

Conclusion

Table for initial and final concentration of ${\text{Na}}^{\text{+}}{\text{,OH}}^{\text{-}}\text{and NaOH}$ was given as,

$\begin{array}{l}{\text{Na}}^{\text{+}}{\text{+OH}}^{\text{-}}\stackrel{}{\rightleftharpoons }\text{NaOH}\\ \overline{\underset{\_}{\begin{array}{l}Initialconcentration:\text{1}\text{1}\text{0}\\ Finalconcentration:1-x1-xx\end{array}}}\end{array}$

The fraction of $\text{Na}$ in the form of ${\text{NaOH}}_{\left(\text{aq}\right)}$ was calculated as $6.85$ or 0.15