Chapter 21, Problem 21.202QP

Interpretation Introduction

Interpretation:

The percentage of sodium acetate in a mixture of sodium acetate and sodium chloride has to be determined.

Concept introduction:

• Base dissociation constant ${\text{K}}_{\text{b}}$ is defined as measure of base strength in a given solution
• ${\text{K}}_{\text{b}}\text{=}\frac{{\text{[OH}}^{\text{-}}{\text{][HB}}^{\text{+}}\text{]}}{\text{[B]}}$ ,  ${\text{K}}_{\text{b}}\text{= }\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}$
• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$  in a solution.
•   $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

Answer to Problem 21.202QP

The percentage of sodium acetate in a mixture of sodium acetate and sodium chloride is $65\%$

Explanation of Solution

To determine: The percentage of sodium acetate in a mixture of sodium acetate and sodium chloride.

Between sodium acetate and sodium chloride, sodium acetate reacts with water.

${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{ + H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\rightleftarrows }{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}_{\left(\text{aq}\right)}{\text{ + OH}}^{\text{-}}{}_{\left(\text{aq}\right)}$

Equilibrium constant for water gives ${\text{K}}_{\text{b}}$,

${\text{K}}_{\text{b}}\text{= }\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\text{=}\frac{\text{1}{\text{.00 x 10}}^{\text{-14}}}{\text{1}{\text{.7 x 10}}^{\text{-5}}}\text{=}\text{5}\text{.882}\times {\text{10}}^{\text{-10}}$

The equilibrium constant expression for above reaction is,

$\begin{array}{l}{\text{K}}_{\text{b}}\text{= }\frac{{\text{[HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{][OH}}^{\text{-}}\text{]}}{{\text{[C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{\text{-}}\text{]}}\\ \\ \text{=}\frac{{\text{(x)}}^{\text{2}}}{{{\text{[C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{\text{-}}\text{]}}_{\text{o}}\text{- x}}\text{=}\text{5}\text{.882}\text{×}{\text{10}}^{\text{-10}}\end{array}$

By using the $\text{pOH}$, the x value is calculated as

$\begin{array}{l}\text{pOH = 14}\text{.00 - pH }\\ \\ \text{= 14}\text{.00 - 9}\text{.03 }\\ \text{= 4}\text{.97}\\ \\ \text{x= }\left[{\text{OH}}^{\text{-}}\right]{\text{ = 10}}^{\text{-pOH}}{\text{= 10}}^{\text{-4}\text{.97}}\text{= 1}\text{.071}\text{×}{\text{10}}^{\text{-5}}\text{M}\end{array}$

By rearranging the equilibrium constant expression we get acetate ion concentration.

$\begin{array}{l}{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{\text{-}}\right]}_{\text{o}}\text{=}\text{x}\text{+ }\frac{{\text{(x)}}^{\text{2}}}{{\text{K}}_{\text{b}}}\\ \\ \text{= 1}\text{.071}\text{×}{\text{10}}^{\text{-5}}\text{M+}\frac{{\text{(1}{\text{.071 × 10}}^{\text{-5}}\text{)}}^{\text{2}}}{\text{5}{\text{.882 × 10}}^{\text{-10}}}\\ \\ \text{= 0}\text{.1952M}\end{array}$

The amount of sodium acetate present in given sample is

$\left(\text{0}\text{.1952M}\right)\text{×}\text{(25}\text{.0}\text{×}{\text{10}}^{\text{-3}}\text{L)}\text{×}\frac{{\text{1 mol NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}}{{\text{1 mol C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{\text{-}}}\text{×}\frac{\text{82}\text{.034 g}}{{\text{1 mol NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}}\text{ = 0}\text{.400 g}$

The percentage of sodium acetate present in given sample is

$=\frac{\text{0}\text{.400 g}}{\text{0}\text{.613 g}}\times 100\%=65\%$

Conclusion

The percentage of sodium acetate in a mixture of sodium acetate and sodium chloride was determined.