Chapter 17, Problem 17.34QP

Interpretation Introduction

Interpretation:

The $\text{pH}$ at the equivalence point of titration $\text{NaOH}\text{Vs}\text{HCOOH}$ has to be calculated.

Concept introduction:

• $\text{pH}$ 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.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• Base dissociation constant ${\text{K}}_{\text{b}}$ is defined as measure of base strength in a given solution
• ${\text{K}}_{\text{b}}\text{=}\frac{{\text{[OH}}^{\text{-}}{\text{][HB}}^{\text{+}}\text{]}}{\text{[B]}}$
• The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.

To find: The $\text{pH}$ of the titration

Answer to Problem 17.34QP

The $\text{pH}$ at the equivalence point of titration $\text{NaOH}\text{Vs}\text{HCOOH}$ is $\text{8}\text{.23}.$

Explanation of Solution

Concentration of $\text{NaOH}$ is $\text{0}\text{.10}\text{M}$

Concentration of $\text{HCOOH}$ is $\text{0}\text{.10}\text{M}$

Find the concentration of sodium formate

$\begin{array}{l}{\text{HCOOH}}_{\text{(aq)}}{\text{ + NaOH}}_{\text{(aq)}}\to {\text{ HCOONa}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\\ \\ \text{Initial concentration(M): }\text{0}\text{.010 }\text{0}\text{0}\\ \text{Change in concentration (M):}\text{-0}\text{.10}\text{-0}\text{.10}\text{+0}\text{.10}\\ \text{Final}\text{concentration (M): }\text{0}\text{0}\text{0}\text{.10}\\ \\ \text{Both the concentrations ofare same and to reach equivalence point, equal amount of }\\ \text{solution should be needed}\text{. Therefore, the volume of solution is twofolded}\text{.}\\ \\ \text{The concentration of sodium formate is}\\ \text{=}\frac{\text{0}\text{.10 mol}}{\text{2}\text{L}}\text{=}\text{0}\text{.050M}\end{array}$

The concentration of sodium formate can be calculated from the final concentration of sodium formate in equilibrium table.  The solution gets doubled and the strength of sodium formate is divided by two.

Calculate the $\text{pH}$ of the titration

$\begin{array}{l}{\text{HCOO}}^{\text{-}}{}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{ HCOOH}}_{\text{(aq)}}{\text{ + OH}}^{\text{-}}{}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{0}\text{.050 }\text{0}\text{0}\\ \text{Change in concentration (M):}\text{-x}+\text{x}\text{+x}\\ \text{Equilibrium}\text{concentration (M): }\text{0}\text{.050-x}\text{x}\text{x}\\ \\ {\text{K}}_{\text{a}}\text{ value for HCOOH is 1}\text{.7 ×}{\text{10}}^{\text{-4}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\text{=}\frac{\text{1}\text{.0 ×}{\text{10}}^{\text{-14}}}{\text{1}\text{.7 ×}{\text{10}}^{\text{-4}}}\text{=}\text{5}\text{.9×}{\text{10}}^{\text{-11}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[HCOOH][OH}}^{\text{-}}\text{]}}{{\text{[HCOO}}^{\text{-}}\text{]}}\\ \text{5}\text{.9}\text{×}{\text{10}}^{\text{-11}}\text{=}\frac{{\text{(x}}^{\text{2}}\text{)}}{\text{(0}\text{.050-x)}}\\ \text{x}\text{is}\text{very}\text{small}\text{and}\text{neglect}\text{it,}\\ {\text{x = [OH}}^{\text{-}}\text{] = 1}\text{.7 ×}{\text{10}}^{\text{-6}}\text{M}\\ \text{pOH}\text{=}{\text{-log[OH}}^{\text{-}}\text{]}\\ \text{=}\text{-log(1}\text{.7 ×}{\text{10}}^{\text{-6}}\text{)}\\ \text{pOH}\text{=}\text{5}\text{.77}\\ \text{pH = 14}\text{.00 -5}\text{.77 = 8}\text{.23}\end{array}$

The concentration of hydroxide can be calculated from the final concentration of hydroxide in equilibrium table using base dissociation constant.  The $\text{pOH}$ is determined by taking negative logarithm of hydroxide ion concentration.  From the obtained $\text{pOH}$, the $\text{pH}$ is calculated.

Conclusion

The $\text{pH}$ at the equivalence point of titration $\text{NaOH}\text{Vs}\text{HCOOH}$ was calculated.