Chapter 16, Problem 16.120QP

a Draw a pH titration curve that represents the titration of 25.0 mL of 0.15 M propionic acid. CH₃CH₂COOH, by the addition of 0.15 M KOH from a buret. Label the axes and put a scale on each axis. Show where the equivalence point and the buffer region are on the titration curve. You should do calculations for the 0%, 50%, 60%, and 100% titration points. b Is the solution neutral, acidic, or basic at the equivalence point? Why?

Interpretation Introduction

Interpretation:

For titration of 25.0 mL of 0.15 M propionic acid, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}$ by the addition of 0.15 M $\text{KOH}$,

A pH titration curve showing the equivalence point and buffer region has to be drawn

1. (a) The pH of the titration points for the 0%, 50%, 60% and 100% has to be calculated
2. (b) Whether the solution at the equivalence point is neutral, acidic or basic has to be explained

Concept Introduction:

Equivalence point:

The equivalence point in titration is the point where the amount of standard titrant solution (in moles) and the unknown concentration analyte solution (in moles) becomes equal.

In other words, the equivalence point is the point obtained in a titration once a stoichiometric amount of reactant has been added.

Relationship between pH and pOH:

$\text{pH + pOH = 14}$

Answer to Problem 16.120QP

A pH titration curve showing the equivalence point and buffer region is given in Figure 1 as follows,

Figure 1

(a)

The pH at the 0% titration point is 2.85

The pH at the 50% titration point is 4.89

The pH at the 60% titration point is 5.06

The pH at the 100% titration point is 8.88

Explanation of Solution

To Draw: A pH titration curve showing the equivalence point and buffer region

The pH titration curve for the titration of 0.15 M propionic acid  with 0.15 M $\text{KOH}$ showing the equivalence point is given in the below Figure 1,

Figure 1

(a)

To Calculate: The pH of the titration points for the 0%, 50 %, 60% and 100%

Given data:

Titration of 0.15 M propionic acid with 0.15 M $\text{KOH}$

pH at the 0% titration point:

Construct an equilibrium table with x as unknown concentration

Consider propionic acid as $\text{HA}$ and propionate anion as ${\text{A}}^{\text{-}}$

	${\text{HA + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ H}}_3{\text{O}}^{\text{+}}{\text{ + A}}^{-}$
Initial $(M)$	0.15 $-x$ 0.15-x	0.00	0.00
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		x	x

Substitute equilibrium concentrations into the equilibrium-constant equation.

The ${\text{K}}_a$ for propionic acid is $1.3\times {10}^{-5}$

Assume x is negligible compared to 0.15 M

$\begin{array}{l}{\text{ K}}_a=\frac{{\text{[H}}_3{\text{O}}^{+}{\text{][A}}^{-}]}{\text{[HA]}}\\ =\frac{{\text{(}x\text{)}}^{\text{2}}}{(0.15-x)}\\ \text{1}\text{.3}\times {10}^{-5}\approx \frac{{\text{(}x\text{)}}^2}{0.15}\\ \text{Rearrange and solve for }x\\ x=\sqrt{(0.15)(\text{1}\text{.3}\times {10}^{-5})}\\ =1.40\times {10}^{-3}\text{ M}\end{array}$

Therefore, the concentration of hydronium ion ${\text{[H}}_3{\text{O}}^{+}]$ is $1.40\times {10}^{-3}\text{ M}$

In the end, pH is calculated as follows,

$\begin{array}{l}\text{pH}\text{=}-\log {\text{[H}}_3{\text{O}}^{+}]\\ =-\log (1.40\times {10}^{-3})\\ =2.85\end{array}$

Therefore, the pH at the 0% titration point is 2.85

pH at the 50% titration point:

$\begin{array}{l}{\text{[H}}_3{\text{O}}^{+}\text{]}={\text{[A}}^{-}]\\ \text{Therefore,}\\ {\text{K}}_a=\frac{{\text{[H}}_3{\text{O}}^{+}{\text{][A}}^{-}]}{\text{[HA]}}\\ {\text{[H}}_3{\text{O}}^{+}\text{] }=\text{1}\text{.3}\times {10}^{-5}\text{ M}\end{array}$

The pH is calculated as follows,

$\begin{array}{l}\text{pH}\text{=}-\log {\text{[H}}_{\text{3}}{\text{O}}^{+}]\\ =-\log (\text{1}\text{.3}\times {10}^{-5})\\ =4.89\end{array}$

Therefore, the pH at the 50% titration point is 4.89

pH at the 60% titration point:

For convenience, express the concentrations as percents.

$\begin{array}{l}\text{[HA}]\approx 40\%\\ {\text{[A}}^{-}]\approx 60\%\end{array}$

Substitute the concentrations into the equilibrium expression.

$\begin{array}{l}{\text{ K}}_a=\frac{{\text{[H}}_3{\text{O}}^{+}{\text{][A}}^{-}]}{\text{[HA]}}\\ =\frac{\text{(60\%)(}x\text{)}}{(40\%)}\\ \text{1}\text{.3}\times {10}^{-5}=\frac{\text{(60\%)(}x\text{)}}{(40\%)}\\ \text{Rearrange and solve for }x\\ x=\text{1}\text{.3}\times {10}^{-5}\times \frac{40\%}{60\%}\\ =8.67\times {10}^{-6}\text{ M}\end{array}$

Therefore, the concentration of hydronium ion ${\text{[H}}_3{\text{O}}^{+}]$ is $8.67\times {10}^{-6}\text{ M}$

In the end, pH is calculated as follows,

$\begin{array}{l}\text{ pH}\text{=}-\log {\text{[H}}_3{\text{O}}^{+}]\\ =-\log (8.67\times {10}^{-6})\\ =5.06\end{array}$

Therefore, the pH at the 60% titration point is 5.06

pH at the 100% titration point:

The salt that got produced has undergone a twofold dilution.

Therefore, ${\text{[A}}^{-}\text{]=0}\text{.0750 M}$

Construct an equilibrium table with x as unknown concentration

	${\text{A}}^{-}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ HA + OH}}^{-}$
Initial $(M)$	0.0750 $-x$ 0.0750-x	0.00	0.00
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		x	x

Now, calculate ${\text{K}}_b$ using ${\text{K}}_w$ as follows,

$\begin{array}{l}{\text{K}}_b=\frac{{\text{K}}_w}{{\text{K}}_a}\\ =\frac{1.0\times {10}^{-14}}{1.3\times {10}^{-5}}\\ =7.69\times {10}^{-10}\end{array}$

Substitute into the equilibrium constant expression.

$\begin{array}{l}{\text{K}}_b=\frac{{\text{[HA][OH}}^{-}]}{{\text{[A}}^{-}\text{]}}\\ 7.69\times {10}^{-10}=\frac{{(x)}^2}{(0.0750-x)}\\ \text{Assume }x\text{ is very small compared to 0}\text{.0500 and neglect it}\\ \cong \frac{{(x)}^2}{(0.0750)}\\ \text{solve for }x,\\ x=\sqrt{(7.69\times {10}^{-10})(0.0750)}\\ =7.59\times {10}^{-6}\text{ M}\end{array}$

Here, x gives the concentration of hydroxide ion, ${\text{[OH}}^{-}]=7.59\times {10}^{-6}\text{ M}$

The pH is calculated as follows,

$\begin{array}{l}\text{pOH}\text{=}-\log {\text{[OH}}^{-}]\\ =-\log (7.59\times {10}^{-6})\\ =5.120\\ \text{pH + pOH }\text{= 14}\\ \text{pH}\text{=14 - 5}\text{.120}\\ \text{=8}\text{.88}\end{array}$

Therefore, the pH at the 100% titration point is 8.88

Conclusion

The pH at the 0% titration point was calculated as 2.85

The pH at the 50% titration point was calculated as 4.89

The pH at the 60% titration point was calculated as 5.06

The pH at the 100% titration point was calculated as 8.88

Interpretation Introduction

Interpretation:

For titration of 25.0 mL of 0.15 M propionic acid, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}$ by the addition of 0.15 M $\text{KOH}$,

A pH titration curve showing the equivalence point and buffer region has to be drawn

1. (a) The pH of the titration points for the 0%, 50%, 60% and 100% has to be calculated
2. (b) Whether the solution at the equivalence point is neutral, acidic or basic has to be explained

Concept Introduction:

Equivalence point:

The equivalence point in titration is the point where the amount of standard titrant solution (in moles) and the unknown concentration analyte solution (in moles) becomes equal.

In other words, the equivalence point is the point obtained in a titration once a stoichiometric amount of reactant has been added.

Relationship between pH and pOH:

$\text{pH + pOH = 14}$

Answer to Problem 16.120QP

The solution at the equivalence point is basic

Explanation of Solution

To Explain: Whether the solution at the equivalence point is neutral, acidic or basic

As a result of titration Potassium propionate salt is produced.

Potassium propionate is the salt of a weak acid and a strong base.

Propionate ion reacts with water to produce hydroxide ions.

Therefore, the given solution is basic

Conclusion

The solution at the equivalence point was found as basic