Chapter 16, Problem 16.54QA

Interpretation Introduction

To find:

The color of red cabbage juice at equivalence point when $25\mathrm{ }\mathrm{m}\mathrm{L}$ of a $0.10\mathrm{ }\mathrm{M}$ solution of acetic acid is titrated with $0.10\mathrm{M}\mathrm{ }\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}.$

Answer to Problem 16.54QA

Solution:

Red cabbage juice would have yellow color since pH of the solution at equivalence point is in the alkaline region.

Explanation of Solution

1) Concept:

 We are asked to find the color of red cabbage juice at equivalence point when a weak acid, acetic acid and a strong base, sodium hydroxide are titrated. Since red cabbage juice is a sensitive acid-base indicator, we need to find pH of the solution at equivalence point. Red cabbage juice changes its color as the pH of the solution changes. It shows red color at acidic $\mathrm{p}\mathrm{H}$ and yellow at alkaline $\mathrm{p}\mathrm{H}.$

The equivalence point of a titration is the point at which chemically equivalent quantities of base and acids have been mixed. So, to reach equivalence point $25 mL$ of $0.10\mathrm{M} \mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}$ is required. We can calculate the moles of the weak acid present and the moles of ${\mathrm{O}\mathrm{H}}^{-}$ added using their respective volumes and molarities. As the $\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}$ is added to the acetic acid solution, the following reaction occurs:

${CH}_3{COOH}_{ }\left(aq\right)+ {OH}_{ }^{-}\left(aq\right) \to {CH}_3{COO}_{ }^{-}\left(aq\right)+H_{2}O\left(aq\right)$

At equivalence point,  neither acetic acid nor $NaOH$ remains. The only species present at the equivalence point will be sodium acetate $\left(C{H}_{3}COONa\right)$. From the dissociation reaction of sodium acetate and creating an ICE table, we can find the $pH$ of the solution.

We can calculate $\left[{\mathrm{C}\mathrm{H}}_3{\mathrm{C}\mathrm{O}\mathrm{O}}^{-}\right]$ and then use equilibrium reaction of acetate with water to determine the $\mathrm{p}\mathrm{H}$ of the solution. The ${\mathrm{K}}_{\mathrm{a}}$ for acetic acid, which can be found in Appendix A5.1, equals $1.74 \times {10}^{-5}$. The $pH$ at equivalence point is greater than $7.00$ because the salt sodium acetate is a basic salt made up of a weak acid and a strong base.

2) Formula:

i) $K_b= \frac{K_w}{K_a}$

ii) $pH+pOH=14$

iii) $pH= -\mathrm{l}\mathrm{o}\mathrm{g}\left[{\mathrm{H}}^{+}\right]$

3) Given:

i) Volume of acetic acid is $25\mathrm{ }\mathrm{m}\mathrm{L}=0.025\mathrm{L}$

ii) Molarity of acetic acid is $0.10\mathrm{M}$

iii) Molarity of $\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}$ is $0.10\mathrm{M}$

4) Calculations:

The initial quantity of ${CH}_{3}COOH$ in the sample is

$25 mL \times \frac{1 L}{1000 mL} \times \frac{0.10 mol {CH}_{3}COOH }{L}=0.0025 mol {CH}_{3}COOH$

At equivalence point, $mol C{H}_{3}COOH=mol NaOH=0.0025 mol$

The quantity of $\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}$ required to reach the equivalence point is:

$\frac{0.0025 mol NaOH }{0.1 mol/L}=0.025L=25.0 mL$

We can use a modified RICE table to determine how many moles of ${CH}_{3}COOH$ will remain and how many moles of ${CH}_3{COO}^{-}$ have been produced.

Reaction	${\mathit{C}\mathit{H}}_3{\mathit{C}\mathit{O}\mathit{O}\mathit{H}}_{\mathit{ }}\left(\mathit{a}\mathit{q}\right)$+ ${\mathit{C}\mathit{H}}_3\mathit{C}\mathit{O}\mathit{O}\mathit{H}\mathit{ }\left(\mathit{ }\mathit{M}\mathit{o}\mathit{l}\right)$	${\mathit{O}\mathit{H}}^{-}$ ${\mathit{O}\mathit{H}}^{-}\left(\mathit{M}\mathit{o}\mathit{l}\right)$	$\to$${\mathit{C}\mathit{H}}_3{\mathit{C}\mathit{O}\mathit{O}}_{\mathit{ }}^{-}\left(\mathit{a}\mathit{q}\right)$ ${\mathit{C}\mathit{H}}_3{\mathit{C}\mathit{O}\mathit{O}}^{-}\mathit{ }\left(\mathit{ }\mathit{M}\mathit{o}\mathit{l}\right)$
Initial	$0.0025$	$0.0025$	$0$
Change	$-0.0025$	$-0.0025$	$+0.0025$
Final	$0$	$0$	$0.0025$

The total sample volume is $\left( 25mL+25mL\right)=50mL or 0.0500L$

The concentration of acetic acid and acetate ion is

$\left[{CH}_3{COO}^{-}\right]= \frac{0.0025 mol}{0.050 L}=0.050M$

Because the product of the neutralization reaction is a weak base, we must consider the reaction of the weak base with water to calculate $H_3{O}^{+}$ at equilibrium, and thus, the $pH$ of the solution.

The equilibrium reaction of acetate with water is as follows:

${CH}_3{COO}_{ }^{-}\left(aq\right)+H_2{O}_{ }\left(l\right) \rightleftharpoons {CH}_3{COOH}_{ }\left(aq\right)+ {OH}_{ }^{-}\left(aq\right)$

The equilibrium constant for this reaction is $K_b= \frac{K_w}{K_a}$ where $K_a$ is the acid ionization constant of acetic acid.  We define $x$ as $\left[{\mathrm{O}\mathrm{H}}^{-}\right]$ produced by the reaction of acetate with water. The complete RICE table of concentrations is as follows:

Reaction	${\mathit{C}\mathit{H}}_3{\mathit{C}\mathit{O}\mathit{O}}_{\mathit{ }}^{-}\left(\mathit{a}\mathit{q}\right)+{\mathit{H}}_2{\mathit{O}}_{\mathit{ }}\left(\mathit{l}\right)\mathit{ }\rightleftharpoons$ ${[\mathit{C}\mathit{H}}_3{\mathit{C}\mathit{O}\mathit{O}}^{-}]$	${\mathit{C}\mathit{H}}_3{\mathit{C}\mathit{O}\mathit{O}\mathit{H}}_{\mathit{ }}\left(\mathit{a}\mathit{q}\right)$+ ${[\mathit{C}\mathit{H}}_3\mathit{C}\mathit{O}\mathit{O}\mathit{H}]$	${\mathit{O}\mathit{H}}^{-}\mathit{ }\left(\mathit{a}\mathit{q}\right)$ ${[\mathit{O}\mathit{H}}^{-}]$
Initial	$0.050$	0	$0$
Change	$-x$	$+x$	$+x$
Equilibrium	$0.050-x$	$x$	$x$

$K_b= \frac{{[CH}_{3}COOH\left] {[OH}^{-}\right] }{{[CH}_3{COO}^{-}]}= \frac{\left(x\right)\left(x\right) }{0.050-x}~ \frac{x^2}{0.050}$

$K_b= \frac{K_w}{K_a}= \frac{1.00 \times {10}^{-14}}{1.74 \times {10}^{-5}} =5.747 \times {10}^{-10}= \frac{x^2}{0.050}$

On solving this equation we get $x=0.00005385$. Thus ${[OH}^{-}]=0.000053605M$

$pOH= -\log {[OH}^{-}]$

$pOH= -\mathrm{l}\mathrm{o}\mathrm{g}\left(0.00005385\right)$

$pOH=5.27$

$pH+pOH=14$

$pH=14-5.27$

$pH=8.73$

$pH=8.73$

Since $pH$ of the solution at equivalence point is in the alkaline region red cabbage juice would have yellow color at this $pH$.

Conclusion:

As expected for the titration of a weak acid and a strong base, the $\mathrm{p}\mathrm{H}$ at the equivalence point is greater than $7.00$ because the salt formed in the reaction is a basic salt which then reacts with water to produce ${\mathrm{O}\mathrm{H}}^{-}$.