Chapter 16, Problem 16.126QA

Interpretation Introduction

To find:

a) The $pH$ at the equivalence point

b) The suitable indicator for the titration referring figure 16.5

c) The $pH$ for the titration after $0, 2.5, 5.0, 7.5, 9.8, \mathrm{10,10.2,10.6,10.8,11,12.5,15,17.5} and 20 mL$ of base have been added

d) Sketch the titration curve $(pH vs mL$ of base added) for this titration, and indicate which major species are present in the solution when $pH=p{K}_a$, at the equivalence point and when the titration is complete.

Answer to Problem 16.126QA

Solution:

a) The calculated $pH$ at the equivalence point is $11.40$.

b) The suitable indicator for the titration, referring figure 16.5, is Alizarin Yellow R.

c) The $pH$ for the titration after

1) $0 mL NaOH added is 5.41$

2) $2.5 mL NaOH added is 9.39$

3) $5.0 mL NaOH added is 9.86$

4) $7.5 mL NaOH added is 10.31$

5) $9.8 mL NaOH added is 11.17$

6) $10 mL NaOH added is 11.39$

7) $10.2 mL NaOH added is 11.82$

8) $10.6 mL NaOH added is 11.34$

9) $10.8 mL NaOH added is 11.57$

10) $11 mL NaOH added is 11.72$

11) $12.5 mL NaOH added is 12.2$

12) $15 mL NaOH added is 12.5$

13) $17.5 mL NaOH added is 12.66$

14) $20 mL NaOH added is 12.77$

d) In the sketched titration curve $(pH vs mL$ of base added) for this titration, the major species at the half equivalence point are $=C_6{H}_5-ONa and C_6{H}_5-OH$ when $pH=p{K}_a$

The major species at the equivalence point is $C_6{H}_5-ONa$.

The major species at the end of titration is $NaOH$.

Explanation of Solution

1) Concept:

To calculate the pH of a weak acid with a strong base, we need to use reaction stoichiometry and RICE table. First, we will calculate the equivalence point from the volume and molarity of base and acid. We will calculate the pH at equivalence point by using the RICE table. At the equivalence point, moles of phenol and $NaOH$ are equal, so the only species present is phenoxide ion. Using the hydrolysis of phenoxide ions and RICE table, we can find the pH at the equivalence point. Suitable indicator for this titration is decided by comparing the pH at equivalence point and the pH range of indicators given in Figure 16.5. We will sketch the titration curve based on the pH of the solution and mL of  a strong base added to the weak acid.

2) Formulae:

i) $p{K}_a= -\log K_a$

ii) $pH=p{K}_a+\log \frac{\left[base\right]}{\left[acid\right]}$

iii) $n=M\times V$

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

3) Given:

i) Volume of phenol = $21.5 mL=0.0215 L$

ii) Molarity of phenol $=0.120 M$

iii) $p{K}_a$ of phenol = $9.89$

iv) Molarity of $NaOH=0.250M$

4) Calculations:

a) Calculating $pH$ of the solution at the equivalence point:

$C_6{H}_5-OH \left(aq\right)+NaOH \left(aq\right)\to C_6{H}_5-ONa\left(aq\right)+H_{2}O$

Step 1)

Finding the volume of $NaOH$ required for the equivalence point by finding the moles of phenol:

At the equivalence point, moles of phenol = moles of $NaOH$

$n_{phenol}=M_{phenol}\times V_{phenol}=0.120 M \times 0.0215 L=0.00258 mol phenol$

At the equivalence point, moles of phenol and $NaOH$ are equal. So, there are $0.00258 mol$ of $NaOH$.

Calculating the volume of $NaOH$ at the equivalence point from the given molarity and calculated moles as

$n_{NaOH}=M_{NaOH}\times V_{NaOH}$

$\frac{n_{NaOH}}{M_{NaOH}}=V_{NaOH}=\frac{0.00258 mol }{0.250 M}=0.01032 L$

So, at the equivalence point, the volume of $NaOH$ required is $0.01032 L=10.32 mL$.

So, the total volume of solution is $=21.5+10.32=31.85 mL=0.03185 L$

At the equivalence point, sodium salt of phenol is formed, and its moles will be equal to moles of phenol and $NaOH$ consumed $=0.00258 mol$

Calculating the molarity of sodium phenoxide:

$M=\frac{n}{V}= \frac{0.00258 mol}{0.03185 L}=0.08108 M$

Step 2)

Creating an RICE table for the dissociation of sodium phenoxide $(C_6{H}_5-ONa)$:

$C_6{H}_5-ONa \left(aq\right)+H_{2}O\left(aq\right)\rightleftharpoons C_6{H}_5-OH \left(aq\right)+O{H}^{-}\left(aq\right)$
RICE	$C_6{H}_5-ONa$	$C_6{H}_5-OH$	$O{H}^{-}$
Initial (M)	$0.08108 M$	$0.0$	$0.0$
Change (M)	$-x$	$+x$	$+x$
Equilibrium (M)	$(0.08108-x)$	$x$	$x$

$p{K}_a$ for phenol $=9.89$

$K_a$ phenol = ${10}^{-p{K}_a}={10}^{-9.89}= 1.288 \times {10}^{-10}$

$K_b=\frac{K_w}{K_a}=\frac{1\times {10}^{-14}}{1.288 \times {10}^{-10}}=7.764 \times {10}^{-5}$

Writing the $K_b$ expression for the reaction and calculating $pH$:

$K_b=7.764 \times {10}^{-5}= \frac{{\left(x\right)}^2}{\left(0.08108-x\right)}$

$x^2= 7.764 \times {10}^{-5}\times 0.08108=6.292 \times {10}^{-6}$

$x= \sqrt{6.292 \times {10}^{-6}}=0.002508 M=\left[O{H}^{-}\right]$

$pOH= -\log \left[O{H}^{-}\right]= -\log 0.00250876=2.600$

$pH=14-2.600=11.40$

The calculated $pH$ at the equivalence point comes out to be $11.40,$ which falls under the basic range of $pH.$ This is reasonable since the salt of phenol (weak acid) and $NaOH$(strong base), will be basic.

b) Choosing the indicator, which is suitable for the titration referring to figure 16.5:

A $pH$ indicator is useful within a range of one $pH$ unit above and below the $p{K}_a$ value of the indicator. The $pH$ at the equivalence point is $11.60$. The indicator Alizarin yellow R is in the range $10-12$, which will be the suitable indicator for this titration.

c) Calculating the $pH$ for the titration after volume of $NaOH$ added to $21.5 mL$ of weak acid:

The calculated volume of $NaOH$ at the equivalence point is $10.32 mL$, so, before reaching this volume, the $NaOH$ would be the limiting reactant, and we can use the Henderson-Hasselbalch equation to find the $pH$.

1) Addition of 0.0 mL NaOH:

At 0.0mL NaOH added, the $pH$ of the phenol is solely because of phenol. The dissociation reaction of phenol is written as

$C_6{H}_5-OH \left(aq\right) +H_{2}O\left(aq\right)\rightleftharpoons C_6{H}_5-O^{-} \left(aq\right) +{ H}_3{O}^{+}\left(aq\right)$
RICE	$C_6{H}_5-OH$$\left(M\right)$	$C_6{H}_5-O^{-}\left(M\right)$	${ H}_3{O}^{+}$$\left(M\right)$
Initial (M)	$0.120M$	$0.0$	$0.0$
Change (M)	$-x$	$+x$	$+x$
Equilibrium (M)	$(0.120-x)$	$x$	$x$

$K_a= 1.288 \times {10}^{-10}= \frac{x^2}{(0.120-M)}$

$x^2=1.288 \times {10}^{-10} \times 0.120=1.5456 \times {10}^{-11}$

$x= \sqrt{1.5456 \times {10}^{-11}}=3.9314 \times {10}^{-6}=\left[H_3{O}^{+}\right]$

$pH= -\log \left[H_3{O}^{+}\right]$= $-\log \left(3.9314 \times {10}^{-6}\right) =5.41$

So, the $pH,$ when $0.0 mL$$NaOH$ were added, is $5.41$

2) Addition of 2.5 mL NaOH:

Moles of phenol:$n_{phenol}=M_{phenol}\times V_{phenol}=0.120 M \times 0.0215 L=0.00258 mol phenol$

Moles of $NaOH$ for $2.5 mL=0.0025L$ volume:

$n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.250 M \times 0.0025 L= 0.000625 mol NaOH$

We create a modified RICE table to determine how many moles of acid remain and how many moles of conjugate base have been produced:

$C_6{H}_5-OH \left(aq\right) + NaOH\left(aq\right)\to C_6{H}_5-O^{-} N{a}^{+}\left(aq\right) +H_{2}O\left(l\right)$ $C_6{H}_5-OH\left(Mol\right)$ $NaOH\left(mol\right)$ $C_6{H}_5-O^{-}\left(mol\right)$
Initial (mol)	$0.00258$	$0.000625$	$0.0$
Change (mol)	$-0.000625$	$-0.000625$	$+0.000625$
Final (mol)	$0.001955$	$0$	$0.000625$

The total sample volume is $21.5 mL + 2.5 mL = 24.0 mL or 0.0240 L$, and the concentrations of the base and conjugate acid is

$\left[C_6{H}_5-OH\right]= \frac{0.001955 mol}{0.0240 L}=0.08146 M$

$\left[C_6{H}_5-O^{-}\right]= \frac{0.000625 mol}{0.0240 L}=0.02604 M$

Applying the Henderson-Hasselbalch equation for the above buffer system formed as

$pH=p{K}_a+\log \frac{\left[base\right]}{\left[acid\right]}$

$pH=9.89+\log \frac{\left[0.02604\right]}{\left[0.08146\right]}$

$pH=9.39$

3) Addition of 5.0 mL NaOH:

Moles of phenol:$n_{phenol}=M_{phenol}\times V_{phenol}=0.120 M \times 0.0215 L=0.00258 mol phenol$

Moles of $NaOH$ for $5.0 mL=0.005L$ volume:

$n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.250 M \times 0.005 L= 0.00125 mol NaOH$

We create a modified RICE table to determine how many moles of acid remain and how many moles of conjugate base have been produced:

$C_6{H}_5-OH \left(aq\right) + NaOH\left(aq\right)\to C_6{H}_5-O^{-} N{a}^{+}\left(aq\right) +H_{2}O\left(l\right)$ $C_6{H}_5-OH\left(Mol\right)$ $NaOH\left(mol\right)$ $C_6{H}_5-O^{-}\left(mol\right)$
Initial (mol)	$0.00258$	$0.000125$	$0.0$
Change (mol)	$-0.00125$	$-0.00125$	$+0.00125$
Final (mol)	$0.00133$	$0$	$0.00125$

The total sample volume is $21.5 mL + 5.0 mL = 26.5 mL or 0.0265 L,$ and the concentrations of the base and conjugate acid is

$\left[C_6{H}_5-OH\right]= \frac{0.001330 mol}{0.0265 L}=0.05019 M$

$\left[C_6{H}_5-O^{-}\right]= \frac{0.00125 mol}{0.0265 L}=0.04717 M$

Applying the Henderson-Hasselbalch equation for the above buffer system formed as

$pH=p{K}_a+\log \frac{\left[base\right]}{\left[acid\right]}$

$pH=9.89+\log \frac{\left[0.04717\right]}{\left[0.05019\right]}$

$pH=9.86$

4) Addition of 7.5 mL NaOH:

Moles of phenol:$n_{phenol}=M_{phenol}\times V_{phenol}=0.120 M \times 0.0215 L=0.00258 mol phenol$

Moles of $NaOH$ for $7.5 mL=0.0075L$ volume:

$n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.250 M \times 0.0075L\mathrm{ }= 0.001875 mol NaOH$

We create a modified RICE table to determine how many moles of acid remains and how many moles of conjugate base have been produced.

$C_6{H}_5-OH \left(aq\right) + NaOH\left(aq\right)\to C_6{H}_5-O^{-} N{a}^{+}\left(aq\right) +H_{2}O\left(l\right)$ $C_6{H}_5-OH\left(Mol\right)$ $NaOH\left(mol\right)$ $C_6{H}_5-O^{-}\left(mol\right)$
Initial (mol)	$0.00258$	$0.001875$	$0.0$
Change (mol)	$-0.001875$	$-0.001875$	$+0.001875$
Final (mol)	$0.000705$	$0$	$0.001875$

The total sample volume is $21.5 mL + 7.5 mL = 29.0 mL or 0.029 L,$ and the concentrations of the base and conjugate acid is

$\left[C_6{H}_5-OH\right]= \frac{0.000705 mol}{0.029 L}=0.02431 M$

$\left[C_6{H}_5-O^{-}\right]= \frac{0.001875 mol}{0.029 L}=0.06466 M$

Applying the Henderson-Hasselbalch equation for the above buffer system formed as

$pH=p{K}_a+\log \frac{\left[base\right]}{\left[acid\right]}$

$pH=9.89+\log \frac{\left[0.06466 \right]}{\left[0.02431\right]}$

$pH=10.31$

5) Addition of 9.8 mL NaOH:

Moles of phenol:$n_{phenol}=M_{phenol}\times V_{phenol}=0.120 M \times 0.0215 L=0.00258 mol phenol$

Moles of $NaOH$ for $9.8 mL=0.0098L$ volume:

$n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.250 M \times 0.0098L\mathrm{ }= 0.00245 mol NaOH$

We create a modified RICE table to determine how many moles of acid remain and how many moles of conjugate base have been produced:

$C_6{H}_5-OH \left(aq\right) + NaOH\left(aq\right)\to C_6{H}_5-O^{-} N{a}^{+}\left(aq\right) +H_{2}O\left(l\right)$ $C_6{H}_5-OH\left(Mol\right)$ $NaOH\left(mol\right)$ $C_6{H}_5-O^{-}\left(mol\right)$
Initial (mol)	$0.00258$	$0.00245$	$0.0$
Change (mol)	$-0.00245$	$-0.00245$	$+0.00245$
Final (mol)	$0.000130$	$0$	$0.00245$

The total sample volume is $21.5 mL + 9.8 mL = 31.3 mL or 0.0313 L,$ and the concentrations of the base and conjugate acid is

$\left[C_6{H}_5-OH\right]= \frac{0.000130 mol}{0.0313 L}=0.00415 M$

$\left[C_6{H}_5-O^{-}\right]= \frac{0.00245 mol}{0.0313 L}=0.07827 M$

Applying the Henderson-Hasselbalch equation for the above buffer system formed as

$pH=p{K}_a+\log \frac{\left[base\right]}{\left[acid\right]}$

$pH=9.89+\log \frac{\left[0.07827 \right]}{\left[0.00415\right]}$

$pH=11.17$

6) Addition of 10 mL NaOH:

Moles of phenol:$n_{phenol}=M_{phenol}\times V_{phenol}=0.120 M \times 0.0215 L=0.00258 mol phenol$

Moles of $NaOH$ for $10 mL=0.01L$ volume:

$n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.250 M \times 0.010L\mathrm{ }= 0.0025 mol NaOH$

We create a modified RICE table to determine how many moles of acid remain and how many moles of conjugate base have been produced:

$C_6{H}_5-OH \left(aq\right) + NaOH\left(aq\right)\to C_6{H}_5-O^{-} N{a}^{+}\left(aq\right) +H_{2}O\left(l\right)$ $C_6{H}_5-OH\left(Mol\right)$ $NaOH\left(mol\right)$ $C_6{H}_5-O^{-}\left(mol\right)$
Initial (mol)	$0.00258$	$0.0025$	$0.0$
Change (mol)	$-0.0025$	$-0.0025$	$+0.0025$
Final (mol)	$0.00008$	$0$	$0.0025$

The total sample volume is $21.5 mL + 10.0 mL = 31.5 mL or 0.0315 L,$ and the concentrations of the base and conjugate acid is

$\left[C_6{H}_5-OH\right]= \frac{0.000080 mol}{0.0315 L}=0.00254 M$

$\left[C_6{H}_5-O^{-}\right]= \frac{0.0025 mol}{0.0315 L}=0.07937 M$

Applying the Henderson-Hasselbalch equation for the above buffer system formed as

$pH=p{K}_a+\log \frac{\left[base\right]}{\left[acid\right]}$

$pH=9.89+\log \frac{\left[0.07937 \right]}{\left[0.00254\right]}$

$pH=11.39$

7) Addition of 10.2 mL NaOH:

Moles of phenol:$n_{phenol}=M_{phenol}\times V_{phenol}=0.120 M \times 0.0215 L=0.00258 mol phenol$

Moles of $NaOH$ for $10.2 mL=0.0102 L$ volume:

$n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.250 M \times 0.0102L\mathrm{ }= 0.00255 mol NaOH$

We create a modified RICE table to determine how many moles of acid remain and how many moles of conjugate base have been produced:

$C_6{H}_5-OH \left(aq\right) + NaOH\left(aq\right)\to C_6{H}_5-O^{-} N{a}^{+}\left(aq\right) +H_{2}O\left(l\right)$ $C_6{H}_5-OH\left(Mol\right)$ $NaOH\left(mol\right)$ $C_6{H}_5-O^{-}\left(mol\right)$
Initial (mol)	$0.00258$	$0.00255$	$0.0$
Change (mol)	$-0.00255$	$-0.00255$	$+0.00255$
Final (mol)	$0.00003$	$0$	$0.00255$

The total sample volume is $21.5 mL + 10.2 mL = 31.7 mL or 0.0317 L,$ and the concentrations of the base and conjugate acid is

$\left[C_6{H}_5-OH\right]= \frac{0.00003 mol}{0.0317 L}=0.00095 M$

$\left[C_6{H}_5-O^{-}\right]= \frac{0.00255 mol}{0.0317 L}=0.08044 M$

Applying the Henderson-Hasselbalch equation for the above buffer system formed as

$pH=p{K}_a+\log \frac{\left[base\right]}{\left[acid\right]}$

$pH=9.89+\log \frac{\left[0.08044 \right]}{\left[0.00095\right]}$

$pH=11.82$

8) Addition of 10.6 mL NaOH:

The contribution of $O{H}^{-}$ ions from sodium pehnoxide is much smaller as compared to the contribution of $O{H}^{-}$ ions from $NaOH.$ The pH of the solution will be governed by solely $NaOH$.

Moles of $NaOH$ for $10.6 mL=0.0106 L$ volume:

$n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.250 M \times 0.0106L\mathrm{ }= 0.00265 mol NaOH$

$C_6{H}_5-OH \left(aq\right) + NaOH\left(aq\right)\to C_6{H}_5-O^{-} N{a}^{+}\left(aq\right) +H_{2}O\left(l\right)$ $C_6{H}_5-OH\left(Mol\right)$ $NaOH\left(mol\right)$ $C_6{H}_5-O^{-}\left(mol\right)$
Initial (mol)	$0.00258$	$0.00265$	$0.0$
Change (mol)	$-0.00258$	$-0.00258$	$+0.00258$
Final (mol)	$0$	$0.00007$	$0.00258$

Total volume of solution = $21.5+10.6=32.1 mL=0.0321L$

Molarity of $O{H}^{-}$ is

$\left[O{H}^{-}\right]=\frac{\left(0.00007 mol\right)}{0.0321 L }= 0.0021807 M$

$pOH= -\log \left[O{H}^{-}\right]= -\log 0.00218065=2.6614$

$pH=14-2.6614=11.34$

9) Addition of 10.8 mL NaOH $10.8 mL$$NaOH$

Moles of $NaOH$ for $10.8 mL=0.0108 L$ volume:

$n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.250 M \times 0.0108L\mathrm{ }= 0.00270 mol NaOH$

$C_6{H}_5-OH \left(aq\right) + NaOH\left(aq\right)\to C_6{H}_5-O^{-} N{a}^{+}\left(aq\right) +H_{2}O\left(l\right)$ $C_6{H}_5-OH\left(Mol\right)$ $NaOH\left(mol\right)$ $C_6{H}_5-O^{-}\left(mol\right)$
Initial (mol)	$0.00258$	$0.00270$	$0.0$
Change (mol)	$-0.00258$	$-0.00258$	$+0.00258$
Final (mol)	$0$	$0.00012$	$0.00258$

Total volume of solution = $21.5+10.8=32.3 mL=0.0323L$

Molarity of $O{H}^{-}$ is

$\left[O{H}^{-}\right]=\frac{\left(0.00012 mol\right)}{0.0323 L }= 0.003715 M$

$pOH= -\log \left[O{H}^{-}\right]= -\log 0.003715 =2.4300$

$pH=14-2.4300=11.57$

10) Addition of 11 mL NaOH

Moles of $NaOH$ for $11 mL=0.011 L$ volume:

$n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.250 M \times 0.011L\mathrm{ }= 0.00275 mol NaOH$

$C_6{H}_5-OH \left(aq\right) + NaOH\left(aq\right)\to C_6{H}_5-O^{-} N{a}^{+}\left(aq\right) +H_{2}O\left(l\right)$ $C_6{H}_5-OH\left(Mol\right)$ $NaOH\left(mol\right)$ $C_6{H}_5-O^{-}\left(mol\right)$
Initial (mol)	$0.00258$	$0.00275$	$0.0$
Change (mol)	$-0.00258$	$-0.00258$	$+0.00258$
Final (mol)	$0$	$0.00017$	$0.00258$

Total volume of solution = $21.5+11=32.5 mL=0.0325L$

Molarity of $O{H}^{-}$ is

$\left[O{H}^{-}\right]=\frac{\left(0.00017 mol\right)}{0.0325 L }= 0.005231 M$

$pOH= -\log \left[O{H}^{-}\right]= -\log 0.005231=2.2814$

$pH=14-2.2814=11.72$

11) Addition of 12.5 mL NaOH

Moles of $NaOH$ for $12.5 mL=0.0125 L$ volume:

$n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.250 M \times 0.0125L\mathrm{ }= 0.003125 mol NaOH$

$C_6{H}_5-OH \left(aq\right) + NaOH\left(aq\right)\to C_6{H}_5-O^{-} N{a}^{+}\left(aq\right) +H_{2}O\left(l\right)$ $C_6{H}_5-OH\left(Mol\right)$ $NaOH\left(mol\right)$ $C_6{H}_5-O^{-}\left(mol\right)$
Initial (mol)	$0.00258$	$0.003125$	$0.0$
Change (mol)	$-0.00258$	$-0.00258$	$+0.00258$
Final (mol)	$0$	$0.000545$	$0.00258$

Total volume of solution = $21.5+12.5=34 mL=0.034L$

Molarity of $O{H}^{-}$ is

$\left[O{H}^{-}\right]=\frac{\left(0.000545 mol\right)}{0.034 L }= 0.016029 M$

$pOH= -\log \left[O{H}^{-}\right]= -\log \left(0.016029 \right)=1.79508$

$pH=14-1.79508=12.20$

12) Addition of 15 mL NaOH

Moles of $NaOH$ for $15 mL=0.0150 L$ volume:

$n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.250 M \times 0.0150L\mathrm{ }= 0.00375 mol NaOH$

$C_6{H}_5-OH \left(aq\right) + NaOH\left(aq\right)\to C_6{H}_5-O^{-} N{a}^{+}\left(aq\right) +H_{2}O\left(l\right)$ $C_6{H}_5-OH\left(Mol\right)$ $NaOH\left(mol\right)$ $C_6{H}_5-O^{-}\left(mol\right)$
Initial (mol)	$0.00258$	$0.00375$	$0.0$
Change (mol)	$-0.00258$	$-0.00258$	$+0.00258$
Final (mol)	$0$	$0.00117$	$0.00258$

Total volume of solution = $21.5+15=36.5 mL=0.0365L$

Molarity of $O{H}^{-}$ is

$\left[O{H}^{-}\right]=\frac{\left(0.00117 mol\right)}{0.0365 L }= 0.03205 M$

$pOH= -\log \left[O{H}^{-}\right]= -\log 0.03205=1.4941$

$pH=14-1.4941=12.51$

13) Addition of 17.5 mL NaOH:

Moles of $NaOH$ for $17.5 mL=0.0175 L$ volume:

$n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.250 M \times 0.0175L\mathrm{ }= 0.004375 mol NaOH$

$C_6{H}_5-OH \left(aq\right) + NaOH\left(aq\right)\to C_6{H}_5-O^{-} N{a}^{+}\left(aq\right) +H_{2}O\left(l\right)$ $C_6{H}_5-OH\left(Mol\right)$ $NaOH\left(mol\right)$ $C_6{H}_5-O^{-}\left(mol\right)$
Initial (mol)	$0.00258$	$0.004375$	$0.0$
Change (mol)	$-0.00258$	$-0.00258$	$+0.00258$
Final (mol)	$0$	$0.001795$	$0.00258$

Total volume of solution = $21.5+17.5=39 mL=0.039 L$

Molarity of $O{H}^{-}$ is

$\left[O{H}^{-}\right]=\frac{\left(0.001795 mol\right)}{0.039 L }= 0.046026 M$

$pOH= -\log \left[O{H}^{-}\right]= -\log 0.046026=1.3370$

$pH=14-1.3370=12.66$

14) Addition of $20 mL$$NaOH$

Moles of $NaOH$ for $20 mL=0.020L$ volume:

$n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.250 M \times 0.020L\mathrm{ }= 0.005 mol NaOH$

$C_6{H}_5-OH \left(aq\right) + NaOH\left(aq\right)\to C_6{H}_5-O^{-} N{a}^{+}\left(aq\right) +H_{2}O\left(l\right)$ $C_6{H}_5-OH\left(Mol\right)$ $NaOH\left(mol\right)$ $C_6{H}_5-O^{-}\left(mol\right)$
Initial (mol)	$0.00258$	$0.005$	$0.0$
Change (mol)	$-0.00258$	$-0.00258$	$+0.00258$
Final (mol)	$0$	$0.00242$	$0.00258$

Total volume of solution = $21.5+20=41.5 mL=0.0415L$

Molarity of $O{H}^{-}$ is

$\left[O{H}^{-}\right]=\frac{\left(0.00242 mol\right)}{0.0415 L }= 0.058313 M$

$pOH= -\log \left[O{H}^{-}\right]= -\log 0.058313=1.2342$

$pH=14-1.2342=12.77$

d) Graph of $pH vs mL of NaOH added$ is

The above curve is a plot of $pH vs. mL of NaOH added$.

The volume of $NaOH$ at half equivalence point is $10.3 mL$, so the half equivalence point is around $5.15 mL$ of $NaOH$. If we draw a vertical line from $5.15 mL$ to the curve, it meets at around $9.89$. So, the $pH$ at half equivalence point is $9.89$. This value matches with the given $p{K}_a$ of $9.89$. At the half equivalence point, the conjugate base formed and the acid present are equal in concentrations, so the Henderson-Hasselbalch equation reduces to $pH=p{K}_a$.

So, at half equivalence point, the major species would be $C_6{H}_5-ONa and C_6{H}_5-OH$.

At the equivalence point, moles of acid initially present are equal to moles of added base. So, there will not be phenol or $NaOH$ present at the equivalence point. The salt of phenol, that is, sodium phenoxide, $\left(C_6{H}_5-ONa\right),$ is a major species that is present at the equivalence point.

At the end of the titration, the only major species present will be $NaOH$.

Conclusion:

In the titration of a weak acid and strong base, the resultant salt formed is a basic salt. Hence, the $pH$ at the equivalence point of the titration is greater than $7$.