Q19.56P - Chemistry: Molecular Nature Of Matter And Change

Question

Find the $\mathbf{\text{pH}}$ during the titration of $\mathbf{\text{20.00 mL}}$ of $\mathbf{\text{0.1000M}}$butanoic acid, ${\mathbf{\text{CH}}}_{\mathbf{\text{3}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}\mathbf{\text{COOH}}\mathbf{\left(}{\mathbf{\text{K}}}_{\mathbf{\text{a}}}{\mathbf{\text{=1.54×10}}}^{\mathbf{\text{-5}}}\mathbf{\right)}$ with $\mathbf{\text{1000MNaOH}}$  solution after the following additions of titrant:

(a)  $\mathbf{\text{0 mL}}$

(b)  $\mathbf{\text{10.00 mL}}$

(c)  $\mathbf{\text{15.00 mL}}$

(d)  $\mathbf{\text{19.00 mL}}$

(c)  $\mathbf{\text{19.95 mL}}$

(f)  $\mathbf{\text{20.00 mL}}$

(g)  $\mathbf{\text{20.05 mL}}$

(h)  $\mathbf{\text{25.00 mL}}$

Step-by-Step Solution

Answer

The required pH values during the titration of  $\text{20.00 mL}$of  $\text{0.1000M}$ butanoic acid are:

$\begin{array}{l}\text{a) }2.91\\ \text{b) }4.81\\ \text{c) }5.29\\ \text{d) }6.09\\ \text{e) }7.41\\ \text{f) }8.76 \\ \text{g})10.10\\ \text{h) }12.05\end{array}$

Step 1: Concept introduction

The pH of water is a measurement of how acidic or basic it is. The scale runs from 0 to 14, with 7 being the neutral value. Acidity is indicated by a pH less than 7, while a pH greater than 7 indicates a base. A pH is a measurement of the proportion of free hydrogen and hydroxyl ions in water.

Step 2: Calculation of pH using the Henderson-Hasselbalch equation

a. We have a weak acid that is titrated by a strong base in this problem. To solve this problem, we must first calculate the weak acid, then calculate the buffer solution using the Henderson-Hasselbalch equation, then calculate the weak base at the equivalence point, followed by the calculation of the strong base excess. 

 ${\text{20ml0.1MCH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{COOH}}^{\text{-1}}\text{ and 0.1MNaOHKa = 1.54}$$\backslash \mathrm{times}10\{\wedge \{-5\left\}\right\}.$  We only have the butanoic acid in the solution because no strong base has been added, thus we must utilize the weak acid equation. calculation with $\text{pH}$ :

 $\begin{array}{c}\mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\\ =-\log \sqrt{\mathrm{Ka}\times \left[\mathrm{HA}\right]}\\ =-\log \sqrt{1.54\times {10}^{-5}\times 0.1\mathrm{M}}\\ =\mathrm{2.91.}\end{array}$

Therefore, the pH value is  $2.91$.

Step 3: Calculating pH using the Henderson-Hasselbalch equation

b. We added  $10\text{\hspace{0.17em}ml}$ of a strong base, thus the weak acid is greater than  ,and we can compute the pH using the Henderson-Hasselbalch equation:

 $\begin{array}{c}\text{pH = -logKa+log}\frac{\text{nNaOH}}{\text{nHBu}}\\ {\text{ = -log1.54×10}}^{\text{-5}}\text{+log}\frac{\text{0.01l×0.1M}}{\text{0.02l×0.1M-0.01l×0.1M}}\\ \text{ = 4.81.}\end{array}$

Hence, the pH value is  $\mathrm{4.81.}$

Step 4: Calculating pH using the Henderson-Hasselbalch equation

c. We added $15\mathrm{ml}$ of a strong base, thus the weak acid is greater than  $5\mathrm{ml}$, and we can compute the pH using the Henderson-Hasselbalch equation:

 $\begin{array}{c}\mathrm{pH}=-\mathrm{logKa}+\log \frac{\mathrm{nNaOH}}{\mathrm{nHBu}}\\ =-\log 1.54\times {10}^{-5}+\log \frac{0.015\mathrm{l}\times 0.1\mathrm{M}}{0.02\mathrm{l}\times 0.1\mathrm{M}-0.015\mathrm{l}\times 0.1\mathrm{M}}\\ =\mathrm{5.29.}\end{array}$

Therefore, the pH value is  $\mathrm{5.29.}$

Step 5: Calculating pH using the Henderson-Hasselbalch equation

d. We added $19\text{\hspace{0.17em}ml}$ of a strong base, thus the weak acid is greater than  $1\text{\hspace{0.17em}ml}$,and we can compute the pH using the Henderson-Hasselbalch equation:

$\begin{array}{c}\text{pH\hspace{0.17em}= -logKa+log}\frac{\text{nNaOH}}{\text{nHBu}}\\ {\text{ = -log1.54×10}}^{\text{-5}}\text{+log}\frac{\text{0.019l×0.1M}}{\text{0.02l×0.1M-0.019l×0.1M}}\\ \text{ = 6.09.}\end{array}$

 

Such that, the pH value is  $6.09$.

Step 6: Calculating pH using the Henderson-Hasselbalch equation

e. 

We added $19.95\text{\hspace{0.17em}ml}$  of a strong base, therefore the weak acid is greater than $0.05\text{\hspace{0.17em}ml}$ , and we can determine the pH using the Henderson-Hasselbalch equation:

 $\begin{array}{c}\text{pH = -logKa+log}\frac{\text{nNaOH}}{\text{nHBu}}\\ {\text{ = -log1.54×10}}^{\text{-5}}\text{+log}\frac{\text{0.01995l×0.1M}}{\text{0.02l×0.1M-0.01995l×0.1M}}\\ \text{ = 7.41.}\end{array}$

Therefore, the pH value is $7.41$ .

Step 7: Calculating pH using the Henderson-Hasselbalch equation:

f.  We added $20\text{\hspace{0.17em}ml}$ of a strong base, and the equivalence point was attained because the molarities and the volumes were the same. We can answer this problem by first computing the $\text{pOH.}$ At this point, all of the weak acid has been transformed to a weak base, and we only have a weak base in the solution.

$\begin{array}{l}\text{ Total V value = 0.02l+0.02l}\\ \text{ = 0.04}\\ \\ \left[{\text{OH}}^{\text{-}}\right]\text{ = }\sqrt{\text{Kb×}\frac{\text{0.02l×0.1M}}{\text{0.04l}}}\\ \text{ =}\sqrt{\frac{\text{Kw}}{\text{Ka}}\text{×}\frac{\text{0.02l×0.1M}}{\text{0.04l}}}{\text{=5.7×10}}^{\text{-6}}\\ \\ \text{pH = -log}\frac{\left[{\text{OH}}^{\text{-}}\right]}{\text{Kw}}\\ \text{ = 8.76.}\end{array}$

As a result, the pH value is  $8.76$.

Step 8: Calculating PH value

g. 

We are past the equivalence point in this example, and we added a  $20.05\text{\hspace{0.17em}ml}$ of a strong base, therefore, the strong base is in excess, and we solve the problem by computing the  $\left[{\text{OH}}^{-}\right]:$

 $\begin{array}{c}\text{TotalVvalue = 0.02l+0.02005l}\\ \text{ = 0.04005\hspace{0.17em}l.}\\ \\ \left[{\text{OH}}^{\text{-}}\right]\text{ = }\frac{\text{0.02005l×0.1M-0.02l×0.1M}}{\text{0.04005l}}\\ \text{=0.000124\hspace{0.17em}M}\\ \\ \text{pH = -log}\frac{\text{Kw}}{\left[{\text{OH}}^{\text{-}}\right]}\\ \text{=10.10.}\end{array}$

Therefore, the pH value is  $10.10$.

Step 9: Calculating pH using the Henderson-Hasselbalch equation

h.

We are past the equivalence point in this example, and we added $25\text{\hspace{0.17em}ml}$ of a strong base, therefore, the strong base is in excess. We solve this problem by calculating the  $\left[{\text{OH}}^{-}\right]:$

 

$\begin{array}{c}\text{Total\hspace{0.17em}\hspace{0.17em}value = 0.02l+0.025l}\\ \text{ = 0.045l}\\ \\ \left[{\text{OH}}^{\text{-}}\right]\text{ = }\frac{\text{0.025l×0.1M-0.02l×0.1M}}{\text{0.045l}}\\ \text{= 0.0111M.}\\ \\ \text{pH = -log}\frac{\text{Kw}}{\left[{\text{OH}}^{\text{-}}\right]}\\ \text{= 12.05.}\end{array}$

 

Hence, the pH value is $12.05$.