91Ó°ÊÓ

A solution's $\mathbf{\text{pH}}$ value that measures the concentration of hydrogen ions, reveal whether a solution is acidic or alkaline.

We have two titrations with a strong base in this problem, one with monoprotic acid and one with diprotic acid. Because it is monoprotic, the first has one equivalent point, but the second has two equivalence points because it is a diprotic acid.

a)

$0.0372\text{²Ñ â¶Ä‰}\mathrm{NaOH}$and $42.2\text{³¾±ô â¶Ä‰}0.0520{\text{ ²Ñ â¶Ä‰C±á}}_3\text{COOH}$.

First, we can calculate the mols of the acetic acid:

$0.0422\text{l}×0.0520\text{M}=0.0022\text{mol}$.

Because these two substances react in molar ratio 1:1, we can use the number of moles of the acetic acid to calculate number of moles of NaOH:

$\begin{array}{c}1:1=0.0022\text{″¾´Ç±ô}:\mathrm{x}\\ \mathrm{x}=0.0022\text{″¾´Ç±ô NaOH}\end{array}$

Now, we can calculate number of $\text{ml}$of $\text{NaOH}$required to reach an equivalence point:

$\begin{array}{l}\mathrm{V}=\frac{\mathrm{n}}{\mathrm{c}}\\ \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}=\frac{0.0022\text{mol}}{0.0372\text{M}}\\ \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}=59.1\text{³¾±ô â¶Ä‰N²¹°¿±á.}\end{array}$

Now we have to calculate the pH value at the equivalence point:

$\begin{array}{c}\mathrm{pH}=−\log \left[{\mathrm{H}}_3{\mathrm{O}}^{†}\right]\\ \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰}=−\log \frac{\mathrm{Kw}}{\left[\mathrm{OH}\right]}.\end{array}$

Because all of the acetic acid has converted to the acetate ion—its weak conjugate base, at the equivalence point, we can use the calculation for a weak base and the Kb:

$\begin{array}{c}\mathrm{Ka}\text{ }\left(\text{²¹³¦±ð³Ù¾±³¦â€‰â¶Ä‰a³¦¾±»å}\right)=1.8×{10}^5\\ \mathrm{Kw}=1×{10}^{14}\\ \left[\mathrm{OH}\right]=\sqrt{\begin{array}{l}\frac{\mathrm{Kw}}{\mathrm{Ka}}×\frac{0.0422\mathrm{l}×0.0520\mathrm{M}}{0.0422\mathrm{l}+0.0591\mathrm{l}}\\ \end{array}}\\ \mathrm{pH}=8.54\end{array}$

Therefore, the required pH is 8.54.

b)

First equivalence point:

Because the acid and the base react in a 1:1 mol ratio when they reach the first equivalence point, we can first calculate ${\text{mols}}^2$of ${\text{H}}_2{\text{SO}}_3$:

$\begin{array}{c}0.0289\mathrm{l}×0.0850\mathrm{M}=0.0025\text{mol}\\ 1:1=0.0025:\mathrm{x}\\ \mathrm{x}=0.0025\text{molNaOH.}\end{array}$

Now, we can figure out how many mL of NaOH we'll need to reach the first equivalence point:

$\begin{array}{c}\mathrm{V}=\frac{\mathrm{n}}{\mathrm{c}}\\ \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}=67\text{ }\mathrm{ml}\text{ â¶Ä‰}\mathrm{NaOH}\end{array}$

Now, we have to calculate the pH value of the first equivalence point:

$\begin{array}{l}\mathrm{Ka}1=1.3×{10}^2\\ \mathrm{Ka}2=6×{10}^8\end{array}$

The pH calculation for amphoteric substances $\left({\text{HSO}}_3\right)$is done as follows:

Second equivalence point:

Because the number of moles of the sulfuric acid at the first equivalence point is the same as the number of moles of HSO at the second equivalence point, and we need exactly the same amount of NaOH, the number of ml of NaOH required in the second equivalence point is double the value of ml in the first equivalence point.

$\begin{array}{c}\mathrm{V}=67\mathrm{ml}+67\mathrm{ml}\\ =134\text{ }\mathrm{ml}\text{ â¶Ä‰}\mathrm{NaOH}\end{array}$

The pH value at the equivalence point must now be calculated.

$\begin{array}{c}\mathrm{pH}=−\log \left[{\mathrm{H}}_3{\mathrm{O}}^{†}\right]\\ \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰}=−\log \frac{\mathrm{Kw}}{\left[\mathrm{OH}\right]}.\end{array}$

At the equivalence point, all of the ${\text{HSO}}_3$has transformed into ${\text{SO}}_3^2$, its weak conjugate base, so we can apply the calculation for a weak base and the Kb:

$\begin{array}{c}\mathrm{Ka}2=6×{10}^8\\ \mathrm{Kw}=1×{10}^{14}\\ \left[\mathrm{OH}\right]=\sqrt{\frac{\mathrm{Kw}}{\mathrm{Ka}2}×\frac{0.0289\mathrm{l}×0.0850\mathrm{M}}{0.0289\mathrm{l}+0.134\mathrm{l}}}\\ \mathrm{pH}=8.33.\end{array}$

Therefore, the required pH is $8.33$.