Q19.59P - Chemistry: Molecular Nature Of Matter And Change

Question

Find the pH of the equivalence point(s) and the volume (mL) of $\mathbf{0}\mathbf{.}\mathbf{0588}\mathbf{\text{M}}\mathbf{ }\mathbf{ }\mathbf{\text{KOH}}$ needed to reach it in titrations of

(a) $\mathbf{23}\mathbf{.}\mathbf{4}\mathbf{ }\mathbf{\text{mL}}$ of $\mathbf{0}\mathbf{.}\mathbf{0390}\mathbf{ }\mathbf{\text{M}}\mathbf{ }\mathbf{ }\mathbf{\text{HNO}}\mathbf{2}$

(b) $\mathbf{17}\mathbf{.}\mathbf{3}\mathbf{ }\mathbf{\text{mL}}$ of $\mathbf{0}\mathbf{.}\mathbf{130}\mathbf{ }\mathbf{\text{M}}\mathbf{ }\mathbf{ }{\mathbf{\text{H}}}_{\mathbf{2}}{\mathbf{\text{CO}}}_{\mathbf{3}}$ (two equivalence points)

Step-by-Step Solution

Answer

a) The pH of the equivalence points and the volume (mL) of $0.0588\text{M} \text{KOH}$ needed to reach it in titrations of $23.4 \text{mL}$of $0.0390\text{M} {\text{HNO}}_2$:

  $$\begin{gathered} \text{V}=15,5 \text{ml } \text{KOH} \\ \text{pH}=7.76. \end{gathered}$$

b)The pH of the equivalence points and the volume (mL) of $0.0588 \text{M} \text{KOH}$ needed to reach it in titrations of $23.4 \text{mL}$ of $0.0390\text{M} {\text{HNO}}_2$:

$17.3 \text{mL}$ of $0.130 \text{M} {\text{H}}_2{\text{CO}}_3$ (two equivalence points):

  $$\begin{gathered} {\text{ V}}_1=38.2 \text{ml KOH} \\ \text{pH}=8.34 \\ {\text{V}}_2=76.4 \text{ml KOH} \\ \text{pH}=10.7. \end{gathered}$$

Step 1: Definition of pH

Solution's pH value, which measures the concentration of hydrogen ions, reveals whether it is acidic or alkaline.

Step 2: Find the pH of the equivalence point(s) and the volume (mL) of needed to reach it in titrations of 0 . 588   m  KOH needed to reach it in titrations of 23 . 4   mL of 0 . 0390   M     HNO 2

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) $0.0588\text{M} \text{KOH}$ and $23.4\text{ml} 0.0390 \text{M} {\text{HNO}}_2$.

First, we can calculate mols of  ${\text{HNO}}_2$:

 $0.023 \text{l}\times 0.0390 \text{M}=0.00091 \text{mol}$.

We may use the number of mols of ${\text{HNO}}_2$to compute the number of mols of KOH because these two chemicals react in a 1: 1 molar ratio.

 $$\begin{gathered} 1:1=0.00091 \text{mol}:x \\ x=0.00091\text{mol KOH} \end{gathered}$$

Now, we can calculate number of ml of KOH required to reach equivalence point:

 $$\begin{gathered} V=\frac{n}{c} \\ V=\frac{0.00091 \text{mol}}{0.0588\text{M}} \\ =15.5\text{ml KOH} \end{gathered}$$

We must now determine the pH value at the equivalence point:

 $$\begin{gathered} \mathrm{pH}=-\log \left[{\text{H}}_3{\text{O}}^{†}\right] \\ \mathrm{pH}=-\log \frac{\mathrm{Kw}}{\left[\text{OH}\right]} \end{gathered}$$

At the equivalence point all of the ${\text{HNO}}_2$ has transformed into ${\text{HNO}}_2{\text{NO}}_2$, its weak conjugate base, so we can apply calculation for weak base and Kb:

 $$\begin{gathered} \mathrm{Ka}\left({\text{HNO}}_2\right)=4\times {10}^4 \\ \mathrm{Kw}=1\times {10}^{14} \\ \left[\mathrm{OH}\right]=\sqrt{\frac{\mathrm{Kw}}{\mathrm{Ka}}\times \frac{0.0234 \mathrm{l}\times 0.0390 \mathrm{M}}{0.0234 \mathrm{l}+0.0155 \mathrm{l}}} \\ \mathrm{pH}=7.81. \end{gathered}$$

Therefore, the required $\mathrm{pH}=7.81$.

Step 3: Find the pH of the equivalence point(s) and the volume (mL) of needed to reach it in titrations of 0 . 588   m   KOH needed to reach it in titrations of 17 . 3   mL of 0 . 130   M     H 2 CO 3

b) 

First equivalence point:

$$\begin{gathered} 0.0173 \text{l}\times 0.130 \text{M}=0.002249 \text{mol} \\ 1:1=0.002249:x \\ x=0.002249 \text{mol KOH.} \end{gathered}$$

 

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

 $$\begin{gathered} V=\frac{n}{c} \\ V=38.2\text{ml NaOH} \end{gathered}$$

 

The pH value of the first equivalency point must now be calculated:

$$\begin{gathered} \mathrm{Ka}1=4.3\times {10}^7 \\ \mathrm{Ka}2=4.8\times {10}^{-11} \end{gathered}$$

pH calculation for amphoteric substances $\left({\text{HCO}}_3\right)$:

 $$\begin{gathered} \mathrm{pH}=\frac{1}{2}(\mathrm{pKa}1+\mathrm{pKa}2) \\ \mathrm{pH}=8.34 \end{gathered}$$

Second equivalence point:

The number of ml of KOH required in the second equivalence point is double that of the first equivalence point because the amount of mols of ${\text{H}}_2{\text{CO}}_3$ acid at the first equivalence point is the same as the number of mols of ${\text{HCO}}_3$ at the second equivalence point and we need exact same amount of KOH.

 $$\begin{gathered} V=38.2 \mathrm{ml}+38.2 \mathrm{ml} \\ =76.4 \mathrm{ml} \mathrm{NaOH} . \end{gathered}$$

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

 $$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}_3\mathrm{O}\right] \\ \mathrm{pH}=-\log \frac{\mathrm{Kw}}{\left[\mathrm{OH}\right]} \end{gathered}$$

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

$$\begin{gathered} \mathrm{Ka}2=4.8\times {10}^{11} \\ \mathrm{Kw}=1\times {10}^{14} \\ \left[\mathrm{OH}\right]=\sqrt{\frac{\mathrm{Kw}}{\mathrm{Ka}2}\times \frac{0.0173 \mathrm{l}\times 0.0130 \mathrm{M}}{0.0173 \mathrm{l}+0.0764 \mathrm{l}}} \\ \mathrm{pH}=10.7 \end{gathered}$$

 

Therefore, the required $\mathrm{pH}=10.7.$