Question

Question:When a diprotic acid ${\mathbf{H}}_{\mathbf{2}}{\mathbf{A}}_{\mathbf{2}}$ is titrated with NaOH, the protons on the diprotic acid are generally removed one at a time, resulting in a pH curve that has the following generic shape:

a. Notice that the plot has essentially two titration curves. if the first equivalence point occurs at 100.0 mL NaOH added, what volume of NaOH added corresponds to the second equivalence point?

b. For the following volumes of NaOH added, list the major species present after the ${\mathbf{\text{OH}}}^{\mathbf{-}}$reacts completely

i. 0 mL NaOH added

ii. Between 0 and 100.0 mL of NaOH added

iii. 100.0 mL NaOH added

iv. Between 100.0 and 200.0 mL NaOH added

v 200.0 mL NaOH added

vi. After 200.0 mL NAOH added

c. If the pH at 50.0 mL NaOH added is 4.0 and the pH at 150.0 mL NaOH added is 8.0, determine the values ${\mathbf{K}}_{\mathbf{a}\mathbf{1}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{K}}_{\mathbf{a}\mathbf{2}}$the diprotic acid.

Short answer

a. The volume of NaOH added corresponds to the second equivalence point is 100 Ml.

b.

i. The major species present in the solution is${\mathrm{H}}_2\mathrm{A}$

ii. The major species present between 0 and 100.0 mL, NAOH added are${\mathrm{H}}_2\mathrm{A}$ and HA- and${\mathrm{H}}_2\mathrm{O}$

iii. The major species present is HA- and ${\mathrm{H}}_2\mathrm{O}$

iv. The major species present are ${\mathrm{HA}}^{-}$,${\mathrm{H}}_2\mathrm{O}$, ${\mathrm{A}}^{2-}$

v. The major species present are role="math" localid="1649140323190" ${\mathrm{Na}}^{+},{\mathrm{H}}_2\mathrm{O},{\mathrm{A}}^{2-}$

vi. The major species present in the solution are ${\mathrm{A}}^{2-},{\mathrm{Na}}^{+},{\mathrm{H}}_2\mathrm{O}\mathrm{and}{\mathrm{OH}}^{-}$

c. The values${\mathrm{K}}_{\mathrm{a}1}\mathrm{and}{\mathrm{K}}_{\mathrm{a}2}$ for the diprotic acid is:

${\mathrm{K}}_{\mathrm{a}1}=1.0\times {10}^{-4}$

${\mathrm{K}}_{\mathrm{a}2}=1.0\times {10}^{-8}$

Step-by-step solution

Definition of titration curve

A graphical representation of the pH of a solution during titration is defined as a titration curve.

Explanation

a. When a polyprotic acid is titrated with a base, the protons of the acid are removed one at a time.

$$\begin{gathered} {\mathrm{H}}_2\mathrm{A}+\mathrm{NaOH}\to \mathrm{NaHA}+{\mathrm{H}}_2\mathrm{O} \\ \mathrm{NaHA}+\mathrm{NaOH}\to {\mathrm{Na}}_2\mathrm{A}+{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

The first equivalence point is achieved when all ${\mathrm{H}}_2\mathrm{A}$ molecules are converted into HA- molecules. If there were X moles of ${\mathrm{H}}_2\mathrm{A}$at the first equivalence point, there are X moles of HA- present.

In the second equivalence point, these X moles of HA- are converted into X moles of ${\mathrm{A}}^2$The volume of NaOH that add corresponds to the second equivalence point is 100.0 Ml.

b.

i. 0 mL NaOH added

No NaOH is added; the major species present in the solution is ${\mathrm{H}}_2\mathrm{A}$.

ii. Between 0 and 100.0 mL of NaOH added

This is below the first equivalence point. Before the first equivalence point, ${\mathrm{H}}_2\mathrm{A}$is present, and when NaOH is added, ${\mathrm{H}}_2\mathrm{A}$is converted to HA- and ${\mathrm{H}}_2\mathrm{O}$. But not all ${\mathrm{H}}_2\mathrm{A}$has been reacted with NaOH. The major species present between 0 and 100.0 mL NAOH added are ${\mathrm{H}}_2\mathrm{A}$, HA- and ${\mathrm{H}}_2\mathrm{O}$.

iii. 100.0 mL NaOH added

This is the first equivalence point. At the first equivalence point, all ${\mathrm{H}}_2\mathrm{A}$ molecules has converted to HA-. So the major species present is HA- and ${\mathrm{H}}_2\mathrm{O}$.

iv. Between 100.0 and 200.0 mL NaOH added

This is in between the first equivalence point and the second equivalence point. Between the two equivalence points, added NaOH converts ${\mathrm{HA}}^{-}$ into ${\mathrm{A}}^{2-}$. So the major species present are ${\mathrm{HA}}^{-}$,${\mathrm{H}}_2\mathrm{O}$, ${\mathrm{A}}^{2-}$.

v. 200.0 mL NaOH added

This is the second equivalence point. At this point, all the ${\mathrm{HA}}^{-}$molecules have converted into ${\mathrm{A}}^{2-}$. So major species present are ${\mathrm{Na}}^{+},{\mathrm{H}}_2\mathrm{O},{\mathrm{A}}^{2-}.$

vi. After 200.0 mL NaOH added

This is beyond the second equivalence point. After the second equivalence point, no net reaction takes place. So the major species present in the solution are ${\mathrm{A}}^{2-},{\mathrm{Na}}^{+},{\mathrm{H}}_2\mathrm{O}\mathrm{and}{\mathrm{OH}}^{-}$.

c. When 50.0 mL NaOH is added, it is the halfway point to the first equivalence point. At this point $\left[{\mathrm{H}}_2\mathrm{A}\right]=\left[{\mathrm{HA}}^{-}\right]$

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}1}+\log \frac{\left[{\mathrm{HA}}^{-}\right]}{\left[{\mathrm{H}}_2\mathrm{A}\right]} \\ 4.0={\mathrm{pK}}_{\mathrm{a}1} \\ {\mathrm{K}}_{\mathrm{a}1}=\mathrm{antilog}(-4) \\ {\mathbf{K}}_{\mathbf{a}\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}} \end{gathered}$$

When 150.0 mL NaOH is added, it is the halfway point to the second equivalence point. At this point, $\left[{\mathrm{HA}}^{-}\right]=\left[{\mathrm{A}}^{2-}\right]$

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}2}+\log \frac{\left[{\mathrm{HA}}^{-}\right]}{\left[{\mathrm{H}}_2\mathrm{A}\right]} \\ 8.0={\mathrm{pK}}_{\mathrm{a}2} \\ {\mathrm{K}}_{\mathrm{a}2}=\mathrm{antilog}(-8) \\ {\mathbf{K}}_{\mathbf{a}\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}} \end{gathered}$$