Q19.43P - Chemistry: Molecular Nature Of Matter And Change

Question

The scenes below depict the relative concentrations of  ${\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{PO}}}_{\mathbf{\text{4}}}{\mathbf{\text{,H}}}_{\mathbf{\text{2}}}\mathbf{\text{P}}{{\mathbf{\text{O}}}_{\mathbf{\text{4}}}}^{\mathbf{\text{ - }}}$, and $\mathbf{\text{HP}}{{\mathbf{\text{O}}}_{\mathbf{\text{4}}}}^{\mathbf{\text{2 - }}}$  during a titration with aqueous  $\mathbf{\text{NaOH}}$, but they are out of order. (Phosphate groups are purple, hydrogens are blue, and ${\mathbf{\text{Na}}}^{\mathbf{\text{ + }}}$ ions and water molecules are not shown.) 

(a) List the scenes in the correct order. 

(b) What is the pH  in the correctly ordered second scene (see Appendix C)? 

(c) If it requires $\mathbf{10}\mathbf{.00}\mathbf{\text{ mL}}$  of the   solution to reach this scene, how much more is needed to reach the last scene?

Step-by-Step Solution

Answer

1. The correct order is  $\text{1.C 2.B 3.D 4.A}$.
2. The value of  $\text{pH = 2.14}$.
3. $\text{20 ml}$ is required from scene B to scene A.

Step 1: Solution

A solution is a mixture of two or more components that is homogenous. In any step, a solution can exist. A solution is made up of two components: a solute and a solvent. The substance that is dissolved in the solvent is referred to as the solute.

Step 2: Subpart (a)

 We can see that in jug C, there are six molecules of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ (purple dots with three blue dots around each), indicating that this is the first scene, as none of the ${\text{H}}^{\text{ + }}$ ions have reacted with $\text{NaOH}$.

• in jug B, there are three purple dots with three blue dots around each and three purple dots with two blue dots around each, indicating that half of the ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ has reacted with $\text{NaOH}$ and formed ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}$, which means this is the second scene.

           ${\mathrm{OH}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{aq}\right)>>{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\left(\mathrm{aq}\right)$

• Three purple dots with two blue dots around each and three purple dots with one blue dot around each can be found in jug D, indicating that half of the ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}$ has interacted with $\text{NaOH}$ to generate ${\text{HPO}}_{\text{4}}^{\text{2 - }}$, indicating that this is the third scene.

          ${\mathrm{OH}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\left(\mathrm{aq}\right)>>{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{HPO}}_4^{2-}\left(\mathrm{aq}\right)$

• Only purple dots with one blue dot around each may be seen in jug A, indicating that only ${\text{HPO}}_{\text{4}}^{\text{2 - }}$ ions are present, and this is the fourth scene.

Therefore, $\text{1.C 2.B 3.D 4.A}$ is the correct order.

Step 3: Subpart (b)

We have some ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ and some ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}$ in the second scene (scene B), which are conjugate acid-base pairs. The Henderson-Hasselbalch equation can be used to get the pH value:

 $\text{pH = pKa + log}\frac{\left[{\text{ A}}^{-}\right]}{\left[\text{HA}\right]}$

As we have three ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ molecules and three ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}$ molecules in this situation, the concentration ratio is one.

$$\begin{gathered} \mathrm{pH}=\mathrm{pKa}+\log 1 \\ \mathrm{pH}=\mathrm{pKa} \\ \mathrm{pKa}\left({\mathrm{H}}_3{\mathrm{PO}}_4\right)=2.14 \\ \mathrm{pH}=2.14 \end{gathered}$$

Therefore,  $\text{pH = 2.14}$.

Step 4: Subpart (c)

If $\text{10 ml}$ of $\text{NaOH}$ is required to reach scene B (scene with $3$ molecules of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$  and $3$ molecules of  ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}$), another $\text{10 ml}$  is required to reach scene D ( $3$ molecules of ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}$  and $3$  molecules of ${\text{HPO}}_{\text{4}}^{\text{2 - }}$), and another $\text{10 ml}$ is required to reach scene A. ( $\text{6}$ molecules of ${\text{HPO}}_{\text{4}}^{\text{2 - }}$). As a result, we'll need an extra $\text{20 ml}$ of $\text{NaOH}$ to get from scene B to scene A.