Question

Consider a solution prepared by mixing equal moles of a weak acid HA, HCl, and NaA. Which of the following best describes what happens?

a. The${\mathbf{\text{H}}}^{\mathbf{\text{+}}}$ from the$\mathbf{\text{HCl}}$ reacts completely with the ${\mathbf{\text{A}}}^{\mathbf{\text{-}}}$from the$\mathbf{\text{NaA}}$ . Then the $\mathbf{\text{HA}}$dissociates to some extent.

b. The ${\mathbf{\text{H}}}^{\mathbf{\text{+}}}$from the $\mathbf{\text{HCl}}$ reacts with the ${\mathbf{\text{A}}}^{\mathbf{\text{-}}}$ from the$\mathbf{\text{NaA}}$ to make , whereas the $\mathbf{\text{HA}}$ is dissociating. Eventually you have equal amounts of everything.

c. The ${\mathbf{\text{H}}}^{\mathbf{\text{+}}}$ from the $\mathbf{\text{HCl}}$ reacts with the ${\mathbf{\text{A}}}^{\mathbf{\text{-}}}$ from the $\mathbf{\text{NaA}}$ to make , whereas the$\mathbf{\text{HA}}$ is dissociating. Eventually all the reactions have equal rates.

d. The ${\mathbf{\text{H}}}^{\mathbf{\text{+}}}$ from the $\mathbf{\text{HCl}}$reacts completely with the${\mathit{A}}^{\mathbf{-}}$ from the$\mathbf{\text{NaA}}$ . Then the HA dissociates until “too much” ${\mathbf{\text{H}}}^{\mathbf{\text{+}}}$and are formed, so the ${\mathbf{\text{H}}}^{\mathbf{\text{+}}}$ and ${\mathbf{\text{A}}}^{\mathbf{\text{-}}}$ react to form$\mathbf{\text{HA}}$ , and so on. Eventually equilibrium is reached.

Justify the best choice. For those you did not choose, explain why they are incorrect.

Short answer

The best choice is explained in d.

Step-by-step solution

Conceptual Introduction

Any acid that is not entirely dissociated in a solution is a weak acid. The amount of dissociation determines how powerful an acid is; the more the dissociation, the stronger the acid.

Explanation for part a

This explanation does a good job of summarising the issue, but since part d's explanation specifies that thedissociates until equilibrium is attained, part d's explanation is more accurate than part a’sand, thus, superior.

Therefore, the best method is not explained in option a.

Explanation for part b

This is an inadequate description of the scenario, since$\text{HCl}$is a strong acid that totally dissociates and reacts with the${\text{A}}^{\text{-}}$in$\text{NaA}$,to create$\text{HA}$up to the point at which either$\text{HCl}$or$\text{NaA}$are exhausted.

Therefore, option b is not the best method explained.

 Step 4: Explanation for part c

The fact that$\text{HCl}$ is a strong acid that totally dissociates and reacts with the${\text{A}}^{\text{-}}$ in$\text{NaA}$to create$\text{HA}$ ,until either$\text{HCl}$or$\text{NaA}$are exhausted, makes this a poor description of the situation.

Therefore, option c is not the best method explained.

Explanation for part d

The method is best explained by this, in fact. Since$\text{HCl}$is a strong acid, it totally dissociates and creates the weak acid$\text{HA}$while leaving the${\text{NA}}^{\text{+}}$and${\text{Cl}}^{\text{-}}$ions in the solution. The weak acid$\text{HA}$will then begin to dissociate partiallyuntil equilibrium is established.

Therefore, option d is the best method explained.