Consider a solution prepared by mixing a weak acid HA, HCl, and NaA. Which of the following statements best describes what happens? a. The \(\mathrm{H}^{+}\) from the HCl reacts completely with the \(\mathrm{A}^{-}\) from the NaA. Then the HA dissociates somewhat. b. The \(\mathrm{H}^{+}\) from the HCl reacts somewhat with the A- from the NaA to make HA, while the HA is dissociating. Eventually you have equal amounts of everything. c. The \(\mathrm{H}^{+}\) from the HCl reacts somewhat with the \(\mathrm{A}^{-}\) from the NaA to make HA while the HA is dissociating. Eventually all the reactions have equal rates. d. The \(\mathrm{H}^{+}\) from the \(\mathrm{HCl}\) reacts completely with the \(\mathrm{A}^{-}\) from the NaA. Then the HA dissociates somewhat until "too much "H' and \(\mathrm{A}^{-}\) are formed, so the \(\mathrm{H}^{+}\) and \(\mathrm{A}^{-}\) react to form HA, and so on. Eventually equilibrium is reached. Justify your choice, and for choices you did not pick, explain what is wrong with them.

Short Answer

Expert verified
The correct statement is (a): The H+ from the HCl reacts completely with the A- from the NaA. Then the HA dissociates somewhat. This statement takes into account that HCl is a strong acid and dissociates completely, while HA is a weak acid that dissociates partially. The other statements do not properly account for the properties of strong and weak acids.

Step by step solution

01

Recall the properties of strong and weak acids

Strong acids, like HCl, dissociate completely in water, meaning they release all their hydrogen ions (H+) into the solution. Weak acids, like HA, only partially dissociate in water and establish an equilibrium between the acid and its conjugate base (A-).
02

Analyze the statements

The four given statements are all related to the behavior of H+, A-, HA, and HCl in the mixture. The correct statement will consider that HCl is a strong acid and dissociates completely in water. It will also consider that HA is a weak acid and dissociates partially, reaching an equilibrium with A-.
03

Choose the correct statement and justify the choice

The correct statement is (a), which says: The H+ from the HCl reacts completely with the A- from the NaA. Then the HA dissociates somewhat. This statement aligns with the properties of strong acids like HCl, which dissociate completely to release H+ ions into the solution. It also considers that weak acids like HA only partially dissociate.
04

Explain what's wrong with the other statements

• Statement (b) is wrong because it says the H+ from the HCl reacts somewhat with the A- from the NaA to make HA, while the HA is dissociating. This suggests equal amounts of all species, which isn't true because the HCl dissociates completely. • Statement (c) is wrong for the same reason as statement (b); it claims that H+ from the HCl reacts somewhat with the A- from the NaA while the HA is dissociating. Strong acids like HCl don't just react "somewhat," they dissociate completely. • Statement (d) is incorrect because it says the HA dissociates somewhat until "too much" H+ and A- are formed and then react to form HA. This suggests a dynamic between H+ and A- that doesn't accurately represent the behavior of strong and weak acids. In conclusion, statement (a) correctly describes the behavior of the mixture, considering that HCl is a strong acid and dissociates completely and that HA is a weak acid that only dissociates partially. The other statements fail to properly account for the properties of strong and weak acids.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the pH of the following solutions: a. 1.2\(M \mathrm{CaBr}_{2}\) b. 0.84$M \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}\left(K_{\mathrm{b}} \text { for } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}=3.8 \times 10^{-10}\right)$ c. 0.57$M \mathrm{KC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}} \text { for } \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}=6.4 \times 10^{-5}\right)$

Use the Lewis acid-base model to explain the following reaction. $$ \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3}(a q) $$

A \(0.20-M\) sodium chlorobenzoate $\left(\mathrm{NaC}_{7} \mathrm{H}_{4} \mathrm{ClO}_{2}\right)\( solution has \)\mathrm{a} \mathrm{pH}\( of \)8.65 .$ Calculate the \(\mathrm{pH}\) of a \(0.20-M\) chlorobenzoic acid \(\left(\mathrm{HC}_{7} \mathrm{H}_{4} \mathrm{ClO}_{2}\right)\) solution.

Are solutions of the following salts acidic, basic, or neutral? For those that are not neutral, write balanced equations for the reactions causing the solution to be acidic or basic. The relevant \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) values are found in Tables 14.2 and \(14.3 .\) $\begin{array}{ll}{\text { a. } \operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}} & {\text { d. } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{ClO}_{2}} \\ {\text { b. } \mathrm{NH}_{4} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}} & {\text { e. } \mathrm{NH}_{4} \mathrm{F}} \\ {\text { c. } \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{O} \mathrm{l}} & {\text { f. } \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{CN}}\end{array}$

Place the species in each of the following groups in order of increasing base strength. Give your reasoning in each case. a. \(\mathrm{IO}_{3}^{-}, \mathrm{BrO}_{3}^{-}\) b. \(\mathrm{NO}_{2}^{-}, \mathrm{NO}_{3}^{-}\) c. \(\mathrm{OCl}^{-}, \mathrm{OI}^{-}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free