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Chapter 15: Problem 1

What are the major species in solution after \(\mathrm{NaHSO}_{4}\) is dissolved in water? What happens to the \(\mathrm{pH}\) of the solution as more \(\mathrm{NaHSO}_{4}\) is added? Why? Would the results vary if baking soda \(\left(\mathrm{NaHCO}_{3}\right)\) were used instead?

Short Answer

Expert verified
When NaHSO4 is dissolved in water, the major species in the solution are Na+, HSO4-, H+ (due to HSO4-'s weak acid dissociation), and SO42-. As more NaHSO4 is added, the solution becomes more acidic due to the increased concentration of H+ ions from HSO4- dissociation. In contrast, if NaHCO3 were used, the solution would be less acidic because HCO3- has a weaker acid dissociation and also acts as a buffer by functioning as a weak base. Thus, the pH would not decrease as much when more NaHCO3 is added compared to NaHSO4.

Step by step solution

01

Dissolution of NaHSO4 in water

First, let's understand what happens when NaHSO4 is dissolved in water. The compound dissociates into ions as follows: \[ \mathrm{NaHSO}_{4} \longrightarrow \mathrm{Na}^{+} + \mathrm{HSO}_{4}^{-} \]
02

Analyze the dissociation of HSO4- in water

The HSO4- ion can also dissociate in water, functioning as a weak acid. HSO4- releases a hydrogen ion (H+) and forms the sulfate ion (SO42-) as shown below: \[ \mathrm{HSO}_{4}^{-} \longleftrightarrow \mathrm{H}^{+} + \mathrm{SO}_{4}^{2-} \]
03

Determine the major species in solution

Considering the above dissociations, when NaHSO4 is dissolved in water, the major species in the solution will be Na+, HSO4-, H+ (due to the dissociation of HSO4-), and SO42-.
04

Effect of adding more NaHSO4 on the pH of the solution

When more NaHSO4 is added to the solution, the concentration of HSO4- ions will increase. As HSO4- acts as a weak acid, it will increase the hydrogen ion concentration (H+), thereby lowering the pH of the solution. In other words, the solution becomes more acidic with the addition of more NaHSO4.
05

Compare the results with NaHCO3

Now, let's analyze the case with NaHCO3 (baking soda). When NaHCO3 is dissolved in water, it dissociates into ions as shown below: \[ \mathrm{NaHCO}_{3} \longrightarrow \mathrm{Na}^{+} + \mathrm{HCO}_{3}^{-} \] The HCO3- ion can also dissociate in water, functioning as a weak acid, forming CO32- (carbonate ion) and releasing a hydrogen ion (H+) as shown below: \[ \mathrm{HCO}_{3}^{-} \longleftrightarrow \mathrm{H}^{+} + \mathrm{CO}_{3}^{2-} \] However, it can also function as a weak base by accepting a hydrogen ion and converting back to H2CO3 (carbonic acid), thus acting as a buffer to changes in pH. In conclusion, if NaHCO3 were used instead of NaHSO4, the solution would be less acidic as HCO3- possesses buffering effects and its acid dissociation is weaker than that of HSO4-. The pH of the solution would not decrease as much when more NaHCO3 is added compared to the case with NaHSO4.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acid-Base Reactions
Understanding acid-base reactions is crucial for interpreting chemical processes in solutions. These reactions are exchanges where acids donate protons (hydrogen ions, H+) to bases. In the dissolution of NaHSO4, this compound dissociates in water releasing a Na+ ion and an HSO4- ion. The HSO4- can further act as a weak acid and release an H+, increasing the hydrogen ion concentration in the solution.

As more NaHSO4 is added to the solution, the acidity increases due to the release of additional H+ ions from HSO4-. This is a classic example of an acid-base reaction driving a change in the chemical composition of a solution and ultimately affecting its pH.
pH Level
The pH level is a measure of how acidic or basic a solution is, on a scale from 0 to 14, with 7 being neutral. pH is logarithmically related to the concentration of hydrogen ions in a solution, meaning each unit change in pH corresponds to a tenfold change in H+ concentration. When NaHSO4 is dissolved in water, the pH will decrease as the concentration of H+ ions increases, indicating the solution is becoming more acidic. If more NaHSO4 is added, the pH will continue to decrease. This occurs because HSO4- ions are able to donate H+ ions, contributing to the acidity of the solution.

Conversely, using NaHCO3 (baking soda), which has a buffering effect, the pH does not change as dramatically with the addition of more substance. This is due to the bicarbonate ion's (HCO3-) ability to react as both an acid and a base, moderating the pH level of the solution.
Buffer Solutions
A buffer solution resists changes in its pH when acids or bases are added. Buffers typically consist of a weak acid and its conjugate base, or a weak base and its conjugate acid. In the context of our exercise, when NaHCO3 is added to water, it forms a buffer system with its conjugated acids and bases (HCO3- and CO32-). This buffer system can absorb excess H+ or donate H+ when needed, maintaining the pH level of the solution.

Hence, compared to NaHSO4, which solely acts as an acid and lowers the pH, NaHCO3 creates a buffering environment that can counteract pH changes, demonstrating how buffer solutions play a significant role in maintaining chemical equilibrium within a system.

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Most popular questions from this chapter

Acid-base indicators mark the end point of titrations by "magically" turning a different color. Explain the "magic" behind acid-base indicators.

The active ingredient in aspirin is acetylsalicylic acid. A 2.51-g sample of acetylsalicylic acid required \(27.36 \mathrm{~mL}\) of \(0.5106 M \mathrm{NaOH}\) for complete reaction. Addition of \(13.68 \mathrm{~mL}\) of \(0.5106 M \mathrm{HCl}\) to the flask containing the aspirin and the sodium hydroxide produced a mixture with \(\mathrm{pH}=3.48\). Determine the molar mass of acetylsalicylic acid and its \(K_{\mathrm{a}}\) value. State any assumptions you must make to reach your answer.

a. Calculate the \(\mathrm{pH}\) of a buffered solution that is \(0.100 \mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\) (benzoic acid, \(K_{\mathrm{a}}=6.4 \times 10^{-5}\) ) and \(0.100 \mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{Na}\). b. Calculate the \(\mathrm{pH}\) after \(20.0 \%\) (by moles) of the benzoic acid is converted to benzoate anion by addition of a strong base. Use the dissociation equilibrium $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}^{+}(a q) $$ to calculate the \(\mathrm{pH}\). c. Do the same as in part b, but use the following equilibrium to calculate the \(\mathrm{pH}\) : \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{OH}^{-}(a q)\) d. Do your answers in parts \(\mathrm{b}\) and \(\mathrm{c}\) agree? Explain.

Consider the following two acids: $$ \mathrm{p} K_{\mathrm{a}_{1}}=2.98 ; \mathrm{p} K_{\mathrm{a}_{2}}=13.40 $$ $$ \begin{array}{l} \mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} \\ \text { Adipic acid } \quad \mathrm{p} K_{\mathrm{a}_{1}}=4.41 ; \mathrm{p} K_{\mathrm{a}_{2}}=5.28 \end{array} $$ In two separate experiments the pH was measured during the titration of \(5.00\) mmol of each acid with \(0.200 M \mathrm{NaOH}\). Each experiment showed only one stoichiometric point when the data were plotted. In one experiment the stoichiometric point was at \(25.00 \mathrm{~mL}\) added \(\mathrm{NaOH}\), and in the other experiment the stoichiometric point was at \(50.00 \mathrm{~mL} \mathrm{NaOH}\). Explain these results. (See Exercise \(113 .\) )

Could a buffered solution be made by mixing aqueous solutions of \(\mathrm{HCl}\) and \(\mathrm{NaOH}\) ? Explain. Why isn't a mixture of a strong acid and its conjugate base considered a buffered solution?
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