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Chapter 1: Problem 33

Which of the following could be added to an aqueous solution of weak acid HF to increase the percent dissociation? (A) \(\operatorname{NaF}(s)\) (B) \(\mathrm{H}_{2} \mathrm{O}(l)\) (C) \(\mathrm{NaOH}(\mathrm{s})\) (D) \(\mathrm{NH}_{3}(a q)\)

Short Answer

Expert verified
The substance that could be added to an aqueous solution of weak acid HF to increase the percent dissociation is \(\mathrm{NaOH}(\mathrm{s})\)

Step by step solution

01

Understand the Reaction of Weak Acid

HF is a weak acid which partially dissociates in water giving the following equilibrium reaction: \(HF(aq) \rightleftharpoons H^+(aq) + F^-(aq)\). In this case, anything that would encourage the dissociation (forward reaction) would increase the percent dissociation.
02

Analyze Each Substance

(A) \(\operatorname{NaF}(s)\): This is a salt of the weak acid. When in solution, it dissociates completely to give \(\operatorname{Na}^+\) and \(\operatorname{F}^-\) ions. According to Le Chatelier's principle, adding more \(\operatorname{F}^-\) ions will shift the equilibrium to the left, hindering further dissociation of HF. (B) \(\mathrm{H}_{2} \mathrm{O}(l)\): While water is the solvent in which HF is dissolved, adding more water does not necessarily change the percent dissociation as it doesn't change the equilibrium directly. (C) \(\mathrm{NaOH}(\mathrm{s})\): This is a strong base and will react with HF to consume the \(\mathrm{H}^+\) ions (NaOH + HF 鈫� NaF + H2O), which would shift the equilibrium to the right causing more HF to dissociate. (D) \(\mathrm{NH}_{3}(a q)\): Ammonia is a weak base and will react with \(\mathrm{H}^+\) ions to a lesser extent than NaOH, but it will still cause a shift in the equilibrium towards right and increase the dissociation of HF. However, the influence of NH3 is considerably less than NaOH.
03

Select the Right Substance

Considering all the analysis, NaOH(s) would cause the greatest increase in the percent dissociation of HF, because it is a strong base that would shift the equilibrium to the right, causing more dissociation of HF. Despite NH3 also causing a similar effect, it won't be as impactful as NaOH due to its weak base nature.

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Key Concepts

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

Weak Acid
A weak acid is a type of acid that doesn't completely dissociate in water. Unlike strong acids which fall apart completely when dissolved, weak acids only partially dissociate. This means that in a solution, only some of the weak acid molecules turn into their ions. For example, HF, or hydrofluoric acid, dissociates partially in water to produce hydrogen ions \(H^+\) and fluoride ions \(F^-\).The dissociation of a weak acid like HF in water can be represented as follows:
  • HF(aq) \(\rightleftharpoons\) H\(^+\)(aq) + F\(^-\)(aq).
Here, the double-headed arrow indicates that the reaction can go both ways 鈥� the forward reaction represents the acid dissociating into ions, while the reverse reaction shows the ions recombining to form the undissociated acid.The partial dissociation creates an equilibrium between the undissociated acid and the ions produced, which is why the concept of chemical equilibrium is crucial for understanding the behavior of weak acids.
Percent Dissociation
Percent dissociation is a measure of how much a weak acid dissociates in solution compared to the initial concentration. It's expressed as a percentage, showing the portion of acid molecules that dissociate into ions.- For an acid like HF, percent dissociation can be calculated using the formula:\[\text{Percent Dissociation} = \left(\frac{\text{Concentration of dissociated }H^+\text{ ions}}{\text{Initial concentration of HF}}\right) \times 100\%\]A higher percent dissociation indicates a stronger dissociation tendency, which generally means more hydrogen ions are produced.Factors that can increase percent dissociation include:
  • Reducing the concentration of \(H^+\) ions, as this shifts the equilibrium towards more dissociation.
  • Adding a strong base like NaOH, which reacts with \(H^+\) ions, effectively removing them and encouraging more acid molecules to dissociate into the solution.
Percent dissociation helps us understand how effective a weak acid is at ionizing in a given environment.
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemistry used to predict how a change in conditions can affect a chemical equilibrium. It states that if a dynamic equilibrium is disturbed by changing the conditions, the system adjusts itself to minimize the effect of that change and re-establish equilibrium.In the context of acid-base equilibrium, this principle can help us understand how the addition of various substances will affect the dissociation of a weak acid like HF. For example:
  • Adding \(F^-\) ions through a substance like NaF will shift the equilibrium to the left, reducing dissociation as the system attempts to offset the increase in \(F^-\) concentration.
  • Conversely, removing \(H^+\) ions by adding a base like NaOH shifts the equilibrium to the right, promoting the formation of more \(H^+\) ions through further dissociation of HF molecules.
By applying Le Chatelier's Principle, we can predict these shifts and manipulate reaction conditions to achieve the desired outcome, such as increasing the percent dissociation of a weak acid.
Chemical Equilibrium
Chemical equilibrium in an acid-base system refers to the state where the rate of the forward reaction equals the rate of the reverse reaction. This means that the concentrations of reactants and products remain constant over time, though they are not necessarily equal.For the weak acid HF, the equilibrium expression is:\[HF(aq) \rightleftharpoons H^+(aq) + F^-(aq)\]As equilibrium is reached, HF molecules dissociate into \(H^+\) and \(F^-\) at the same rate that these ions recombine to form HF, resulting in no net change in concentrations. Key factors tightly controlling equilibrium include:
  • Concentration of reactants and products.
  • Temperature of the system.
  • Presence of a catalyst or additional reacting substances.
Understanding chemical equilibrium is essential for predicting and controlling the behavior of weak acids and their ionizing patterns in various solutions, providing essential insights for chemists in both academic and practical applications.

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Most popular questions from this chapter

Which substance would have the highest boiling point? (A) Ethanol, because it is the most asymmetrical (B) Acetone, because of the double bond (C) Ethylene glycol, because it has the most hydrogen bonding (D) All three substances would have very similar boiling points because their molar masses are similar.

150 \(\mathrm{mL}\) of saturated \(\mathrm{SrF}_{2}\) solution is present in a 250 \(\mathrm{mL}\) beaker at room temperature. The molar solubility of \(\mathrm{SrF}_{2}\) at 298 \(\mathrm{K}\) is \(1.0 \times 10^{-3} \mathrm{M}\) . How could the concentration of \(\mathrm{Sr}^{2+}\) ions in solution be decreased? (A) Adding some \(\operatorname{NaF}(s)\) to the beaker (B) Adding some \(\operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}(s)\) to the beaker (C) By heating the solution in the beaker (D) By adding a small amount of water to the beaker, but not dissolving all the solid

A bottle of water is left outside early in the morning. The bottle warms gradually over the course of the day. What will happen to the pH of the water as the bottle warms? (A) Nothing; pure water always has a pH of 7.00. (B) Nothing; the volume would have to change in order for any ion concentration to change. (C) It will increase because the concentration of \(\left[\mathrm{H}^{+}\right]\) is increasing. (D) It will decrease because the auto-ionization of water is an endothermic process.

A solution of \(\mathrm{Co}^{2+}\) ions appears red when viewed under white light. Which of the following statements is true about the solution? (A) A spectrophotometer set to the wavelength of red light would read a high absorbance. (B) If the solution is diluted, the amount of light reflected by the solution will decrease. (C) All light with a frequency that is lower than that of red light will be absorbed by it. (D) Electronic transmissions within the solution match the wavelength of red light.

\(\begin{array}{ll}{\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-390 \mathrm{kJ} / \mathrm{mol}} \\\ {\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-290 \mathrm{kJ} / \mathrm{mol}} \\ {2 \mathrm{C}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)} & {\Delta H^{\circ}=+230 \mathrm{kJ} / \mathrm{mol}}\end{array}\) Based on the information given above, what is \(\Delta H^{\circ}\) for the following reaction? $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \\ \text { (A) }-1,300 \mathrm{kJ} \\ \text { (B) }-1,070 \mathrm{kJ} \\ \text { (C) }-840 \mathrm{kJ} \\ \text { (D) }-780 \mathrm{kJ} \end{aligned} $$
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