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The fluoride ion reacts with water to produce HF. (a) Write out the chemical equation for this reaction. (b) Will a concentrated solution of NaF in water be acidic, basic, or neutral? (c) Is fluoride acting as a Lewis acid or as a Lewis base when reacting with water?

Short Answer

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(a) The balanced chemical equation for the reaction between fluoride ion (F-) and water (H2O) is: \(F^- + H_2O \rightarrow HF + OH^-\). (b) A concentrated solution of NaF in water will be basic due to the presence of hydroxide ions (OH-). (c) Fluoride ion (F-) is acting as a Lewis base in this reaction, as it donates an electron pair to form a bond with the hydrogen ion (H+) from water.

Step by step solution

01

Write the balanced chemical equation for the reaction

The given reaction is between fluoride ions (F-) and water (H2O) to produce hydrogen fluoride (HF). The balanced chemical equation for this reaction is: F- + H2O -> HF + OH- The fluoride ions combine with a hydrogen ion from water, forming hydrogen fluoride and leaving hydroxide ions.
02

Determine if the resulting solution is acidic, basic, or neutral

The presence of an excess of either hydrogen ions (H+) or hydroxide ions (OH-) determines whether the solution is acidic, basic or neutral. In our given reaction, we can see that hydroxide ions (OH-) are produced, which are basic in nature. Therefore, a concentrated solution of NaF in water will be basic.
03

Identify if fluoride ion is acting as a Lewis acid or Lewis base in this reaction

A Lewis acid is a substance that can accept an electron pair, while a Lewis base is a substance that can donate an electron pair. In our reaction, F- donates an electron pair to form a bond with the hydrogen ion (H+) from water. As fluoride ion is donating an electron pair, it is acting as a Lewis base in this reaction.

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Key Concepts

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

Chemical Equation
Understanding chemical reactions often begins with a balanced chemical equation, which offers a clear schematic of how substances interact. In the textbook exercise, fluoride ion (F-) reacts with water (H2O) to produce hydrogen fluoride (HF) and hydroxide ion (OH-), represented by the equation \[ F^- + H_2O \rightarrow HF + OH^- \]. Let's break down this equation. The fluoride ion, carrying a negative charge, interacts with a neutral water molecule. This leads to the creation of hydrogen fluoride, a molecule consisting of hydrogen and fluorine, along with a hydroxide ion that remains in solution. The equation signifies that for every fluoride ion, one hydroxide ion is produced, ensuring charge balance. This chemistry is pivotal in explaining the behavior of fluoride in water and has practical significance in fields such as dentistry and water treatment.

To assist students further, referencing atomic charges and the conservation of charges and atoms might provide a clearer insight into why the ions combine in such a manner. Visual aids or illustrations depicting the ion interactions before and after the reaction could also enhance comprehension.
Acidic Basic Neutral Solutions
When a chemical species like fluoride ion reacts with water, it can disrupt the balance of hydrogen ions (H+) and hydroxide ions (OH-) in the solution, which in turn affects its pH. But what tells us about the acidity or basicity of the solution? It's the presence of excess hydrogen ions or hydroxide ions. In our discussed reaction, the production of OH- indicates an increase in the solution's basicity. Therefore, a concentrated solution of sodium fluoride (NaF) in water will have a higher concentration of OH- ions, making it basic in nature. Had the reaction resulted in an excess of H+ ions, the solution would have been acidic. If neither ion were present in excess, the solution would be neutral. For a student grappling with this concept, it might be useful to examine the pH scale, which ranges from 0 (acidic) to 14 (basic), with 7 being neutral. The importance of understanding solutions' acidity or basicity is vast – from biological systems to industrial processes, it shapes outcomes and reactions.
Lewis Acid Base Theory
Going beyond the traditional definition of acids and bases that focuses on H+ and OH-, the Lewis acid-base theory provides a broader perspective. It defines an acid as an electron pair acceptor and a base as an electron pair donor. This theory is highly useful for reactions where no protons are transferred.

In the textbook exercise, the fluoride ion (F-) donates an electron pair to form a covalent bond with a hydrogen ion (H+) from water, a process not involving direct proton transfer. From this perspective, F- is a Lewis base because it is providing an electron pair. A helpful approach to make this concept more accessible is to use visual diagrams showing electron pair donation and acceptance, providing a visual representation of the Lewis theory in action. By distinguishing between the behavior of molecules and ions in reactions beyond mere proton transfer, students can expand their understanding of complex chemical reactions.

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Most popular questions from this chapter

If a neutral solution of water, with \(\mathrm{pH}=7.00\) , is cooled to \(10^{\circ} \mathrm{C},\) the ph rises to \(7.27 .\) Which of the following three statements is correct for the cooled water: (i) \(\left[\mathrm{H}^{+}\right]>\left[\mathrm{OH}^{-}\right],\) (ii) \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right],\) or (iii) \(\left[\mathrm{H}^{+}\right]<\left[\mathrm{OH}^{-}\right] ?\)

(a) Using dissociation constants from Appendix D, determine the value for the equilibrium constant for each of the following reactions. \((\mathrm{i}) \mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{co}_{3^{-}(a q)}+\mathrm{H}_{2} \mathrm{O}(l)\) (ii) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \rightleftharpoons \mathrm{NH}_{3}(a q)+\mathrm{HCO}_{3}^{-}(a q)\) (b) We usually use single arrows for reactions when the for- ward reaction is appreciable \((K\) much greater than 1\()\) equilibrium is never established. If we follow this convention, which of these equilibria might be written with a single arrow?

The average \(\mathrm{pH}\) of normal arterial blood is \(7.40 .\) At normal body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{w}=2.4 \times 10^{-14} .\) Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right],\) and \(\mathrm{pOH}\) for blood at this temperature.

Benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) and aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\) are both derivatives of benzene. Benzoic acid is an acid with \(K_{a}=6.3 \times 10^{-5}\) and aniline is a base with \(K_{a}=4.3 \times 10^{-10} .\) (a) What are the conjugate base of benzoic acid and the conjugate acid of aniline? (b) Anilinium chloride \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\right)\) is a strong electrolyte that dissociates into anilinium ions \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\right)\) and chloride ions. Which will be more acidic, a 0.10\(M\) solution of benzoic acid or a 0.10 M solution of anilinium chloride? (c) What is the value of the equilibrium constant for the following equilibrium? $$\begin{array}{c}{\mathrm{C}_{6} \mathrm{H}_{5} \operatorname{COOH}(a q)+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}(a q) \rightleftharpoons} \\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad {\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}(a q)}\end{array}$$

Ammonia, \(\mathrm{NH}_{3},\) acts as an Arrhenius base, a Bronsted-Lowry base, and a Lewis base, in aqueous solution. Write out the reaction \(\mathrm{NH}_{3}\) undergoes with water and explain what properties of ammonia correspond to each of the three definitions of "base."

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