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The equilibrium constant for the reaction \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{NH}_{3}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)\) is \(5.6 \times 10^{-10}\) (a) Is a solution of ammonium ion very acidic or only slightly acidic? (b) Is water acting as an acid or a base according to the Bronsted-I owry definition? Explain.

Short Answer

Expert verified
(a) A solution of ammonium ion is slightly acidic, as the pH of the solution depends on the initial concentration of ammonium ion and the small equilibrium constant K鈧 shows that very few NH鈧勨伜 ions undergo the reaction to form acidic H鈧僌鈦 ions. (b) In this reaction, water is acting as a base according to the Bronsted-Lowry definition, as it accepts a proton from the ammonium ion to form hydronium ion (H鈧僌鈦).

Step by step solution

01

Write down the reaction and its equilibrium constant expression

The given reaction is: \[ NH_{4}^{+}(aq)+H_{2}O(l) \rightleftarrows NH_{3}(aq)+H_{3}O^{+}(aq) \] The equilibrium constant expression for this reaction is: \[ K_a = \frac{[NH_{3}][H_{3}O^{+}]}{[NH_{4}^{+}]} \]
02

Establish the relationship between equilibrium concentration and initial concentration given by ICE table

Let the initial concentration of NH鈧勨伜 be C鈧. Initially, the concentrations of all species in the reaction will be: [ NH鈧勨伜 ] = C鈧 [ NH鈧 ] = 0 [ H鈧僌鈦 ] = 0 At equilibrium, the concentrations will change according to the stoichiometry of the reaction: [ NH鈧勨伜 ] = C鈧 - x [ NH鈧 ] = x [ H鈧僌鈦 ] = x Using these expressions, we can rewrite the equilibrium constant expression: \[ K_a = \frac{x^2}{C_{0}-x} \]
03

Solve for x, the concentration of H鈧僌鈦 ions at equilibrium

We're given the value of K鈧 as \(5.6 * 10^{-10}\). Therefore: \[ 5.6 \times 10^{-10} = \frac{x^2}{C_{0}-x} \] As K鈧 is very small, we can assume that x is much smaller than C鈧, so we can approximate \(C_{0}-x \approx C_{0}\): \[ 5.6 \times 10^{-10} = \frac{x^2}{C_{0}} \] Solving for x, we get: \[ x = \sqrt{5.6 \times 10^{-10} \times C_{0}} \] This x value represents the concentration of H鈧僌鈦 ions in the solution at equilibrium.
04

Determine the pH of the solution

To determine the pH of the solution, we can use the formula: \[ pH = -\log_{10} [H_{3}O^{+}] \] Plugging in the value of x, we get the pH in terms of the initial concentration of C鈧: \[ pH = -\log_{10} \sqrt{5.6 \times 10^{-10} \times C_{0}} \]
05

Assess the acidity of the ammonium ion solution

The pH of the solution is dependent on the initial concentration of ammonium ion. However, as we can see from the formula, the equilibrium constant K鈧 is very small. This indicates that only very few NH鈧勨伜 ions in the solution undergo the reaction and form acidic H鈧僌鈦 ions. Therefore, we can conclude that a solution of ammonium ion is slightly acidic.
06

Determine if water is acting as an acid or a base according to the Bronsted-Lowry definition

In the reaction, water accepts a proton from the ammonium ion to form hydronium ion (H鈧僌鈦): \[ NH_{4}^{+}(aq)+H_{2}O(l) \rightleftarrows NH_{3}(aq)+H_{3}O^{+}(aq) \] According to the Bronsted-Lowry definition, a base is a substance that accepts a proton, while an acid is a substance that donates a proton. Since water accepts the proton from the ammonium ion, it acts as a base in this reaction. In conclusion: (a) A solution of ammonium ion is slightly acidic. (b) In this reaction, water is acting as a base according to the Bronsted-Lowry definition.

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Key Concepts

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

Bronsted-Lowry theory
The Bronsted-Lowry theory is a fundamental concept in understanding acid-base reactions. According to this theory, an acid is a substance that donates a proton (H鈦 ion), while a base is a substance that accepts a proton. In the context of the reaction between ammonium ion (\( \text{NH}_4^+ \)) and water (\( \text{H}_2\text{O} \)), we can clearly see the application of the Bronsted-Lowry theory. Here, the ammonium ion donates a proton to water, making it a Bronsted-Lowry acid. Consequently, water acts as a base by accepting the proton to become hydronium ion (\( \text{H}_3\text{O}^+ \)). This theory is especially useful because it broadens the definition of acids and bases compared to other models, like the Arrhenius model, which only considers substances that create or accept hydroxide or hydrogen ions in water.
acid-base reactions
Acid-base reactions are chemical processes involving the transfer of protons between reactants. They are fundamental in chemistry, controlling processes in various environments, from biological systems to industrial applications. In our example reaction of \( \text{NH}_4^+ \)and \( \text{H}_2\text{O} \), we witness a straightforward acid-base reaction. The ammonium ion, acting as the acid, donates a proton to water. As a result, water, the base, accepts this proton, resulting in the formation of ammonia (\( \text{NH}_3 \)) and hydronium ion (\( \text{H}_3\text{O}^+ \)).Understanding acid-base reactions helps predict product formation and allows calculation of chemical equilibria, which are paramount for many applications like drug formulation and water treatment.
pH calculation
The pH is a measure of the acidity or alkalinity of a solution, determined by the concentration of hydronium ions (\( [\text{H}_3\text{O}^+] \)) in the solution. Calculating pH is crucial because it helps evaluate the degree of acidity of a solution and predict how it will behave in different chemical conditions.The formula used for calculating pH is:\[ pH = -\log_{10} [\text{H}_3\text{O}^+] \]In the case of our ammonium ion solution, the small equilibrium constant \( K_a \) reveals that the concentration of \( \text{H}_3\text{O}^+ \) formed is very low, indicating a slightly acidic solution. Solving for \( x \) gives us the concentration of \( \text{H}_3\text{O}^+ \), which is substituted back into the pH formula to find the result.The pH calculation is vital for chemists to assess solution conditions, important in areas like agriculture, medicine, and environmental science.
ammonium ion acidity
Ammonium ion (\( \text{NH}_4^+ \)) is an interesting subject in acid-base chemistry. It is the conjugate acid of ammonia (\( \text{NH}_3 \)) and plays an important role in various chemical and biological systems by influencing the pH of a solution.When dissolved in water, ammonium ion can donate a proton to produce a small amount of hydronium (\( \text{H}_3\text{O}^+ \)) and ammonia. However, due to the low value of the equilibrium constant \( K_a \) \( (5.6 \times 10^{-10}) \), only a tiny fraction of ammonium ions react, which means solutions of \( \text{NH}_4^+ \) are only slightly acidic.This mild acidity is essential in contexts such as soil chemistry, where ammonium can act as a source of nitrogen for plants without overly acidifying the soil, or in buffering systems in biochemical research.

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Most popular questions from this chapter

Write a chemical equation for the reaction between each pair of reactants, using single or double arrows as appropriate: (a) \(\mathrm{HNO}_{3}\) and \(\mathrm{OH}^{-}\) (b) \(\mathrm{HF}\) and \(\mathrm{OH}^{-}\) (c) \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{HCO}_{3}^{-}\) and \(\mathrm{H}_{2} \mathrm{O}\)

Pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\), is a weak base. (a) Write the chemical equation for the reaction between pyridine and water. (b) List all species present in an aqueous solution of pyridine in order of concentration, highest to lowest. (c) In a 0.100 M pyridine solution, \(3.2 \%\) of the pyridine has reacted with water to form products. Calculate the concentration of all species present (except water) in \(1 \mathrm{~L}\) of a \(0.100\) M pyridine solution. (d) What is the pH of this solution?

Indicate whether each compound is an electrolyte or a nonelectrolyte in water: (a) \(\mathrm{N}_{2}\) (b) \(\mathrm{NH}_{4} \mathrm{ClO}_{4}\) (c) \(\mathrm{HI}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (propane) (e) \(\mathrm{Li}_{3} \mathrm{PO}_{4}\) (f) \(\mathrm{AgNO}_{3}\) (g) \(\mathrm{CCl}_{4}\)

An aqueous solution has an \(\mathrm{H}_{3} \mathrm{O}\) ' concentration of \(1.0 \mathrm{M}\). What is the OH concentration? Is this solution acidic or basic? Justify your answer.

If you dissolve \(0.250 \mathrm{~g}\) of \(\mathrm{Ba}(\mathrm{OH})_{2}\) in \(3.00 \mathrm{~L}\) of water, what is the \(\mathrm{pH}\) of the solution?

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