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Predict whether aqueous solutions of the following compounds are acidic, basic, or neutral: \((\mathbf{a})\mathrm{NH}_{4} \mathrm{Br},(\mathbf{b}) \mathrm{FeCl}_{3}\) \((\mathbf{c}) \mathrm{Na}_{2} \mathrm{CO}_{3},(\mathbf{d}) \mathrm{KClO}_{4},(\mathbf{e}) \mathrm{NaHC}_{2} \mathrm{O}_{4}\)

Short Answer

Expert verified
The predicted behaviors of the given aqueous solutions are: a) NH鈧凚r: acidic b) FeCl鈧: acidic c) Na鈧侰O鈧: basic d) KClO鈧: neutral e) NaHC鈧侽鈧: acidic

Step by step solution

01

(Step 1: Identify Cations and Anions)

In this step, we will identify the cations and anions in each of the given compounds. This will help us understand how they will interact with water. a) NH鈧凚r: Cation: NH鈧勨伜 (ammonium) Anion: Br鈦 (bromide) b) FeCl鈧: Cation: Fe鲁鈦 (ferric) Anion: Cl鈦 (chloride) c) Na鈧侰O鈧: Cation: Na鈦 (sodium) Anion: CO鈧兟测伝 (carbonate) d) KClO鈧: Cation: K鈦 (potassium) Anion: ClO鈧勨伝 (perchlorate) e) NaHC鈧侽鈧: Cation: Na鈦 (sodium) Anion: HC鈧侽鈧勨伝 (hydrogen oxalate)
02

(Step 2: Determine Cation and Anion Behavior in Water)

In this step, we will determine whether the cations and anions will act as acids or bases when they react with water. NH鈧勨伜: acidic (ammonium ion behaves as a weak acid) Br鈦: neutral (conjugate base of a strong acid, HBr) Fe鲁鈦: acidic (transition metal ions with positive charges tend to be acidic because they can hydrolyze water) Cl鈦: neutral (conjugate base of a strong acid, HCl) Na鈦: neutral (alkali metal ions do not react with water) CO鈧兟测伝: basic (conjugate base of a weak acid, H鈧侰O鈧) K鈦: neutral (alkali metal ions do not react with water) ClO鈧勨伝: neutral (conjugate base of a strong acid, HClO鈧) Na鈦: neutral (alkali metal ions do not react with water) HC鈧侽鈧勨伝: acidic (hydrogen oxalate is an acidic ion since it can donate an H鈦 ion)
03

(Step 3: Predict Solution Behavior)

Based on the behavior of the cations and anions in water, we can predict whether the overall aqueous solution will be acidic, basic, or neutral. a) NH鈧凚r: NH鈧勨伜 (acidic) + Br鈦 (neutral) 鈫 acidic solution b) FeCl鈧: Fe鲁鈦 (acidic) + Cl鈦 (neutral) 鈫 acidic solution c) Na鈧侰O鈧: Na鈦 (neutral) + CO鈧兟测伝 (basic) 鈫 basic solution d) KClO鈧: K鈦 (neutral) + ClO鈧勨伝 (neutral) 鈫 neutral solution e) NaHC鈧侽鈧: Na鈦 (neutral) + HC鈧侽鈧勨伝 (acidic) 鈫 acidic solution So the final predictions are as follows: a) NH鈧凚r: acidic b) FeCl鈧: acidic c) Na鈧侰O鈧: basic d) KClO鈧: neutral e) NaHC鈧侽鈧: acidic

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Key Concepts

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

Cation and Anion Identification
Cations and anions are positively and negatively charged ions found in compounds. Identifying them is the first step in predicting how a compound will behave in a solution. In each compound, the cation is usually listed first and tends to lose electrons to form a positive charge. Anions, on the other hand, gain electrons to form a negative charge. Understanding the nature of these ions helps in predicting how they will react with water.

For example, in the compound ammonium bromide ( ext{NH}_4 ext{Br}), the cation is ext{NH}_4^+ (ammonium), a positively charged ion. The anion is ext{Br}^- (bromide), a negatively charged ion. Identifying these ions allows us to anticipate their potential interactions when dissolved in water. Cations like ext{Fe}^{3+} (in ext{FeCl}_3) show strong tendencies to interact due to their higher positive charge, indicating a possible acidic behavior. Similarly, distinguishing whether an anion such as ext{CO}_3^{2-} (in ext{Na}_2 ext{CO}_3) is basic or neutral further guides us in predicting the pH of the solution.
Aqueous Solutions
An aqueous solution refers to a substance dissolved in water. Water acts as a solvent, facilitating the dissociation of ionic compounds into their constituent cations and anions. This dissociation is crucial because it determines how the solution behaves, whether acidic, basic, or neutral. When a compound dissolves, its cations and anions disperse through the solution, each potentially affecting the solution's properties.

This dissociation is determined by the nature of the ions themselves. Remember that ext{Na}^+ and ext{K}^+ ions, often found in salts, typically remain neutral in water and do not affect its acidity or basicity. On the other hand, transition metal ions like ext{Fe}^{3+} can hydrolyze water, increasing the solution's acidity. Understanding these dynamics helps us predict the behavior of the entire solution effectively, allowing us to anticipate whether it will be acidic, basic, or neutral. This is especially important when dealing with chemical reactions in real-world applications.
Solution pH Prediction
Predicting the pH of a solution involves understanding the behavior of both cations and anions upon dissolution. Solutions with acidic behaviors arise when cations act as weak acids and can donate ext{H}^+ ions. An excellent example is the ammonium ion ( ext{NH}_4^+), which releases ext{H}^+ ions, tilting the solution towards acidity when coupled with neutral anions like bromide ( ext{Br}^-).

Conversely, if anions behave as bases, such as ext{CO}_3^{2-} from ext{Na}_2 ext{CO}_3, they will attract ext{H}^+ ions from water, raising the pH and making the solution basic. Neutral solutions, like those formed from ext{KClO}_4, occur when neither ion affects the pH. This balance or imbalance of acidic and basic tendencies in solution is what ultimately defines the solution's pH.
  • Acidic Solution Example: ext{NH}_4 ext{Br} contains ext{NH}_4^+, an ion that makes the solution acidic.
  • Basic Solution Example: ext{Na}_2 ext{CO}_3 releases ext{CO}_3^{2-}, making the solution basic.
  • Neutral Solution Example: ext{KClO}_4 neither increases nor decreases the pH, maintaining neutrality.
With these insights, one can confidently predict how a compound will impact a solution's overall pH, facilitating better planning and execution of chemical processes.

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Most popular questions from this chapter

Indicate whether each of the following statements is true or false. For each statement that is false, correct the statement to make it true. (a) In general, the acidity of binary acids increases from left to right in a given row of the periodic table. (b) In a series of acids that have the same central atom, acid strength increases with the number of hydrogen atoms bonded to the central atom. (c) Hydrotelluric acid \(\left(\mathrm{H}_{2} \mathrm{Te}\right)\) is a stronger acid than \(\mathrm{H}_{2} \mathrm{S}\) because Te is more electronegative than \(\mathrm{S} .\)

At \(50^{\circ} \mathrm{C},\) the ion-product constant for \(\mathrm{H}_{2} \mathrm{O}\) has the value \(K_{w}=5.48 \times 10^{-14}\) . (a) What is the pH of pure water at \(50^{\circ} \mathrm{C} ?\) (b) Based on the change in \(K_{w}\) with temperature, predict whether \(\Delta H\) is positive, negative, or zero for the autoionization reaction of water: $$2 \mathrm{H}_{2} \mathrm{O}(l) \Longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)$$

Indicate whether each of the following statements is correct or incorrect. (a) Every Bronsted-Lowry acid is also a Lewis acid. (b) Every Lewis acid is also a Bronsted-Lowry acid. (c) Conjugate acids of weak bases produce more acidic solutions than conjugate acids of strong bases. (d) \(\mathrm{K}^{+}\) ion is acidic in water because it causes hydrating water molecules to become more acidic. (e) The percent ionization of a weak acid in water increases as the concentration of acid decreases.

Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs: (a) \(\mathrm{NO}_{3}^{-}\) or \(\mathrm{NO}_{2}^{-},(\mathbf{b}) \mathrm{PO}_{4}^{3-}\) or \(\mathrm{AsO}_{4}^{3-}\) \((\mathbf{c}) \mathrm{HCO}_{3}^{-}\) or \(\mathrm{CO}_{3}^{2-}.\)

Calculate the \(\mathrm{pH}\) of each of the following strong acid solutions: (a) \(0.0167 M \mathrm{HNO}_{3},(\mathbf{b}) 0.225 \mathrm{g}\) of \(\mathrm{HClO}_{3}\) in 2.00 \(\mathrm{L}\) of solution, \((\mathbf{c}) 15.00 \mathrm{mL}\) of 1.00 \(\mathrm{M} \mathrm{HCl}\) diluted to \(0.500 \mathrm{L},\) (d) a mixture formed by adding 50.0 \(\mathrm{mL}\) of 0.020 \(\mathrm{MHCl}\) to 125 \(\mathrm{mL}\) of 0.010 \(\mathrm{M} \mathrm{HI} .\)

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