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Chapter 16: Problem 83

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) NH₄Br: acidic b) FeCl₃: acidic c) Na₂CO₃: basic d) KClO₄: neutral e) NaHC₂O₄: 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) NH₄Br: Cation: NH₄⁺ (ammonium) Anion: Br⁻ (bromide) b) FeCl₃: Cation: Fe³⁺ (ferric) Anion: Cl⁻ (chloride) c) Na₂CO₃: Cation: Na⁺ (sodium) Anion: CO₃²⁻ (carbonate) d) KClO₄: Cation: K⁺ (potassium) Anion: ClO₄⁻ (perchlorate) e) NaHC₂O₄: Cation: Na⁺ (sodium) Anion: HC₂O₄⁻ (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. NH₄⁺: acidic (ammonium ion behaves as a weak acid) Br⁻: neutral (conjugate base of a strong acid, HBr) Fe³⁺: acidic (transition metal ions with positive charges tend to be acidic because they can hydrolyze water) Cl⁻: neutral (conjugate base of a strong acid, HCl) Na⁺: neutral (alkali metal ions do not react with water) CO₃²⁻: basic (conjugate base of a weak acid, H₂CO₃) K⁺: neutral (alkali metal ions do not react with water) ClO₄⁻: neutral (conjugate base of a strong acid, HClO₄) Na⁺: neutral (alkali metal ions do not react with water) HC₂O₄⁻: acidic (hydrogen oxalate is an acidic ion since it can donate an H⁺ 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) NH₄Br: NH₄⁺ (acidic) + Br⁻ (neutral) → acidic solution b) FeCl₃: Fe³⁺ (acidic) + Cl⁻ (neutral) → acidic solution c) Na₂CO₃: Na⁺ (neutral) + CO₃²⁻ (basic) → basic solution d) KClO₄: K⁺ (neutral) + ClO₄⁻ (neutral) → neutral solution e) NaHC₂O₄: Na⁺ (neutral) + HC₂O₄⁻ (acidic) → acidic solution So the final predictions are as follows: a) NH₄Br: acidic b) FeCl₃: acidic c) Na₂CO₃: basic d) KClO₄: neutral e) NaHC₂O₄: acidic

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Most popular questions from this chapter

An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the pH of the solution is 8.08. What is the identity of the salt?

Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the reaction arrow: (a) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons\) (b) \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)\) (c) \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{F}(a q) \rightleftharpoons\)

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?

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) \(\mathrm{BrO}^{-}\) or \(\mathrm{ClO}^{-},(\mathbf{b}) \mathrm{BrO}^{-}\) or \(\mathrm{BrO}_{2}^{-}\) (c) \(\mathrm{HPO}_{4}^{2-}\) or \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}.\)
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