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Write a balanced net ionic equation for the principal reaction in solutions of each of the following salts. In each case, identify the Brensted-Lowry acids and bases and the conjugate acid-base pairs. (a) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) (b) \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) (c) \(\mathrm{NaCl}\) (d) \(\mathrm{ZnCl}_{2}\)

Short Answer

Expert verified
(a) \( \mathrm{CO}_{3}^{2-} + \mathrm{H}_{2} \mathrm{O} \leftrightarrow \mathrm{HCO}_{3}^{-} + \mathrm{OH}^{-} \); (b) \( \mathrm{NH}_{4}^{+} + \mathrm{H}_{2} \mathrm{O} \leftrightarrow \mathrm{NH}_{3} + \mathrm{H}_{3} \mathrm{O}^{+} \); (c) No reaction; (d) \( \mathrm{Zn}^{2+} + 2\mathrm{H}_{2} \mathrm{O} \leftrightarrow \mathrm{Zn(OH)}^{+} + \mathrm{H}_{3} \mathrm{O}^{+} \).

Step by step solution

01

Dissociate the Salts (Part a)

For \( \mathrm{Na}_{2} \mathrm{CO}_{3} \), it dissociates into \( 2\mathrm{Na}^{+} \) and \( \mathrm{CO}_{3}^{2-} \) in solution. Since \( \mathrm{CO}_{3}^{2-} \) can react with water, we write the net ionic equation for this interaction: \[ \mathrm{CO}_{3}^{2-} + \mathrm{H}_{2} \mathrm{O} \leftrightarrow \mathrm{HCO}_{3}^{-} + \mathrm{OH}^{-} \]. Here, \( \mathrm{H}_{2} \mathrm{O} \) acts as a Bronsted-Lowry acid, donating a proton to \( \mathrm{CO}_{3}^{2-} \).
02

Identify Conjugate Pairs (Part a)

In the equation \( \mathrm{CO}_{3}^{2-} + \mathrm{H}_{2} \mathrm{O} \leftrightarrow \mathrm{HCO}_{3}^{-} + \mathrm{OH}^{-} \), the conjugate acid-base pairs are \( \mathrm{H}_{2} \mathrm{O} \) and \( \mathrm{OH}^{-} \), and \( \mathrm{CO}_{3}^{2-} \) and \( \mathrm{HCO}_{3}^{-} \). The reaction is a base-acid reaction.
03

Dissociate the Salts (Part b)

\( \mathrm{NH}_{4} \mathrm{NO}_{3} \) dissociates into \( \mathrm{NH}_{4}^{+} \) and \( \mathrm{NO}_{3}^{-} \). The \( \mathrm{NH}_{4}^{+} \) ion can interact with water: \[ \mathrm{NH}_{4}^{+} + \mathrm{H}_{2} \mathrm{O} \leftrightarrow \mathrm{NH}_{3} + \mathrm{H}_{3} \mathrm{O}^{+} \]. \( \mathrm{NH}_{4}^{+} \) is the acid here, donating a proton.
04

Identify Conjugate Pairs (Part b)

In the reaction \( \mathrm{NH}_{4}^{+} + \mathrm{H}_{2} \mathrm{O} \leftrightarrow \mathrm{NH}_{3} + \mathrm{H}_{3} \mathrm{O}^{+} \), the conjugate pairs are \( \mathrm{NH}_{4}^{+} \) and \( \mathrm{NH}_{3} \), \( \mathrm{H}_{2} \mathrm{O} \) and \( \mathrm{H}_{3} \mathrm{O}^{+} \). \( \mathrm{NH}_{4}^{+} \) acts as a Bronsted-Lowry acid.
05

Dissociate the Salts (Part c)

\( \mathrm{NaCl} \) dissociates into \( \mathrm{Na}^{+} \) and \( \mathrm{Cl}^{-} \). Neither ion reacts with water to form another acid or base, so no principal reaction occurs, and both ions are spectator ions.
06

No Reaction for Part c

Since neither \( \mathrm{Na}^{+} \) nor \( \mathrm{Cl}^{-} \) acts as an acid or base, no net ionic equation occurs in this case.
07

Dissociate the Salts (Part d)

\( \mathrm{ZnCl}_{2} \) dissociates into \( \mathrm{Zn}^{2+} \) and \( 2\mathrm{Cl}^{-} \). \( \mathrm{Zn}^{2+} \) can hydrolyze in water, reacting with \( \mathrm{H}_{2} \mathrm{O} \): \[ \mathrm{Zn}^{2+} + 2\mathrm{H}_{2} \mathrm{O} \leftrightarrow \mathrm{Zn(OH)}^{+} + \mathrm{H}_{3} \mathrm{O}^{+} \].
08

Identify Conjugate Pairs (Part d)

In the reaction \( \mathrm{Zn}^{2+} + 2\mathrm{H}_{2} \mathrm{O} \leftrightarrow \mathrm{Zn(OH)}^{+} + \mathrm{H}_{3} \mathrm{O}^{+} \), the pairs are \( \mathrm{H}_{2} \mathrm{O} \) and \( \mathrm{H}_{3} \mathrm{O}^{+} \), and \( \mathrm{Zn}^{2+} \) with \( \mathrm{Zn(OH)}^{+} \). \( \mathrm{Zn}^{2+} \) acts as a Bronsted-Lowry acid.

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Key Concepts

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

Bronsted-Lowry acid
When it comes to acid-base chemistry, a crucial concept is the Bronsted-Lowry theory. This theory describes acids as compounds that donate protons (H\(^+\)) and bases as those that accept protons. For example, in the reaction involving ammonium nitrate, \(\mathrm{NH}_{4}^{+} + \mathrm{H}_{2} \mathrm{O} \leftrightarrow \mathrm{NH}_{3} + \mathrm{H}_{3} \mathrm{O}^{+}\), \(\mathrm{NH}_{4}^{+}\) is the Bronsted-Lowry acid as it donates a proton to water, forming \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{3} \mathrm{O}^{+}\). Similarly, in the reaction with \(\mathrm{ZnCl}_{2}\), \(\mathrm{Zn}^{2+}\) acts as a Bronsted-Lowry acid by donating a proton to water, resulting in zinc hydroxide and hydronium ions.
conjugate acid-base pairs
In every acid-base reaction, conjugate acid-base pairs play a key role. These pairs result from the transfer of a proton from the acid to the base. Once an acid donates its proton, it forms its conjugate base. Conversely, the base that accepts a proton forms its conjugate acid.
For instance:
  • In the reaction \(\mathrm{CO}_{3}^{2-} + \mathrm{H}_{2} \mathrm{O} \leftrightarrow \mathrm{HCO}_{3}^{-} + \mathrm{OH}^{-}\), \(\mathrm{CO}_{3}^{2-}\) forms \(\mathrm{HCO}_{3}^{-}\) as its conjugate acid, and water forms \(\mathrm{OH}^{-}\) as its conjugate base.
  • In \(\mathrm{NH}_{4}^{+} + \mathrm{H}_{2} \mathrm{O} \leftrightarrow \mathrm{NH}_{3} + \mathrm{H}_{3} \mathrm{O}^{+}\), \(\mathrm{NH}_{4}^{+}\) becomes \(\mathrm{NH}_{3}\) and water forms \(\mathrm{H}_{3} \mathrm{O}^{+}\) as conjugate acid-base pairs.
Understanding these pairs helps to predict the direction of acid-base equilibrium reactions.
salt dissociation
Salt dissociation is a process where a salt compound breaks down into its component ions when dissolved in water. This step is crucial in determining whether a reaction will occur in an aqueous solution.
  • For example, baking soda, \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), dissociates into \(2\mathrm{Na}^{+}\) and \(\mathrm{CO}_{3}^{2-}\) ions.
  • Ammonium nitrate, \(\mathrm{NH}_{4} \mathrm{NO}_{3}\), separates into \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{NO}_{3}^{-}\).
  • Table salt, \(\mathrm{NaCl}\), breaks into \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\), although neither reacts further with water in a noteworthy manner.
Sourcing specific ions from a salt enables reactions like those between \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{CO}_{3}^{2-}\) with water to occur.
hydrolysis reactions
When ionic compounds dissolve and react with water, this is known as hydrolysis. Hydrolysis reactions are an interaction where either a cation or anion from a salt reacts with water to affect the solution's pH.
For example, the \(\mathrm{CO}_{3}^{2-}\) ion from Sodium carbonate reacts with water in the equation \(\mathrm{CO}_{3}^{2-} + \mathrm{H}_{2} \mathrm{O} \leftrightarrow \mathrm{HCO}_{3}^{-} + \mathrm{OH}^{-}\). This increases the hydroxide ion concentration, making the solution more basic. Similarly, in the case of \(\mathrm{ZnCl}_{2}\), \(\mathrm{Zn}^{2+}\) undergoes hydrolysis: \(\mathrm{Zn}^{2+} + 2\mathrm{H}_{2} \mathrm{O} \leftrightarrow \mathrm{Zn(OH)}^{+} + \mathrm{H}_{3} \mathrm{O}^{+}\). This reaction introduces more \(\mathrm{H}_{3} \mathrm{O}^{+}\), making the solution more acidic. Hydrolysis explains why some solutions from salt dissolutions become acidic or basic.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) and the concentrations of all species present \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}, \mathrm{HCO}_{3}^{-}, \mathrm{CO}_{3}^{2-}, \mathrm{H}_{3} \mathrm{O}^{+}\right.\), and \(\left.\mathrm{OH}^{-}\right)\) in \(0.010 \mathrm{M} \mathrm{H}_{2} \mathrm{CO}_{3}\left(K_{\Delta 1}=4.3 \times 10^{-7} ; K_{\Delta 2}=5.6 \times 10^{-11}\right)\)

Calculate the pH of the following solutions: (a) \(0.050 \mathrm{M} \mathrm{HClO}_{4}\) (b) \(6.0 \mathrm{MHCl}\) (c) \(4.0 \mathrm{MKOH}\) (d) \(0.010 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\)

The concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions in the runoff from a coal mine is \(1.4 \times 10^{-4} \mathrm{M}\). Calculate the concentration of \(\mathrm{OH}^{-}\) ions, and classify the solution as acidic, neutral, or basic.

Sodium benzoate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{Na}\right)\) is used as a food preservative. Calculate the \(\mathrm{pH}\) and the concentrations of all species present \(\left(\mathrm{Na}^{+}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, \mathrm{H}_{3} \mathrm{O}^{+}\right.\), and \(\left.\mathrm{OH}^{-}\right)\) in \(0.050 \mathrm{M}\) so- dium benzoate; \(K_{a}\) for benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right)\) is \(6.5 \times 10^{-5} .\)

Natural rain has a \(\mathrm{pH}\) of \(5.6\) due to dissolved atmospheric carbon dioxide at a current level of 400 ppm. Various models predict that burning fossil fuels will increase the atmospheric \(\mathrm{CO}_{2}\) concentration to between 500 and 1000 ppm (by volume) by the year \(2100 .\) Calculate the \(\mathrm{pH}\) of rain in a scenario where the \(\mathrm{CO}_{2}\) concentration is \(750 \mathrm{ppm}\). (a) First use Henry's Law to calculate the concentration of dissolved \(\mathrm{CO}_{2} .\) Solubility \(=k \cdot P\) (Section \(\left.12.4\right)\) and the Henry's Law constant \((k)\) for \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(3.2 \times 10^{-2} \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{atm})\) (b) Next calculate the \(\mathrm{pH}\) of the rain. \(\mathrm{CO}_{2}\) reacts with water to product carbonic acid according to the equation: $$ \mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) $$ Assume all the dissolved \(\mathrm{CO}_{2}\) is converted to \(\mathrm{H}_{2} \mathrm{CO}_{3}\). Acid dissociation constants for \(\mathrm{H}_{2} \mathrm{CO}_{3}\) are \(K_{\mathrm{a} 1}=4.3 \times 10^{-7} ; K_{\Delta 2}=5.6 \times 10^{-11}\). (Worked Example 15.11 is a model for this calculation.) (c) Will rising \(\mathrm{CO}_{2}\) levels affect the acidity of rainfall?

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