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Chapter 14: Problem 70

For each salt, predict whether an aqueous solution has a pH less than, equal to, or greater than 7 . Explain your prediction. (a) \(\mathrm{AlCl}_{3}\) (b) \(\mathrm{Na}_{2} \mathrm{~S}\) (c) \(\mathrm{NaNO}_{3}\)

Short Answer

Expert verified
(a) pH < 7, (b) pH > 7, (c) pH = 7.

Step by step solution

01

Analyze the Salt Composition - AlCl3

Consider the ions that make up the salt \( \mathrm{AlCl}_{3} \). It dissociates into \( \mathrm{Al}^{3+} \) and \( \mathrm{Cl}^{-} \) in water. The \( \mathrm{Al}^{3+} \) ion can hydrolyze in water, forming \( \mathrm{Al(OH)}^{2+} \) and releasing \( \mathrm{H}^{+} \), which makes the solution acidic. \( \mathrm{Cl}^{-} \) does not affect the pH as it's a conjugate base of a strong acid (HCl).
02

Conclusion for AlCl3

Based on the hydrolysis of \( \mathrm{Al}^{3+} \), the solution will likely have a pH less than 7 because of the production of \( \mathrm{H}^{+} \) ions.
03

Analyze the Salt Composition - Na2S

Next, consider \( \mathrm{Na}_{2} \mathrm{S} \). It dissociates into \( 2\mathrm{Na}^{+} \) and \( \mathrm{S}^{2-} \) ions in water. The \( \mathrm{S}^{2-} \) ion can react with water in a hydrolysis reaction to form \( \mathrm{HS}^{-} \) and \( \mathrm{OH}^{-} \), making the solution basic. \( \mathrm{Na}^{+} \) does not affect the pH as it is a cation of a strong base (NaOH).
04

Conclusion for Na2S

Due to the formation of \( \mathrm{OH}^{-} \) ions, the aqueous solution of \( \mathrm{Na}_{2} \mathrm{S} \) will have a pH greater than 7.
05

Analyze the Salt Composition - NaNO3

Consider \( \mathrm{NaNO}_{3} \). It dissociates into \( \mathrm{Na}^{+} \) and \( \mathrm{NO}_{3}^{-} \) in water. Both \( \mathrm{Na}^{+} \) and \( \mathrm{NO}_{3}^{-} \) come from a strong base (NaOH) and a strong acid (HNO3), neither affects the pH significantly in water.
06

Conclusion for NaNO3

With no significant hydolysis from \( \mathrm{Na}^{+} \) or \( \mathrm{NO}_{3}^{-} \), the solution is neutral and will have a pH equal to 7.

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Key Concepts

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

Hydrolysis
Hydrolysis refers to a chemical process where a molecule reacts with water, causing it to break down into two or more parts. It鈥檚 a significant concept in predicting the pH of salt solutions. When certain ions undergo hydrolysis, they can change the pH of the solution. For example, when the ion
  • \( \mathrm{Al}^{3+} \) hydrolyzes in water, it forms \( \mathrm{Al(OH)}^{2+} \) and releases \( \mathrm{H}^{+} \) ions, making the solution acidic.
  • Similarly, the ion \( \mathrm{S}^{2-} \) from \( \mathrm{Na}_{2} \mathrm{~S} \) reacts with water to produce \( \mathrm{HS}^{-} \) and \( \mathrm{OH}^{-} \) ions, leading to a basic solution.
Understanding which ions undergo hydrolysis helps predict whether a solution will be acidic or basic. Hydrolysis is especially noticeable in salts where the metal cations or non-metal anions are weak conjugates of their respective acids or bases.
Acidic and Basic Solutions
Solutions are classified as acidic, neutral, or basic based on their pH level.
  • An acidic solution has a pH less than 7, which occurs when there is an excess of \( \mathrm{H}^{+} \) ions. For instance, \( \mathrm{AlCl}_{3} \) in water increases \( \mathrm{H}^{+} \) due to the hydrolysis of \( \mathrm{Al}^{3+} \).
  • On the other hand, a basic solution has a pH greater than 7, resulting from an excess of \( \mathrm{OH}^{-} \) ions. This is the case with \( \mathrm{Na}_{2} \mathrm{S} \), as it generates \( \mathrm{OH}^{-} \) through the hydrolysis of \( \mathrm{S}^{2-} \).
  • A neutral solution has a pH of 7, like pure water. For example, \( \mathrm{NaNO}_{3} \) dissociates completely but does not alter the pH, leading to a neutral solution.
Understanding pH and how various ions affect it allows for accurate predictions of whether a solution will be acidic or basic.
Dissociation in Water
Dissociation is a crucial concept in understanding how salts affect pH. When salts dissolve in water, they separate into their constituent ions.
  • For example, \( \mathrm{AlCl}_{3} \) dissociates into \( \mathrm{Al}^{3+} \) and \( \mathrm{Cl}^{-} \) ions. Some of these ions can react with water molecules.
  • Similarly, \( \mathrm{Na}_{2} \mathrm{~S} \) splits into \( 2\mathrm{Na}^{+} \) and \( \mathrm{S}^{2-} \), and \( \mathrm{NaNO}_{3} \) into \( \mathrm{Na}^{+} \) and \( \mathrm{NO}_{3}^{-} \).
  • This dissociation process is essential for understanding the subsequent hydrolysis reactions that determine the pH of the solution.
By examining which ions are produced from dissociation, one can predict the acidity or basicity of the resulting solution.

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Most popular questions from this chapter

Ascorbic acid (vitamin \(\left.\mathrm{C}, \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\right)\) is a diprotic acid \(\left(K_{\mathrm{a}_{1}}=7.9 \times 10^{-5}, K_{\mathrm{a}_{2}}=1.6 \times 10^{-12}\right) .\) Calculate the \(\mathrm{pH}\) of a solution that contains \(5.0 \mathrm{mg}\) acid per mL water. (Assume that only the first ionization is important in determining pH.)

Write a chemical equation to describe the proton transfer that occurs when each of these bases is added to water. (a) \(\mathrm{PO}_{4}^{3-}\) (b) \(\mathrm{SO}_{3}^{2-}\) (c) \(\mathrm{HPO}_{4}^{2-}\)

Calculate the \(K_{\mathrm{a}}\) of butyric acid if a \(0.025-\mathrm{M}\) butyric acid solution has a pH of 3.21 .

Consider these four solutions: $$ \begin{array}{lcc} \hline \text { Solution } & {\left[\mathrm{H}_{3} \mathrm{O}^{+}\right](\mathrm{M})} & {\left[\mathrm{OH}^{-}\right](\mathrm{M})} \\ \hline \mathrm{D} & 2 \times 10^{-3} & \\ \mathrm{E} & & 2 \times 10^{-7} \\ \mathrm{~F} & 4 \times 10^{-5} & \\ \mathrm{G} & & 5 \times 10^{-11} \\ \hline \end{array} $$ (a) Which solution has the highest \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration? (b) Which solution has the highest \(\mathrm{OH}^{-}\) concentration? (c) Which solution is closest to being a neutral solution?

A solution of benzyl amine, \(\mathrm{C}_{7} \mathrm{H}_{7} \mathrm{NH}_{2}\), has a hydroxide ion concentration of \(2.4 \times 10^{-3} \mathrm{M}\). Calculate the \(\mathrm{pH}\) of the solution. Calculate its \(\mathrm{pOH}\).
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