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Chapter 7: Problem 154

Which of the following solutions will have no effect on \(\mathrm{pH}\) on dilution? (1) \(0.2 \mathrm{M}\) KHS (2) \(0.8 \mathrm{M} \mathrm{NH}_{4} \mathrm{NO}_{3}\) (3) \(0.01 \mathrm{M} \mathrm{NaCl}\) (4) \(1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COONH}_{4}\) (a) 1,3 (b) 1,4 (c) 2,3 (d) 2,4

Short Answer

Expert verified
The correct options are (c) 2,3.

Step by step solution

01

Understanding the Problem

We need to identify which of the given solutions have a pH that remains unchanged upon dilution. To do this, we have to identify solutions that neither significantly produce nor consume hydronium ions when diluted.
02

Role of KHS

Potassium hydrogen sulfate (KHS) can partially dissociate in water to produce HSO4鈦�, which is a weak acid, and K鈦�. Dilution affects the amount of hydronium ions due to the acid-base equilibrium, so KHS solution's pH will change upon dilution.
03

Role of NH4NO3

Ammonium nitrate (NH鈧凬O鈧�) dissociates into NH鈧勨伜 and NO鈧冣伝 ions. NH鈧勨伜 is an acidic ion that can slightly affect the pH when diluted, as it contributes H鈦� ions.
04

Role of NaCl

Sodium chloride (NaCl) is a neutral salt comprising Na鈦� and Cl鈦� ions, neither of which react with water to produce H鈦� or OH鈦� ions. Thus, the pH of a NaCl solution is unaffected by dilution since it remains neutral (pH close to 7).
05

Role of CH3COONH4

Ammonium acetate (CH3COONH4) is a salt of a weak base (CH3COO鈦�) and a weak acid (NH4鈦�). The pH is approximately neutral and does not change significantly with dilution, as both ions offset each other's effect on pH.
06

Identifying Solutions with No pH Change

Based on the analysis, NaCl (option 3) and ammonium acetate (option 4) will not change in pH with dilution because they maintain a balance in ion contribution, keeping the solution neutral.

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Key Concepts

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

Neutral Solutions
When we talk about a neutral solution, we're referring to one where the \( \text{pH} \approx 7 \). This is the same pH as pure water. In such solutions, the amount of hydrogen ions \( \text{(H}^+\text{)} \) and hydroxide ions \( \text{(OH}^-\text{)} \) are balanced.
One common example of a neutral solution is sodium chloride \( \text{(NaCl)} \).
  • NaCl dissolves in water to give Na鈦� and Cl鈦� ions.
  • These ions do not affect the concentration of hydrogen or hydroxide ions in water.
Because of this, the pH remains unchanged, even when you dilute the solution. Another example is ammonium acetate \( \text{(CH}_3\text{COONH}_4) \), which is derived from a weak acid and a weak base. Upon dilution, the contributions of ammonium and acetate ions to the pH balance each other out, thus maintaining a neutral pH.
Salt Hydrolysis
Salt hydrolysis occurs when a salt derived from a weak acid or weak base reacts with water, affecting the hydrogen ion concentration. For example, let's consider \( \text{(CH}_3\text{COONH}_4) \), a salt composed of an acetic acid anion \( \text{(CH}_3\text{COO}^-) \) and an ammonium cation \( \text{(NH}_4^+) \).
In water, these can undergo hydrolysis:
  • Ammonium ions\( \text{(NH}_4^+)\) may release H鈦�, slightly dropping the pH.
  • Acetate ions\( \text{(CH}_3\text{COO}^-)\) may absorb H鈦�, slightly increasing the pH.
These effects cancel each other, leading to no significant pH change. Salts like NaCl do not undergo hydrolysis because they come from strong acids and bases, leaving the pH unchanged upon dilution.
Amphoteric Substances
An amphoteric substance is capable of acting as either an acid or a base, depending on the environment. Water is the most common example because it can donate \( \text{H}^+ \) in a basic environment or accept \( \text{H}^+ \) in an acidic one.
This means:
  • In basic solutions, water can donate a proton turning into hydroxide \( \text{(OH}^-\text{)} \).
  • In acidic solutions, it can accept a proton, becoming hydronium \( \text{(H}_3\text{O}^+) \).
Ammonium acetate can, to some extent, show similar behavior. The ammonium ion can donate a proton, showing acidic behavior, while the acetate ion can accept a proton, showing basic character. This dual ability helps maintain a generally consistent pH when diluted, emphasizing the amphoteric nature in balance.
Electrolyte Dissociation
Electrolyte dissociation refers to the process by which ionic compounds break apart into their constituent ions in solution, helping conduct electricity. For instance, when NaCl dissolves:
  • Sodium ions \( \text{(Na}^+\text{)} \)
  • Chloride ions \( \text{(Cl}^-\text{)} \)
are released into the solution. Electrolytes are classified as strong or weak based on the degree of dissociation.
Strong electrolytes like NaCl dissolve completely, meaning they almost fully ionize in solution.
Weak electrolytes like acetic acid only partially dissociate, maintaining an equilibrium between ions and undissociated molecules. The extent of dissociation - for instance, how ammonium acetate behaves differently from NaCl - determines how a solution adjusts pH or remains stable upon dilution.

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Most popular questions from this chapter

For the reaction, \(\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}\) the equilibrium concentration of \(\mathrm{H}_{2}, \mathrm{I}_{2}\) and \(\mathrm{HI}\) are \(8.0,3.0\) and \(28.0\) mole/litre, respectively, the equilibrium constant is: (a) \(28.34\) (b) \(32.66\) (c) \(34.78\) (d) \(38.88\)

Which of the following change will shift the reaction in forward direction: \(\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 21(\mathrm{~g})\) Take \(\Delta \mathrm{H}^{\circ}=+150 \mathrm{~kJ}\) (a) Increase in concentration of I (b) Increase in total pressure (c) Decrease in concentration of \(\mathrm{I}_{2}\) (d) Increase in temperature

\(9.2 \mathrm{~g}\) of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) is taken in a closed one litre vessel and heated till the following equilibrium is reached \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\). At equilibrium, \(50 \%\) of \(\mathrm{N}_{2} \mathrm{O}_{4}\) (g) is dissociated. What is the equilibrium constant (in \(\left.\mathrm{mol} \mathrm{L}^{-1}\right) ?\) (molecular weight of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is 92 ): (a) \(0.1\) (b) \(0.2\) (c) \(0.4\) (d) 2

In the reaction \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\), the equilibrium concentrations of \(\mathrm{PCl}_{5}\) and \(\mathrm{PCl}_{3}\) are \(0.4\) and \(0.2\) mole/litre respectively. If the value of \(\mathrm{K}_{\mathrm{c}}\) is \(0.5\), what is the concentration of \(\mathrm{Cl}_{2}\) in mole/litre: (a) \(2.0\) (b) \(1.5\) (c) \(1.0\) (d) \(0.5\)

In which of the following reactions, the concentration of reactant is equal to concentration of product at equilibrium \((\mathrm{K}=\) equilibrium constant \()\) : (a) \(\mathrm{A} \rightleftharpoons \mathrm{B} ; \mathrm{K}=0.01\) (b) \(\mathrm{R} \rightleftharpoons \mathrm{P} ; \mathrm{K}=1\) (c) \(\mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}=10\) (d) \(\mathrm{L} \rightleftharpoons \mathrm{J} ;=0.025\)
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