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The correct order of increasing \(\mathrm{H}_{3} \mathrm{O}^{+}\) in the following aqueous solution is: (a) \(0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}<0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}<0.01 \mathrm{M} \mathrm{NaCl}<\) \(0.01 \mathrm{M} \mathrm{NaNO}_{2}\) (b) \(0.01 \mathrm{M} \mathrm{NaCl}^{2}<0.01 \mathrm{M} \mathrm{NaNO}_{2}<0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}<\) \(0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) (c) \(0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}<0.01 \mathrm{M} \mathrm{NaNO}_{3}<0.01 \mathrm{M} \mathrm{NaCl}<\) \(0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) (d) \(0.01 \mathrm{M} \mathrm{NaNO}_{2}<0.01 \mathrm{M} \mathrm{NaCl}<0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}<\) \(0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\)

Short Answer

Expert verified
The correct order is option (d).

Step by step solution

01

Analyze the Acidity of Different Solutions

Start by examining each solution for its ability to produce hydronium ions (\(\mathrm{H}_{3}\mathrm{O}^{+}\)) in water. Strong acids will produce more \(\mathrm{H}_{3}\mathrm{O}^{+}\) ions than weak acids or neutral substances. In this problem:- \(\mathrm{H}_{2}\mathrm{SO}_{4}\) is a strong acid that completely dissociates, producing a high concentration of \(\mathrm{H}_{3}\mathrm{O}^{+}\).- \(\mathrm{H}_{2}\mathrm{S}\) is a weak acid that partially dissociates, producing a lower concentration of \(\mathrm{H}_{3}\mathrm{O}^{+}\).- \(\mathrm{NaCl}\) and \(\mathrm{NaNO}_{2}\) are salts; neither produces \(\mathrm{H}_{3}\mathrm{O}^{+}\). However, NaNO2 is the salt of a weak acid and a strong base, slightly increasing \(\mathrm{OH}^{-}\).
02

Rank Solutions Based on Hydronium Ion Production

Given the acidity analysis, you can rank the solutions by the concentration of \(\mathrm{H}_{3}\mathrm{O}^{+}\):1. \(0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) - Strongest acid, highest \(\mathrm{H}_{3}\mathrm{O}^{+}\) concentration.2. \(0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}\) - Weak acid, moderate \(\mathrm{H}_{3}\mathrm{O}^{+}\) concentration.3. \(0.01 \mathrm{M} \mathrm{NaNO}_{2}\) - Salt that slightly affects pH by slightly raising \(\mathrm{OH}^{-}\) concentration, but it does not provide \(\mathrm{H}_{3}\mathrm{O}^{+}\).4. \(0.01 \mathrm{M} \mathrm{NaCl}\) - Neutral salt without effect on \(\mathrm{H}_{3}\mathrm{O}^{+}\).
03

Identify the Correct Ordering from Provided Options

From the analysis, we should expect an order that places \(\mathrm{H}_{2}\mathrm{SO}_{4}\) at the highest \(\mathrm{H}_{3}\mathrm{O}^{+}\) production, followed by \(\mathrm{H}_{2} \mathrm{~S}\), then \(\mathrm{NaNO}_{2}\), and finally \(\mathrm{NaCl}\). Looking at the options given:Option (d): \(0.01 \mathrm{M} \mathrm{NaNO}_{2}<0.01 \mathrm{M} \mathrm{NaCl}<0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}<0.01 \mathrm{M} \mathrm{H}_{2}\mathrm{SO}_{4}\) matches our expected order.

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Key Concepts

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

Hydronium Ion Production
Hydronium ions ( H_3O^+ ) are a crucial component in acid-base chemistry, serving as an indicator of a solution's acidity. When an acid dissolves in water, it releases hydrogen ions ( H^+ ), which then associate with water molecules to form hydronium ions. The concentration of hydronium ions directly determines the acidity of a solution.

To understand how hydronium ions are produced, consider the example of adding hydrochloric acid ( HCl ) to water. HCl , being a strong acid, completely dissociates into H^+ and Cl^- ions. The H^+ ions quickly bond with water to form a high concentration of H_3O^+ , making the solution highly acidic.

In contrast, acetic acid ( CH_3COOH ) is a weak acid. When CH_3COOH dissolves in water, only some of its molecules dissociate to form H_3O^+ , resulting in a lower hydronium ion concentration compared to a strong acid at the same molarity. This illustrates the difference in H_3O^+ production between strong and weak acids.
Strong and Weak Acids
The categorization of acids into strong and weak stems from how completely they dissociate in water to produce H_3O^+ ions. Understanding this concept helps in predicting a solution's acidity.

Strong Acids:
  • Completely dissociate in water, releasing all their hydrogen ions.
  • Examples include sulfuric acid ( H_2SO_4 ) and nitric acid ( HNO_3 ).
  • Create very high concentrations of H_3O^+ , leading to low pH values.
Weak Acids:
  • Only partially dissociate in water.
  • Common examples are formic acid ( HCOOH ) and hydrogen sulfide ( H_2S ).
  • Produce lower concentrations of H_3O^+ ; hence, they are less acidic but can still affect pH significantly depending on their concentration.
Both strong and weak acids play significant roles in chemistry, influencing not only the pH of solutions but also reactions and mechanisms across various scientific fields.
Salt Solutions and pH
Salts form from the reaction between acids and bases, and their behavior in water can impact the pH of a solution. Not all salt solutions are neutral; their effect on pH depends on the strength of the original acids and bases from which they were derived.

Neutral Salts:
  • Formed from strong acids and strong bases, resulting in no significant change to the H_3O^+ ion concentration.
  • Sodium chloride ( NaCl ) is an example, originating from hydrochloric acid, a strong acid, and sodium hydroxide, a strong base.
Salts from Weak Acids or Bases:
  • Can produce solutions that are either slightly acidic or basic due to the partial dissociation of the contributing weak component.
  • Sodium nitrite ( NaNO_2 ) derives from nitric acid and a strong base, but because nitrous acid ( HNO_2 ) is a weak acid, its resulting solution is slightly basic due to a higher concentration of OH^- ions.
It's important to analyze the components of a salt and their strengths to anticipate the resulting pH of a given salt solution. This understanding helps chemists predict how salts will interact within various systems.

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Most popular questions from this chapter

\(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) in the above reaction \(K_{p}\) and \(K_{c}\) are related as: (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})\) (b) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})^{-1}\) (c) \(K_{c}=K_{p} \times(R T)^{2}\) (d) \(K_{p}=K_{c} \times(R T)^{-2}\)

In which of the following gaseous reaction, \(\mathrm{K}_{\mathrm{p}}\) and \(\mathrm{K}_{\mathrm{c}}\) have the same values: (a) \(2 \mathrm{Hl} \rightleftharpoons \mathrm{H}_{2}+\mathrm{I}_{2}\) (b) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) (c) \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\) (d) \(\mathrm{PCI}_{5} \rightleftharpoons \mathrm{PCI}_{3}+\mathrm{Cl}_{2}^{3}\)

4 moles each of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) gases are allowed to react to form \(\mathrm{SO}_{3}\) in a closed vessel. At equilibrium \(25 \%\) of \(\mathrm{O}_{2}\) is used up. The total number of moles of all the gases at equilibrium is: (a) \(6.5\) (b) \(7.0\) (c) \(8.0\) (d) \(2.0\)

The law of chemical equilibrium was first given by: (a) Guldberg (b) Boyle (c) Waage (d) Both (a) and (c)

At \(700 \mathrm{~K}\), the equilibrium constant \(\mathrm{K}_{\mathrm{p}}\) for the reaction \(2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) is \(1.80 \times 10^{-3}\) What is the numerical value in mole per litre of equilibrium constant \(\mathrm{K}_{\mathrm{c}}\) for this reaction at the same temperature: (a) \(8.1 \times 10^{-8}\) (b) \(9.1 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}\) (c) \(3.1 \times 10^{-7}\) (d) \(6.1 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\)

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