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The \(\mathrm{pH}\) of the neutralization point of \(0.1 \mathrm{~N}\) ammonium hydroxide with \(0.1 \mathrm{NHCl}\) is: (a) 1 (b) 6 (c) 9 (d) 7

Short Answer

Expert verified
The pH at the neutralization point is 7.

Step by step solution

01

Analyze the chemical reaction

Ammonium hydroxide ( \( ext{NH}_4 ext{OH} \)) is a weak base and hydrochloric acid ( \( ext{HCl} \)) is a strong acid. When these two react, they form water ( \( ext{H}_2 ext{O} \)) and ammonium chloride ( \( ext{NH}_4 ext{Cl} \)) according to the equation:\[\text{NH}_4\text{OH} + \text{HCl} \rightarrow \text{NH}_4\text{Cl} + \text{H}_2\text{O}\]

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Key Concepts

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

Ammonium Hydroxide
Ammonium hydroxide, also known as \(\text{NH}_4\text{OH}\), is a solution of ammonia gas dissolved in water. It's an example of a weak base, which means it doesn't completely dissociate in water. As a weak base, ammonium hydroxide exists in equilibrium with its ionized form, ammonia (\(\text{NH}_3\)) and the hydroxide ion (\(\text{OH}^-\)).

It's essential to recognize what a weak base means in chemistry:
  • It does not ionize fully in solution.
  • Only a fraction of the ammonium hydroxide molecules dissociate to form hydroxide ions.
  • It results in a lower concentration of \(\text{OH}^-\) ions compared to a strong base of the same concentration.
In the context of a neutralization reaction, ammonium hydroxide behaves differently than a strong base. Its contribution to the solution’s alkalinity is less intense, which impacts the \text{pH}\ after neutralization.
pH Calculation
Calculating \text{pH}\ in a reaction involving a weak base like ammonium hydroxide requires understanding how \text{pH}\ is impacted by the balance of acid and base. Typically, the \text{pH}\ scale ranges from 0 to 14, where values below 7 indicate an acidic solution, and values above 7 indicate a basic one. Neutral solutions are exactly at \text{pH}\ 7.

For ammonium hydroxide neutralized by hydrochloric acid in equal normality (0.1 \text{N}\ each), the resulting solution will not be neutral with a \text{pH}\ of 7 due to the nature of the weak base. Instead, the \text{pH}\ after complete neutralization tends towards being slightly acidic due to the weak base strong acid combination. Ammonium ion (\(\text{NH}_4^+\)) is a conjugate acid of ammonia, and when formed it slightly lowers the \text{pH}\ of the solution.
Weak Base Strong Acid Neutralization
The neutralization reaction between a weak base like ammonium hydroxide and a strong acid such as hydrochloric acid is an interesting chemical process. During this reaction, each molecule of \(\text{NH}_4\text{OH}\) combines with \(\text{HCl}\) to form \(\text{NH}_4\text{Cl}\) and water (\(\text{H}_2\text{O}\)).

Here’s what typically happens:
  • The strong acid provides more \(\text{H}^+\) ions than needed to neutralize the hydroxide ions from the weak base, driving the formation of an acidic salt.
  • As \(\text{NH}_4^+\) ions form, they shift the equilibrium, slightly lowering the \text{pH}\ of the solution.
  • This results in a neutralization point \text{pH}\ below 7, usually around 5-6 depending on the exact conditions.
This phenomenon emphasizes that the reaction’s outcome doesn’t lead to a perfectly neutral \text{pH}\, due to the limited disassociation and weaker nature of the base.

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Most popular questions from this chapter

In which of the following cases does the reaction go farthest to completion: (a) \(\mathrm{K}=1\) (b) \(\mathrm{K}=10\) (c) \(\mathrm{K}=10^{-2}\) (d) \(\mathrm{K}=10^{2}\)

Equilibrium constant for the reaction: \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})\) is \(81 .\) If the velocity constant of the forward reaction is \(162 \mathrm{~L}\) \(\mathrm{mol}^{-1} \mathrm{sec}^{-1}\), what is the velocity constant (in \(\mathrm{L} \mathrm{mol}^{-1}\) sec. \(^{-1}\) ) for the backward reaction: (a) 13122 (b) 2 (c) 261 (d) 243

\(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) 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) \(K_{p}=K_{c} \times(R T)^{-2}\)

At equilibrium total number of moles for the reaction \(2 \mathrm{HI} \rightleftharpoons \mathrm{H}_{2}+\mathrm{I}_{2}\), if \(\alpha\) is degree of dissociation, are: (a) 2 (b) \(2-\alpha\) (c) 1 (d) \(\mathrm{I}-\alpha\)

If an inert gas is added in the reaction: \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) at constant volume, then its equi- librium (a) Remains unaffected (b) Favours the backward reaction (c) Favours the forward reaction (d) Increases the dissociation of reactants

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