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A solution has \(\mathrm{pH}=5\), it is diluted 100 times, then it will become: a. Neutral b. More acidic c. Basic d. Unaffected

Short Answer

Expert verified
The solution becomes neutral (pH=7).

Step by step solution

01

Understanding Initial pH

The initial pH of the solution is given as 5. The pH scale measures the acidity or basicity of a solution, where a pH less than 7 indicates an acidic solution, a pH of 7 is neutral, and a pH greater than 7 is basic. Therefore, a pH of 5 indicates that the solution is acidic.
02

Concept of Dilution

When a solution is diluted, its concentration of hydrogen ions decreases. Since pH is related to the hydrogen ion concentration \( ext{[H}^+\)", this means dilution will affect the pH.
03

Calculate New Hydrogen Ion Concentration After Dilution

The initial concentration can be expressed as \( ext{[H}^+] = 10^{-5}\) M. When we dilute the solution 100 times, the new concentration becomes \( rac{10^{-5}}{100} = 10^{-7}\) M.
04

Determine New pH

The new concentration of \( ext{[H}^+]\) is \((10^{-7})\) M, which corresponds to a pH of 7. This is calculated as \(pH = - ext{log}(10^{-7}) = 7\).
05

Determine the nature of the solution after dilution

A pH of 7 indicates that the solution becomes neutral after the dilution process.

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Key Concepts

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

Solution Dilution
Solution dilution involves reducing the concentration of a solute in a solution by adding more solvent, typically water. The process of dilution doesn't change the amount of solute present but spreads it out over a larger volume of solvent, thereby decreasing its concentration.
  • When a solution is diluted, the concentration of hydrogen ions, represented as \( \text{[H}^+] \), is affected.
  • This change in concentration results in a change to the pH of the solution.
Consider a solution with a pH of 5. Diluting it 100 times significantly decreases the concentration of hydrogen ions, demonstrating how dilution can shift the pH towards neutrality. After dilution, the new hydrogen ion concentration can be calculated using the formula \( \text{[H}^+]_{\text{new}} = \frac{\text{[H}^+]_{\text{initial}}}{\text{dilution factor}} \).

The result shows how a solution initially acidic becomes less acidic due to dilution, moving towards a neutral pH of 7.
Hydrogen Ion Concentration
The concept of hydrogen ion concentration, denoted as \( \text{[H}^+] \), is central to understanding the pH of a solution. The pH scale measures how acidic or basic a solution is based on its hydrogen ion concentration.
  • Low \( \text{[H}^+] \) implies a higher pH, indicating a basic solution.
  • High \( \text{[H}^+] \) implies a lower pH, indicating an acidic solution.
For a solution with a pH of 5, the concentration of hydrogen ions can be determined using the formula \( \text{[H}^+] = 10^{-\text{pH}} \). If you dilute this solution 100 times, the concentration changes from \( 10^{-5} \) M to \( 10^{-7} \) M.

This shift in concentration directly affects the solution's pH, moving from acidic (pH 5) to neutral (pH 7). Understanding this relation solidifies the understanding of how solutions behave during dilution.
Acidic and Basic Solutions
Acidic and basic solutions are defined based on their pH levels, giving insight into their chemical nature and behavior in reactions.
  • Acidic solutions have a pH less than 7 and contain higher hydrogen ion concentrations.
  • Basic solutions have a pH greater than 7 and contain lower hydrogen ion concentrations.
  • A pH of exactly 7 denotes a neutral solution, like pure water.
Initially, a solution with a pH of 5 is acidic because its hydrogen ion concentration is higher than that of neutral water. As the solution is diluted 100 times, its pH shifts to 7, indicating it has become neutral.

Understanding the classification of solutions based on acidity and basicity helps predict the behavior of the solution during chemical reactions or environmental changes, such as dilution.

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Most popular questions from this chapter

Which metal sulphides can be precipitated from a solution that is \(0.01 \mathrm{M}\) in \(\mathrm{Mn}^{2+}, \mathrm{Zn}^{2+}, \mathrm{Pb}^{2+}\) and \(\mathrm{Cu}^{2+}\) and \(0.10 \mathrm{M}\) in \(\mathrm{H}_{2} \mathrm{~S}\) at a \(\mathrm{pH}\) of \(1.0\) ? Metal sulphide \(\quad \mathrm{Ksp}\) MnS \(3 \times 10^{16}\) ZnS \(3 \times 10^{-2}\) PbS \(3 \times 10^{-7}\) \(\mathrm{CuS} \quad 6 \times 10^{-16}\) a. \(\mathrm{CuS}\) b. \(\mathrm{MnS}\) c. \(\mathrm{ZnS}, \mathrm{PbS}, \mathrm{CuS}\) d. PbS, CuS

Which of the following are homogeneous equilibria? a. \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\) b. \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{X}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HX}(\mathrm{g})\) c. \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) d. \(\mathrm{CaCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\)

Which of the following reaction can have same units of \(\mathrm{K}_{\mathrm{p}}\) ? a. \(\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\) b. \(2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) c. \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}(\mathrm{g})\) d. \(\mathrm{XX}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{XY}(\mathrm{g})+\mathrm{Y}(\mathrm{g})\)

For the decomposition of \(\mathrm{PCl}_{5}\) (g) in a closed vessel which is the correct relation between totalpressure (P), equilibrium constant \(\left(\mathrm{K}_{\mathrm{p}}\right)\) and degree of dissociation \((\alpha)\) a. \(\alpha=\sqrt{\left(K_{p}+P\right)}\) b. \(\alpha=\sqrt{\left[1 /\left(\mathrm{K}_{\mathrm{p}}+\mathrm{P}\right)\right]}\) c. \(\alpha=\sqrt{\left[K_{p} /\left(\mathrm{K}_{\mathrm{p}}+\mathrm{P}\right)\right]}\) d. \(\left.\alpha=\sqrt{[}\left(\mathrm{K}_{\mathrm{n}}+\mathrm{P}\right) / \mathrm{K}_{\mathrm{n}}\right]\)

At \(298 \mathrm{~K}\) the value of ionization constant of anilinium hydroxide is \(4.6 \times 10^{-10}\) and that of ionic product of water \(1 \times 10^{-14}\), the value of degree of hydrolysis constant is nearly? a. \(0.415 \%\) b. \(4.15 \%\) c. \(0.163 \%\) d. \(0.217 \%\)

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