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The \(\mathrm{pH}\) of pure water at \(80^{\circ} \mathrm{C}\) will be: \((\mathrm{a})=7\) (b) \(<7\) (c) \(>7\) (d) None of these

Short Answer

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The \\(\mathrm{pH}\\) of pure water at \\(80^{\circ} \, \mathrm{C}\\) will be \\<7\\.

Step by step solution

01

Understanding Pure Water at Different Temperatures

Pure water at any temperature is neutral, meaning it always maintains an equal concentration of hydrogen ions \(H^+\) and hydroxide ions \(OH^-\). However, the ionic product of water \(K_w\) changes with temperature, influencing the \(\mathrm{pH}\) scale.
02

Knowing Ion Product Constant

At \(25^{\circ} \, \mathrm{C}\), \(K_w\) is \(1.0 \times 10^{-14}\), where \(\mathrm{pH} = 7\). However, at higher temperatures, like \(80^{\circ} \, \mathrm{C}\), \(K_w\) increases, meaning more \(H^+\) and \(OH^-\) are formed in water, thus reducing \(\mathrm{pH}\).
03

Exploring the pH at 80°C

At \(80^{\circ} C\), \(K_w\) will be greater than \(1.0 \times 10^{-14}\). Therefore, \(10^{-7}\), which is the concentration of \(H^+\), becomes larger, making \(\mathrm{pH}\) less than 7 while water remains neutral.
04

Determine the Correct Option

Given the increased \(K_w\) at \(80^{\circ} C\), causing a decrease in \(\mathrm{pH}\), the correct option is (b) \(<7\) since this reflects the changed scale due to temperature effects.

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Key Concepts

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

Ionic Product of Water
The ionic product of water, commonly represented as \( K_w \), is a crucial parameter in understanding the behavior of water at different temperatures. In its simplest form, \( K_w \) is the product of the concentrations of hydrogen ions \( H^+ \) and hydroxide ions \( OH^- \) in water:\[ K_w = [H^+][OH^-] \].At standard laboratory temperature, which is \( 25^{\circ}C \), \( K_w \) is approximately \( 1.0 \times 10^{-14} \). This value indicates that the concentration of \( H^+ \) and \( OH^- \) in pure water are each \( 1.0 \times 10^{-7} \), leading to a neutral \( \mathrm{pH} \) of 7.
  • The equal concentrations of \( H^+ \) and \( OH^- \) keep water neutral.
  • Even though \( K_w \) changes with temperature, neutrality remains because the ion concentrations are always balanced.
When the temperature changes, the value of \( K_w \) adjusts. This is a natural response to changes in temperature, affecting the ionization of water. As temperature increases, \( K_w \) increases, meaning both \( H^+ \) and \( OH^- \) concentrations become higher, but stay equal to each other.
Temperature Effect on pH
Temperature has a significant effect on the \( \mathrm{pH} \) of water despite its neutrality. As temperature rises, the water molecules gain energy, which increases their ability to ionize.This results in higher concentrations of hydrogen ions \( H^+ \) and hydroxide ions \( OH^- \). Consequently, the ionic product of water \( K_w \) becomes larger than \( 1.0 \times 10^{-14} \) at standard conditions.When \( K_w \) is greater, the value of hydrogen ion concentration also increases, causing the \( \mathrm{pH} \) value to drop below 7. This means pure water will have a \( \mathrm{pH} \) less than 7 at temperatures like \( 80^{\circ}C \), but it is still neutral because the hydrogen ions and hydroxide ions are in equal amounts.
  • Higher temperature leads to greater ionization of water.
  • Increased \( H^+ \) implies a lower \( \mathrm{pH} \), but neutrality is maintained.
This shift in \( \mathrm{pH} \) does not mean the water is more acidic. It simply reflects the higher degree of ionization occurring at elevated temperatures.
Neutral Water Concept
The concept of neutral water is foundational in chemistry, referring to a condition where the concentrations of hydrogen ions \( H^+ \) and hydroxide ions \( OH^- \) are equal. Generally, we equate this with a \( \mathrm{pH} \) of 7, but this is a temperature-dependent characteristic.In the context of temperature, neutrality adjusts based on \( K_w \). So, even at different temperatures, water remains neutral if \( [H^+] = [OH^-] \), even if \( \mathrm{pH} \) is less or more than 7.At \( 80^{\circ}C \), for instance, the \( \mathrm{pH} \) is less than 7. Pure water remains neutral because the increase in \( H^+ \) is matched by an increase in \( OH^- \), effectively balancing the solution.
  • Neutral water means \( [H^+] = [OH^-] \).
  • The \( \mathrm{pH} \) number alone doesn't dictate neutrality at varying temperatures.
Understanding this helps in appreciating how water's neutrality remains constant across different conditions, providing an insight into the adaptability and resilience of chemical equilibria in nature.

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Most popular questions from this chapter

In the reaction \(\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}\) at equilibrium, some \(\mathrm{I}_{2}\) is added. What happens to the equilibrium: (a) It gets shifted to the right (b) It remains unchanged (c) It gets shifted to the left (d) First (b) then (c)

The relation between \(K_{n}\) and \(K_{c}\) for the reaction \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g})\) is: (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-1}\) (b) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}\) (c) \(K_{p}^{p}=K_{c} /(R T)^{2}\) (d) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} / \mathrm{RT}\)

For the chemical reaction: \(3 \mathrm{X}(\mathrm{g})+\mathrm{Y}(\mathrm{g}) \rightleftharpoons \mathrm{X}_{3} \mathrm{Y}(\mathrm{g})\), the amount of \(\mathrm{X}_{3} \mathrm{Y}\) at equilibrium is affected by (a) Temperature and pressure (b) Temperature only (c) Pressure only (d) Temperature, pressure and catalyst

What is the correct sequence of active masses in increasing order in gaseous mixture, containing one gram per litre of each of the following: 1\. \(\mathrm{NH}_{3}\) 2\. \(\mathrm{N}_{2}\) 3\. \(\mathrm{H}_{2}\) 4\. \(\mathrm{O}_{2}\) Select the correct answer using the codes given below: (a) \(3,1,4,2\) (b) \(3,4,2,1\) (c) \(2,1,4,3\) (d) \(4,2,1,3\)

The value of \(\mathrm{K}_{\mathrm{p}}\) in the reaction: \(\mathrm{MgCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) is (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{P}\left(\mathrm{CO}_{2}\right)\) (b) \(\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}\left(\mathrm{MgCO}_{3}\right)}{\mathrm{P}\left(\mathrm{CO}_{2}\right) \times \mathrm{P}(\mathrm{MgO})}\)

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