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Chapter 8: Problem 14

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

Short Answer

Expert verified
The pH of pure water at \(80^{\circ} \mathrm{C}\) is (b) \(<7\).

Step by step solution

01

Understand the Relationship between Temperature and Kw

The ion product of water, often denoted as \(K_w\), changes with temperature. At \(25^{\circ} C\), the \(K_w\) is \(1.0 \times 10^{-14}\). However, at higher temperatures such as \(80^{\circ} C\), \(K_w\) increases, indicating more ionization of water molecules, which affects the \(\text{pH}\).
02

Define pH in Terms of Kw

The \(\text{pH}\) is defined as \( -\log[H^+] \). In pure water, \([H^+] = [OH^-]\), so \(K_w = [H^+][OH^-] = [H^+]^2\.\) Therefore, \([H^+] = \sqrt{K_w}\). As \(K_w\) increases with temperature, the concentration \([H^+]\) also increases.
03

Determine the Resulting pH Value

As \([H^+]\) increases while \(K_w\) increases, the \(\text{pH}\) (which equals \(-\log[H^+]\)) decreases. Therefore, the \(\text{pH}\) of pure water at \(80^{\circ} C\) will be less than 7, indicating that water becomes more acidic in terms of numerical pH scale at increased temperatures.
04

Choose the Correct Option

Given that the \(\text{pH}\) decreases with increased temperature, pure water at \(80^{\circ} C\) will have a \(\text{pH}\) less than 7. Thus, the correct option is (b) \(<7\).

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Key Concepts

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

Temperature Effect on \(K_w\)
Water's ionization product, denoted as \(K_w\), is highly dependent on temperature.
At \(25^{\circ} C\), \(K_w\) is commonly known to be \(1.0 \times 10^{-14}\).
As the water temperature rises, such as at \(80^{\circ} C\), \(K_w\) increases, and this alteration is crucial in understanding how variations in temperature affect water's properties.
This increase in \(K_w\) implies more ionization of water, meaning more molecules break up into ions.
A higher \(K_w\) shows increased ionization, which has a direct impact on the calculation and understanding of pH changes when water is heated. Essentially:
  • Higher temperature \(\rightarrow\) higher \(K_w\)
  • More ionization \(\rightarrow\) greater presence of \([H^+]\) and \([OH^-]\) ions
These dynamic changes help explain the variations in pH with temperature shifts in water.
Ionization of Water
The process of ionization in water refers to water molecules dissociating into hydrogen ions \([H^+]\) and hydroxide ions \([OH^-]\).
At room temperature, water is partially ionized, maintaining a balance in its neutral state.
When we discuss ionization at varying temperatures, such as \(80^{\circ} C\), we observe an increase in the number of ions present.
The ionization equation for water is:\[H_2O(l) \leftrightarrows H^+(aq) + OH^-(aq)\]With increased temperature, the equilibrium shifts to favor more ionization, meaning that there are more \([H^+]\) and \([OH^-]\) ions in the water.
Key insights:
  • Increase in temperature spikes the ionization process
  • Explains increased values of \([H^+]\) and \(K_w\) at higher temperatures
This helps students understand why the \(pH\) scale, typically at \(7\) for pure water at room temperature, shifts at higher temperatures.
Relationship between \(K_w\) and pH
The pH is a logarithmic scale used to express the acidity or basicity of a solution, calculated using the formula \(\text{pH} = -\log[H^+]\).
In pure water, the concentrations of hydrogen ions \([H^+]\) and hydroxide ions \([OH^-]\) are equal.
The equation \(K_w = [H^+][OH^-] = [H^+]^2\) confirms this balance.
When \(K_w\) increases, due to factors like rising temperature, the concentration \([H^+]\) will increase as well. This affects the pH value:
  • As \([H^+]\) rises, the pH value lowers
  • A higher \(K_w\) at increased temperatures indicates that pure water becomes more 'acidic' in numerical terms
Thus, even though chemically water remains neutral, its pH value tends to decline below \(7\) at higher temperatures.
Understanding this intricate relationship helps in predicting the behavior of water's pH in various thermal conditions.

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Most popular questions from this chapter

Which of the following are the correct statements (a) The \(\mathrm{pH}\) of blood is same in summer and winter (b) \(\mathrm{pH}\) of an acidic buffer increases if more salt is added (c) \(\mathrm{pH}\) of a basic buffer decreases if more salt is added (d) The term solubility product is only for sparingly soluble salts

Which of the following statements are correct? (a) The conjugate base of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)is \(\mathrm{HPO}_{4}^{2-}\). (b) \(\mathrm{pH}\) of \(1.0 \times 10^{-8} \mathrm{M}\) aqueous solution of \(\mathrm{HCl}\) is 8 . (c) When a weak monoprotic acid solution is treated with a strong base, at half neutralization point, \(\mathrm{pH}=\frac{1}{2} \mathrm{pK}_{\mathrm{a}}\) (d) The autoprotolysis constant of water increases with temperature.

One of the following species acts as both Bronsted acid and base (a) \(\mathrm{H}_{2} \mathrm{PO}_{2}\) (b) \(\mathrm{HPO}_{3}^{2}\) (c) \(\mathrm{HPO}_{4}^{2}\) (d) all of the above

Solubility of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is \(\mathrm{S}\) mol litre \(^{-1} .\) The solubility product (Ksp) under the same condition is [2002] (a) \(4 \mathrm{~S}^{3}\) (b) \(3 \mathrm{~S}^{4}\) (c) \(4 \mathrm{~S}^{2}\) (d) \(\mathrm{S}^{3}\)

In the precipitation titration of \(\mathrm{KCl}\) against \(\mathrm{AgNO}_{3}\), \(\mathrm{K}_{2} \mathrm{CrO}_{4}\) is used as an indicator since, \(\mathrm{AgCl}\) is white coloured. End point is detected by appearance of deep yellow coloured precipitate of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\). The minimum concentration of chromate ion required for detectionof end point is \(\left[\mathrm{K}_{s p}\right.\) of \(\mathrm{AgCl}=2.5 \times 10^{-10}\) and \(\mathrm{K}_{\text {sp }}\) of \(\left.\mathrm{Ag}_{2} \mathrm{CrO}_{4}=1.8 \times 10^{-12}\right]\) (a) \(7.3 \times 10^{-2} \mathrm{M}\) (b) \(5.3 \times 10^{-4} \mathrm{M}\) (c) \(7.3 \times 10^{-3} \mathrm{M}\) (d) \(3.6 \times 10^{-5} \mathrm{M}\)
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