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Chapter 14: Problem 16

Why is the pH of water at \(25^{\circ} \mathrm{C}\) equal to 7.00\(?\)

Short Answer

Expert verified
At 25°C, the autoionization of water results in equal concentrations of hydrogen ions (\(H^+\)) and hydroxide ions (\(OH^-\)), with the ion product constant \(K_w = 1.0 \times 10^{-14} M^2\). The concentration of hydrogen ions in pure water at this temperature is \(1.0 \times 10^{-7}\) M. Since pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration, the pH of water at 25°C is: \[pH = -\log_{10} (1.0 \times 10^{-7}) = 7.00\]

Step by step solution

01

Understanding pH

pH is a measure of how acidic or basic a solution is. It is defined as the negative logarithm (base 10) of the hydrogen ion concentration (\([H^+]\)) in moles per liter (M). In mathematical terms, pH is given as: \[pH = -\log_{10} [H^+]\]
02

Autoionization of water

At 25°C, water undergoes autoionization, a process in which water molecules split into hydrogen ions (\(H^+\)) and hydroxide ions (\(OH^-\)). This can be represented by the following equilibrium equation: \[H_2O \rightleftharpoons H^+ + OH^-\] Under these conditions, the concentration of hydrogen ions and hydroxide ions in pure water is equal, and their product (\([H^+][OH^-]\)) equals the ion product constant for water (\(K_w\)). At 25°C, \(K_w\) is equal to \(1.0 \times 10^{-14} M^2\).
03

Calculating the concentration of hydrogen ions

Since the concentrations of hydrogen and hydroxide ions are equal in pure water at 25°C, we can write the ion product constant equation as: \[[H^+][OH^-] = K_w\] Substitute the value of \(K_w\) and solve for the concentration of hydrogen ions (\([H^+]\)): \[[H^+][H^+] = 1.0 \times 10^{-14}\] \[[H^+]^2 = 1.0 \times 10^{-14}\] \[[H^+] = \sqrt{1.0 \times 10^{-14}}\] \[[H^+] = 1.0 \times 10^{-7}\]
04

Calculating the pH of water

We now know that the hydrogen ion concentration in water at 25°C is \(1.0 \times 10^{-7}\) M. To find the pH value, we use the pH formula: \[pH = -\log_{10} [H^+]\] Substitute the value of \([H^+]\) and solve for pH: \[pH = -\log_{10} (1.0 \times 10^{-7})\] \[pH = 7.00\] So, at 25°C, the pH of pure water is equal to 7.00.

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