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Chapter 14: Problem 14

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

Short Answer

Expert verified
The pH of water at \(25^{\circ} \mathrm{C}\) is equal to 7.00 because the ion product constant for water (Kw) at this temperature is \(1 \times 10^{-14}\). In pure water, the concentrations of hydrogen ions [\(H^+\)] and hydroxide ions [\(OH^-\)] are equal. Thus, each ion concentration is \(1 \times 10^{-7}\) M. Using the pH formula, \(pH = -\log[H^+]\), we get a value of 7.00 for the pH of pure water at \(25^{\circ} \mathrm{C}\).

Step by step solution

01

Understand the definition of pH

The pH of a solution is defined as the negative logarithm of the hydrogen ion concentration: \[ pH = -\log[H^{+}] \] where \(H^{+}\) represents the concentration of hydrogen ions in the solution.
02

Calculate the ion product constant for water (Kw) at 25掳C

In water, a small fraction of the water molecules dissociate into hydrogen ions (H鈦�) and hydroxide ions (OH鈦�). This can be represented by the following chemical equation: \[ H_{2}O \rightleftharpoons H^{+} + OH^{-} \] The ion product constant for water (Kw) is the equilibrium constant for this dissociation, and it is given by the product of the concentrations of hydrogen and hydroxide ions: \[ K_w = [H^{+}][OH^{-}] \] At 25掳C, the value of Kw is equal to \(1 \times 10^{-14}\).
03

Equate the concentrations of hydrogen and hydroxide ions in pure water

In pure water, the concentrations of hydrogen ions and hydroxide ions are equal: \[ [H^{+}] = [OH^{-}] \] Since both concentrations are equal, we can introduce a variable x for both concentrations: \[ [H^{+}] = [OH^{-}] = x \]
04

Solve for the hydrogen ion concentration

Now, we can substitute this value of x into the equation for Kw to find the concentration of hydrogen ions in pure water at 25掳C: \[ K_w = x^2 \] Solving for x, we get: \[ x = \sqrt{K_w} = \sqrt{1 \times 10^{-14}} \] Thus, \(x = 1 \times 10^{-7}\), which is the concentration of hydrogen ions in pure water at 25掳C.
05

Calculate the pH of water at 25掳C

Finally, we can use the definition of pH to calculate the pH of water at 25掳C: \[ pH = -\log[H^{+}] = -\log(1 \times 10^{-7}) \] Therefore, the pH of water at 25掳C is equal to 7.00.

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