Problem 14 - Chemistry

Question

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

Step-by-Step Solution

Answer

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 1: 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.

Step 2: Calculate the ion product constant for water (Kw) at 25°C

In water, a small fraction of the water molecules dissociate into hydrogen ions (H⁺) and hydroxide ions (OH⁻). 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°C, the value of Kw is equal to \(1 \times 10^{-14}\).

Step 3: 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: \[ [H^{+}] = [OH^{-}] = x \]

Step 4: 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°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°C.

Step 5: Calculate the pH of water at 25°C

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