Question

Calculate the \(\mathrm{pH}\) of a \(1.0 \times 10^{-7}-M\) solution of \(\mathrm{NaOH}\) in water.

Short answer

The pH of a \(1.0 \times 10^{-7} \ \text{M}\) solution of NaOH in water is 7.

Step-by-step solution

Write the dissociation reaction of NaOH in water

Write the reaction of NaOH dissociating into its ions in water: NaOH(aq) ⟶ Na⁺(aq) + OH⁻(aq)

Write the equilibrium reaction for self-ionization of water

Write the self-ionization of water reaction where water reacts with itself to form hydronium ions (H₃O⁺) and hydroxide ions (OH⁻): 2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)

Find the concentration of OH⁻

Since NaOH completely dissociates in water, the concentration of OH⁻ is equal to the concentration of NaOH: [OH⁻] = \(1.0 \times 10^{-7} \ \text{M}\)

Find the concentration of H₃O⁺ ions using equilibrium constant (Kw)

The ion product of water, \(K_w\), is the product of the concentration of H₃O⁺ ions and OH⁻ ions in the solution. At 25°C, \(K_w = 1.0 \times 10^{-14}\). \(K_w = [\text{H}_3\text{O}^+][\text{OH}^-]\) Rearrange the equation to find the concentration of H₃O⁺ ions: \([\text{H}_3\text{O}^+] = \dfrac{K_w}{[\text{OH}^-]}\) Insert the given values: \([\text{H}_3\text{O}^+] = \dfrac{1.0 \times 10^{-14}}{1.0 \times 10^{-7}}\) Solve for H₃O⁺ concentration: \([\text{H}_3\text{O}^+] = 1.0 \times 10^{-7} \ \text{M}\)

Calculate the pH of the solution

Use the formula for pH, which is \(pH = -\log_{10} [\text{H}_3\text{O}^+]\), to calculate the pH of the solution: \(pH = -\log_{10} (1.0 \times 10^{-7})\) Solve for pH: \(pH = 7\) So, the pH of a \(1.0 \times 10^{-7} \ \text{M}\) NaOH solution in water is 7.