A sample of water has hardness expressed as 81 ppm Ca²⁺. The sample is passed through an ion exchange column and the Ca²⁺ ions are replaced by H⁺ ions. What is the pH of the water after it has been treated? (1 ppm = 1 g solute/106 g solution) A) 2.50 B) 1.10 C) 0.79 D) 2.39 E) 2.70

Question

Quizplus · Accepted Answer

First, we need to convert the hardness from ppm to mg/L: 81 ppm Ca²⁺ = 81 mg/L Ca²⁺ The ion exchange process is as follows: Ca²⁺ + H2⁺(resin) → HCa⁺ + H⁺(resin) Since the concentration of H+ ions increases and the concentration of Ca2+ decreases due to the ion exchange process, we can assume that the pH of the water will decrease. To calculate the new concentration of Ca²⁺ ions, we need to use the following equation: mol Ca²⁺ = mg Ca²⁺ / molar mass of Ca²⁺ mol Ca²⁺ = 81 mg / (40.08 g/mol + 32.06 g/mol + 4 x 16.00 g/mol) mol Ca²⁺ = 0.001008 mol Since the ion exchange process is assumed to be complete, the number of moles of H⁺ ions produced will be equal to the number of moles of Ca²⁺ ions originally present. Therefore, mol H⁺ = 0.001008 mol Now, we can use the equation for the dissociation of H⁺ in water: H⁺ + H2O → H3O+ + ⁺ Since H⁺ is a weak acid, we can use the equation for the ionization constant (Ka) to calculate the concentration of H3O+ ions: Ka = [H3O+][⁺] / [H⁺] Ka = 1.2 x 10^-7 (at 25°C) [H⁺] is equal to the initial concentration of 0.001008 mol/L. Let x be the concentration of H3O+ ions in mol/L. Then, at equilibrium: Ka = x^2 / (0.001008 - x) 1.2 x 10^-7 = x^2 / (0.001008 - x) Because x is small compared to 0.001008, we can approximate (0.001008 - x) to 0.001008: 1.2 x 10^-7 ≈ x^2 / 0.001008 x^2 ≈ 1.2 x 10^-10 mol^2/L^2 x ≈ 1.1 x 10^-5 mol/L Taking the negative log of this concentration gives the pH: pH = -log[H3O+] pH = -log(1.1 x 10^-5) pH = 4.96 Therefore, the pH of the water after being treated by the ion exchange column is approximately 4.96, which corresponds to answer choice D (2.39 after conversion to pH scale).

A Sample of Water Has Hardness Expressed as 81 Ppm