Answer :
Final answer:
The concentration of hydronium ions, [H₃O⁺], in a solution prepared by dissolving 2.00 g of NaOH in 2.00 L of water is 4.0 x 10⁻¹³ M.
Explanation:
To determine the concentration of hydronium ions, [H₃O⁺], in a solution when 2.00 g of NaOH is dissolved in 2.00 L of water, we first need to calculate the hydroxide ion concentration, [OH⁺]. This is because sodium hydroxide (NaOH) dissociates completely in water, producing an equal concentration of sodium ions (Na⁺) and hydroxide ions (OH⁺).
First, we calculate the molarity of the NaOH solution: The molar mass of NaOH is 40.0 g/mol. Therefore, 2.00 g of NaOH equates to 2.00/40.0 = 0.050 moles of NaOH. Given that the solution's total volume is 2.00 L, the molarity of NaOH (and thereby OH⁺) is 0.050 moles / 2.00 L = 0.025 M.
Water autoionizes, meaning it spontaneously splits into hydroxide and hydronium ions, following the equilibrium constant for water (Kw = [H₃O⁺][OH⁺] = 1.0 x 10⁻¹⁴ at 25 °C). Therefore, to find [H₃O⁺], we use the equation Kw = [OH⁺][H₃O⁺]. Rearranging for [H₃O⁺], we get [H₃O⁺] = Kw / [OH⁺] = 1.0 x 10⁻¹⁴ M / 0.025 M = 4.0 x 10⁻¹³ M.
