Final answer:
The concentration of hydronium ions in a solution where 6.00 g of NaOH has been dissolved in 5.00 L of water is 3.33 × 10-13 M, calculated using the ionic product of water at 25 °C.
Step-by-step explanation:
The student is asking for the concentration of hydronium ions, H3O+, in a solution where 6.00 g of NaOH has been dissolved in 5.00 L of water. Sodium hydroxide is a strong base, and it dissociates completely in water into sodium ions Na+ and hydroxide ions OH-. To find the concentration of hydronium ions, we first calculate the molarity of the NaOH solution.
The molecular weight of NaOH is approximately 40.00 g/mol. Therefore, 6.00 g of NaOH is equal to 6.00 g / 40.00 g/mol = 0.15 mol. Given that this amount is dissolved in 5.00 L of solution, the molarity (M) of NaOH is 0.15 mol / 5.00 L = 0.03 M.
Under conditions where temperature is at 25 °C, water auto-ionizes minimally, and the ionic product of water (Kw) is 1.0 × 10-14. As such, we can calculate the concentration of hydronium ions using the equation [H3O+] × [OH-] = Kw, where [OH-] is the concentration of hydroxide ions, which is equal to 0.03 M. Therefore:
[H3O+] = Kw / [OH-] = 1.0 × 10-14 / 0.03 M = 3.33 × 10-13 M.
The hydronium ion concentration in this NaOH solution is 3.33 × 10-13 M.