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Chapter 15: Problem 134

A solution is prepared by dissolving \(2.00\) moles of \(\mathrm{HNO}_{3}\) in enough water to get \(800.0 \mathrm{~mL}\) of solution. What are the \(\mathrm{H}_{3} \mathrm{O}^{+}\) and the \(\mathrm{OH}\) molar concentrations?

Short Answer

Expert verified
The molarity of HNO₃ is \(M = \frac{2.00 \text{ moles}}{0.8 \text{ L}} = 2.50 \text{ M}\). Since HNO₃ is a strong acid, its concentration is equal to the H₃O⁺ concentration: \([H₃O⁺] = 2.50 \text{ M}\). Using the ion product of water, \(K_w = 1.0 \times 10^{-14}\), we find the OH⁻ concentration: \([OH⁻] = \frac{K_w}{[H₃O⁺]} = \frac{1.0 \times 10^{-14}}{2.50} \approx 4.0 \times 10^{-15} \text{ M}\).

Step by step solution

01

Calculate the molarity of HNO₃ solution

To find the molarity, divide the number of moles of solute by the volume of the solution (in liters). The volume of the solution is 800.0 mL, which is 0.8 L. Molarity (M) = moles of solute / volume of solution (L) M = 2.00 moles / 0.8 L
02

Calculate the H₃O⁺ concentration

Since nitric acid (HNO₃) is a strong acid, it will ionize completely in water: HNO₃ → H⁺ + NO₃⁻ By doing so, it will release H⁺ ions, which will combine with water molecules to form H₃O⁺ ions in the solution. Hence, the concentration of H₃O⁺ ions will be equal to the concentration of HNO₃ in the solution: [H₃O⁺] = M (HNO₃)
03

Calculate the OH⁻ concentration

Using the relation between H₃O⁺ and OH⁻ concentrations, we can calculate the OH⁻ concentration. This relation is given by the ion product of water (Kw): Kw = [H₃O⁺] × [OH⁻] At 25℃, the value of Kw is \(1.0 \times 10^{-14}\). Rearrange the equation to solve for [OH⁻]: [OH⁻] = Kw / [H₃O⁺] Now we can plug in the values for the molarity of HNO₃, H₃O⁺ concentration, and Kw to calculate the OH⁻ concentration.

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