A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are \(50.0 \mathrm{~mL}\) of \(0.100 M\) hydrochloric acid, \(100.0 \mathrm{~mL}\) of \(0.200 M\) of nitric acid, \(500.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M}\) calcium hydroxide, and \(200.0 \mathrm{~mL}\) of \(0.100 M\) rubidium hydroxide. Did the acids and bases exactly neutralize each other? If not, calculate the concentration of excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions left in solution.

To calculate the moles of \(\mathrm{H}^{+}\) ions from the hydrochloric acid and nitric acid solutions, we'll use the formula: moles = molarity × volume. Since both these acids are strong acids, they will dissociate completely in the solution. For hydrochloric acid: Moles of \(\mathrm{H}^{+}\) = molarity × volume = \(0.100\,\mathrm{M}\) × \(50.0\,\mathrm{mL}\) = \(0.100\,\mathrm{M}\) × \(0.050\,\mathrm{L}\) = 0.00500 mol For nitric acid: Moles of \(\mathrm{H}^{+}\) = molarity × volume = \(0.200\,\mathrm{M}\) × \(100.0\,\mathrm{mL}\) = \(0.200\,\mathrm{M}\) × \(0.100\,\mathrm{L}\) = 0.0200 mol Total moles of \(\mathrm{H}^{+}\) ions = 0.00500 mol + 0.0200 mol = 0.0250 mol

To calculate the moles of \(\mathrm{OH}^{-}\) ions from the calcium hydroxide and rubidium hydroxide solutions, we'll use the same formula as before. Since both these compounds are strong bases, they will dissociate completely in the solution. For calcium hydroxide (it produces 2 moles of \(\mathrm{OH}^{-}\) ions for each mole of calcium hydroxide): Moles of \(\mathrm{OH}^{-}\) = (molarity × volume) × 2 = \((0.0100\,\mathrm{M})\times(500.0\,\mathrm{mL})\times2\) = \((0.0100\,\mathrm{M})\times(0.500\,\mathrm{L})\times2\) = 0.0100 mol For rubidium hydroxide: Moles of \(\mathrm{OH}^{-}\) = molarity × volume = \(0.100\,\mathrm{M}\) × \(200.0\,\mathrm{mL}\) = \(0.100\,\mathrm{M}\) × \(0.200\,\mathrm{L}\) = 0.0200 mol Total moles of \(\mathrm{OH}^{-}\) ions = 0.0100 mol + 0.0200 mol = 0.0300 mol

Now, we will compare the moles of \(\mathrm{H}^{+}\) ions and \(\mathrm{OH}^{-}\) ions to determine if the acids and bases neutralize each other completely. Since we have 0.0250 mol of \(\mathrm{H}^{+}\) ions and 0.0300 mol of \(\mathrm{OH}^{-}\) ions, it is clear that the solution is not completely neutralized. The excess ions are \(\mathrm{OH}^{-}\) ions. Moles of excess \(\mathrm{OH}^{-}\) ions = 0.0300 mol - 0.0250 mol = 0.00500 mol

To calculate the concentration of excess \(\mathrm{OH}^{-}\) ions in the solution, we will use the following formula: concentration = moles/volume. First, we need to know the total volume for our mixture: Total volume = volume of HCl + volume of HNO3 + volume of Ca(OH)2 + volume of RbOH = \(50.0\,\mathrm{mL} + 100.0\,\mathrm{mL} + 500.0\,\mathrm{mL} + 200.0\,\mathrm{mL}\) = 850.0 ml = 0.850 L Now, we can calculate the concentration of the excess \(\mathrm{OH}^{-}\) ions: Concentration of excess \(\mathrm{OH}^{-}\) ions = moles/volume = 0.00500 mol / 0.850 L = 0.00588 M (rounded to five significant figures) Thus, the concentration of excess \(\mathrm{OH}^{-}\) ions in the solution after mixing is 0.00588 M.