A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are 50.0 mL of 0.100 M hydrochloric acid, 100.0 mL of 0.200 M of nitric acid, 500.0 mL of 0.0100 M calcium hydroxide, and 200.0 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.

First, we need to calculate the amount (in moles) of \(\mathrm{H}^{+}\) ions from the hydrochloric acid and nitric acid, and the amount of \(\mathrm{OH}^{-}\) ions from calcium hydroxide and rubidium hydroxide. For hydrochloric acid: \(0.0500 \,\text{L} \times 0.100 \,\text{M} = 0.00500 \,\text{mol} \, \mathrm{H}^{+}\) For nitric acid: \(0.100 \,\text{L} \times 0.200 \,\text{M} = 0.0200 \,\text{mol} \, \mathrm{H}^{+}\) For calcium hydroxide: \(0.500 \,\text{L} \times 0.0100 \,\text{M} = 0.00500 \,\text{mol} \, \mathrm{Ca(OH)_2}\) Calcium hydroxide provides 2 moles of \(\mathrm{OH}^{-}\) ions per mole, so the total amount of \(\mathrm{OH}^{-}\) ions is: \(0.00500 \,\text{mol} \times 2 = 0.0100 \,\text{mol} \, \mathrm{OH}^{-}\) For rubidium hydroxide: \(0.200 \,\text{L} \times 0.100 \,\text{M} = 0.0200 \,\text{mol} \, \mathrm{OH}^{-}\) Now, we can find the total amounts of \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) ions. Total amount of \(\mathrm{H}^{+}\) ions: \(0.00500 \,\text{mol} + 0.0200 \,\text{mol} = 0.0250 \,\text{mol}\) Total amount of \(\mathrm{OH}^{-}\) ions: \(0.0100 \,\text{mol} + 0.0200 \,\text{mol} = 0.0300 \,\text{mol}\)

We can see that there is an excess of \(\mathrm{OH}^{-}\) ions: Excess \(\mathrm{OH}^{-}\) ions: \(0.0300 \,\text{mol} - 0.0250 \,\text{mol} = 0.00500 \,\text{mol}\)

Now let's calculate the final volume of the mixed solution, which is the sum of the volumes of the four individual reagents: Total volume: \(0.0500 \,\text{L} + 0.100 \,\text{L} + 0.500 \,\text{L} + 0.200 \,\text{L} = 0.850 \,\text{L}\)

Finally, we can calculate the concentration of excess \(\mathrm{OH}^{-}\) ions left in the solution: Concentration of excess \(\mathrm{OH}^{-}\) ions: \(\frac{0.00500 \,\text{mol}}{0.850 \,\text{L}} = 0.00588 \,\text{M}\) The acids and bases did not exactly neutralize each other. There is a 0.00588 M concentration of excess \(\mathrm{OH}^{-}\) ions left in the solution.