Question

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.

Short answer

The acids and bases did not exactly neutralize each other. The excess ions in the solution are \(\mathrm{OH}^-\) ions with a concentration of 0.00588 M.

Step-by-step solution

Calculate moles of \(\mathrm{H}^{+}\) ions

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

Calculate moles of \(\mathrm{OH}^{-}\) ions

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

Compare moles of \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) ions

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

Calculate the concentration of excess ions

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.

Key concepts

Mole Calculation

The concept of mole calculation is fundamental in chemistry, especially when dealing with reactions like acid-base neutralizations. A mole is a unit that represents a specific number of particles, typically atoms or molecules. In this exercise, we calculate the moles of ions present in the acid and base solutions using a straightforward formula:

• Moles = Molarity × Volume

For example, when provided with molarity (the concentration of a solution in moles per liter) and volume (often in liters), you multiply these values to find the number of moles.
In the context of this problem, both hydrochloric acid and nitric acid are strong acids, implying they completely dissociate in water, releasing hydrogen ions (\( \mathrm{H}^{+} \)).Similarly, calcium hydroxide and rubidium hydroxide are strong bases that fully dissociate, releasing hydroxide ions (\( \mathrm{OH}^{-} \)) in the solution. Accurate mole calculation is crucial because these values dictate whether the solutions neutralize each other perfectly or if there’s an imbalance after mixing.

Solution Concentration

Solution concentration is a measure of how much solute is dissolved in a given solvent to form a solution. It plays a key role in predicting the behavior of reactions, such as how acidic or basic a solution will be.
After calculating the total moles of hydrogen and hydroxide ions separately, we observe if they balance out.
This exercise involves finding the concentration of any excess ions by adjusting our calculations based on the total solution volume after mixing all reagents.

• Total Volume = Sum of individual solution volumes
• Concentration = Moles of excess ions / Total volume

This calculation helps determine whether there are leftover hydrogen (\( \mathrm{H}^{+} \)) or hydroxide (\( \mathrm{OH}^{-} \)) ions in the solution, indicating whether the solution remained neutral (pH 7) or not. For instance, the calculated concentration of the excess \( \mathrm{OH}^{-} \) ions in this scenario is 0.00588 M, demonstrating a basic solution.

Strong Acids and Bases

Understanding strong acids and bases is essential for predicting their behavior in aqueous solutions. Strong acids completely dissociate in water, meaning all acid molecules ionize to produce hydrogen ions (\( \mathrm{H}^{+} \)). Some common strong acids include hydrochloric acid (\( \mathrm{HCl} \)) and nitric acid (\( \mathrm{HNO_3} \)).
Similarly, strong bases fully dissociate to release hydroxide ions (\( \mathrm{OH}^{-} \)). For example, calcium hydroxide (\( \mathrm{Ca(OH)_2} \)) produces two moles of \( \mathrm{OH}^{-} \) ions per mole due to its two hydroxide groups. Rubidium hydroxide (\( \mathrm{RbOH} \)) is another such base.

• Complete dissociation means that strong acids and bases update the solution's pH significantly.

In acid-base reactions, understanding whether an acid or base is strong is crucial, as it influences the outcome—like observing an excess of \( \mathrm{OH}^{-} \) ions in our problem, resulting in a basic solution.