Question

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.

Short answer

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.

Step-by-step solution

Calculate initial amounts of \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) ions contributed by each reagent

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: \(0.0100 \,\text{mol} + 0.0200 \,\text{mol} = 0.0300 \,\text{mol}\)

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

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

Determine final volume of mixed solution

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: \(0.0500 \,\text{L} + 0.100 \,\text{L} + 0.500 \,\text{L} + 0.200 \,\text{L} = 0.850 \,\text{L}\)

Calculate the concentration of excess \(\mathrm{OH}^{-}\) ions

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.

Key concepts

Molarity

Molarity is an essential concept in chemistry, specifically in aqueous solutions. It is a way to express the concentration of a solute in a solution. Molarity is defined as the number of moles of solute per liter of solution. The formula for molarity (M) is given by:\[M = \frac{\text{moles of solute}}{\text{liters of solution}}\]Understanding molarity helps to calculate how much of a substance needs to be dissolved to achieve a desired concentration in a specific volume of solution. This is crucial when preparing solutions for experiments or reactions.
To calculate molarity, know the amount of solute (in moles) and the volume of the solution (in liters). Simply divide the moles by the liters. For example, if you have 0.1 moles of HCl in 1 liter of water, the molarity is 0.1 M.

• Quick way to express concentration
• Useful for calculations in chemical reactions
• Helps determine quantities needed for reactions

Hydrochloric Acid

Hydrochloric acid ( HCl ) is a strong acid commonly used in laboratory settings. When dissolved in water, it disassociates completely to yield hydrogen ions ( H^+ ) and chloride ions ( Cl^- ).
As a monoprotic acid, each molecule of HCl provides one proton ( H^+ ), making it straightforward to calculate its contribution to the solution's acidity.

• Common in laboratories and industries
• Disassociates fully into ions in water
• Useful for pH reduction and cleaning purposes

In the exercise, 50.0 mL of 0.100 M hydrochloric acid is used. This translates to an initial amount of 0.00500 moles of H^+ ions in the mixture, which is required for analyzing acid-base reactions.

Calcium Hydroxide

Calcium hydroxide, often referred to as slaked lime, is represented by the formula Ca(OH)_2 . It is an important base used in various industrial and laboratory applications. In an aqueous solution, each formula unit of calcium hydroxide disassociates into one calcium ion ( Ca^{2+} ) and two hydroxide ions ( OH^- ).
This characteristic makes it a diprotic base, meaning it provides two hydroxide ions per formula unit, affecting calculations of neutralization reactions significantly.

• Used in construction and to neutralize acids
• Provides two OH^- ions per molecule in solution
• Strong base suitable for various chemical processes

In the provided exercise, the 500.0 mL of 0.0100 M calcium hydroxide contributes two moles of OH^- ions per mole of Ca(OH)_2 , resulting in 0.0100 moles of OH^- ions in the solution.

Rubidium Hydroxide

Rubidium hydroxide ( RbOH ) is a less common, yet strong base, known for its complete disassociation into rubidium ions ( Rb^+ ) and hydroxide ions ( OH^- ) in aqueous solutions. As a monobasic hydroxide, each rubidium hydroxide formula unit provides one hydroxide ion in solution.

• Strong base used in specialized applications
• Disassociates fully in water

In the context of the exercise, the solution contains 0.0200 moles of OH^- from 200.0 mL of 0.100 M rubidium hydroxide. This contributes to the overall amount of hydroxide ions in the reaction, essential for determining the final concentration of excess ions after neutralization.