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, \(100.0 \mathrm{mL}\) of \(0.200 \mathrm{M}\) of nitric acid, \(500.0 \mathrm{mL}\) of \(0.0100 \mathrm{M}\) 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 solutions do not exactly neutralize each other. The concentration of excess OH⁻ ions in the solution is 0.00588 M. \( \)

Step-by-step solution

Calculate moles of each reagent

First, let's calculate the moles of each reagent present in the solution (n = concentraction × volume). For HCl: Concentration = 0.100 M, Volume = \(50.0 mL = 0.0500 L\), Moles of HCl = \(0.100 M × 0.0500 L = 0.00500 mol\) For HNO₃: Concentration = 0.200 M, Volume = \(100.0 mL = 0.100 L\), Moles of HNO₃ = \(0.200 M × 0.100 L = 0.0200 mol\) For Ca(OH)₂: Concentration = 0.0100 M, Volume = \(500.0 mL = 0.500 L\), Moles of Ca(OH)₂ = \(0.0100 M × 0.500 L = 0.00500 mol\) For RbOH: Concentration = 0.100 M, Volume = \(200.0 mL = 0.200 L\), Moles of RbOH = \(0.100 M × 0.200 L = 0.0200 mol\)

Calculate moles of H⁺ and OH⁻ ions

H⁺ ions are present in HCl and HNO₃, whereas OH⁻ ions are present in Ca(OH)₂ and RbOH. Since Ca(OH)₂ produces 2 moles of OH⁻ ions for each mole of Ca(OH)₂, while RbOH produces 1 mole of OH⁻ ion for each mole of RbOH. Thus, we need to calculate the moles of H⁺ and OH⁻ ions in each respective reagent. Moles of H⁺ ions in HCl = 0.00500 mol Moles of H⁺ ions in HNO₃ = 0.0200 mol Total moles of H⁺ ions = 0.00500 mol + 0.0200 mol = 0.0250 mol Moles of OH⁻ ions in Ca(OH)₂ = 2 × 0.00500 mol = 0.0100 mol Moles of OH⁻ ions in RbOH = 0.0200 mol Total moles of OH⁻ ions = 0.0100 mol + 0.0200 mol = 0.0300 mol

Determine if the solutions neutralize each other

Now let's compare the moles of H⁺ and OH⁻ ions: Total moles of H⁺ ions = 0.0250 mol Total moles of OH⁻ ions = 0.0300 mol Since the moles of H⁺ ions and OH⁻ ions are not equal, the solutions do not exactly neutralize each other.

Calculate excess ions' concentration

Since there are more moles of OH⁻ ions than H⁺ ions, the solution is slightly basic. We need to find the concentration of excess OH⁻ ions. We can do this by subtracting the moles of H⁺ ions from the moles of OH⁻ ions and then dividing by the total volume of the final mixed solution. Excess moles of OH⁻ ions = 0.0300 mol - 0.0250 mol = 0.00500 mol Total volume of the solution = 0.0500 L + 0.100 L + 0.500 L + 0.200 L = 0.850 L Concentration of excess OH⁻ ions = \(\frac{Excess \ moles \ of \ OH^- \ ions}{Total \ volume} = \frac{0.00500 \ mol}{0.850 \ L} = 0.00588 \, M\) The concentration of excess OH⁻ ions in the solution is 0.00588 M.

Key concepts

Chemical Stoichiometry

Chemical stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It enables chemists and students to predict the amount of substances consumed and produced in a reaction.

In the textbook exercise, we see the concept of stoichiometry in action when the student tries to neutralize acids with bases. The exercise requires the calculation of moles of reactants to understand if equal amounts of acid and base were mixed.

By calculating the moles of hydrochloric acid (HCl), nitric acid (HNO₃), calcium hydroxide (Ca(OH)₂), and rubidium hydroxide (RbOH), you set the stage for determining whether the reactants will completely neutralize each other. Chemical stoichiometry in this context involves using the coefficients in balanced chemical equations (in this case, the 1:1 ratio for H⁺ and OH⁻ ions in a neutralization reaction) to calculate the expected outcome of mixing different solutions.

Concentration Calculation

Concentration calculation involves determining how much of a substance is present within a certain volume of solution. The concentration can be expressed in various ways, but in our exercise, molarity (M) is used, which is moles of solute per liter of solution.

In the problem's context, students need to calculate the concentration of each solution before they can determine the amount of each reactant in moles. For example, the concentration of hydrochloric acid is given as 0.100 M, and its volume is 50.0 mL, which is converted into liters (0.0500 L) before multiplication to find the moles of HCl. This step is crucial for stoichiometry, as it allows the comparison of the amounts of acids and bases on a mole basis.

Furthermore, after finding out whether or not the solutions neutralize each other, the student is asked to calculate the concentration of any excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions. This part of the problem integrates both stoichiometry, as you compare molar amounts of reactants, and concentration calculation to find the final concentration of ions in the mixed solution.

Molarity

Molarity is a measure of the concentration of a solution, defined as the number of moles of a solute dissolved per liter of solution (mol/L). It is an important concept for students to grasp because it's extensively used in chemistry to describe solution concentrations.

In the original exercise, molarity is used to express the concentration of both the acids and the bases. To solve the problem, you first convert the volume of each solution from milliliters to liters to use it in molarity calculations. It's important to ensure the correct conversion because molarity is always expressed in liters.

The calculation of the number of moles of each compound based on molarity is a stepping stone to solving the broader question. Once the moles of \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) ions are calculated, the concept of molarity comes back into play to determine the concentration of the excess ions in the final solution. Understanding molarity helps students make sense of the concentration of the reactants and the products in a chemical reaction, allowing them to predict and understand the outcome of chemical processes.