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 \(0.100 \mathrm{M}\) 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 \(200.0 \mathrm{~mL}\) 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 concentration of excess OH⁻ ions in the solution is approximately \(5.88 \times 10^{-3} \mathrm{M}\).

Step-by-step solution

Calculate moles of each reagent

To calculate the moles of each reagent, use the equation: moles = volume × molarity. For hydrochloric acid (HCl): moles of HCl = (50.0 mL) × (0.100 M) = 5.00 mmol For nitric acid (HNO3): moles of HNO3 = (100.0 mL) × (0.200 M) = 20.0 mmol For calcium hydroxide (Ca(OH)2): moles of Ca(OH)2 = (500.0 mL) × (0.0100 M) = 5.00 mmol Since there are 2 moles of OH- ions in each mole of Ca(OH)2, we multiply the moles of Ca(OH)2 with 2 to find moles of OH-: moles of OH- from Ca(OH)2 = (5.00 mmol) × 2 = 10.0 mmol For rubidium hydroxide (RbOH): moles of RbOH = (200.0 mL) × (0.100 M) = 20.0 mmol

Determine if the moles of acids and bases are equal or if any are in excess

Total moles of acid (H+): 5.00 mmol (HCl) + 20.0 mmol (HNO3) = 25.0 mmol Total moles of base (OH-): 10.0 mmol (Ca(OH)2) + 20.0 mmol (RbOH) = 30.0 mmol Since there are more moles of base than acid, the bases are in excess. We will now calculate how many moles of excess OH- are left.

Calculate excess moles

Excess moles of OH- = Total moles of OH- - Total moles of H+ Excess moles of OH- = 30.0 mmol - 25.0 mmol = 5.00 mmol

Calculate the concentration of excess OH- ions

To find the concentration of excess OH- ions, we need to divide the excess moles of OH- by the total volume of the solution. Total volume of the solution = 50.0 mL (HCl) + 100.0 mL (HNO3) + 500.0 mL (Ca(OH)2) + 200.0 mL (RbOH) = 850.0 mL Now, we can find the concentration of excess OH-. Concentration of excess OH- = Excess moles of OH- / Total volume of solution Concentration of excess OH- = \( \frac{5.00 \mathrm{~mmol}}{850.0 \mathrm{~mL}} \) = \( 5.88 \times 10^{-3} \mathrm{M} \) The concentration of excess OH- ions in the solution is approximately \(5.88 \times 10^{-3} \mathrm{M}\).

Key concepts

Molarity Calculation

Molarity, often represented by the symbol M, is a measure of concentration that tells us how many moles of a solute are present in a liter of solution. It's a fundamental concept in chemistry and a vital part of understanding how to conduct and interpret experiments, such as acid-base titrations.

The formula for molarity is:\[\begin{equation}Molarity (M) = \frac{moles of solute}{volume of solution in liters}\end{equation}\]To calculate the moles of solute, one must know the molecular weight of the solute and have a precise measurement of its mass. The volume of the solution should be in liters for the molarity to be accurate. Importantly, molarity can be influenced by temperature, as the volume of a solution can expand or contract with changes in temperature.

When you're confronted with a problem that involves different volumes and molarities—for example, mixing reagents of different concentrations—it's crucial to remember to convert all volumes to liters and calculate the moles of each reactant individually before combining them to get the overall reaction molarity. This process is essential for accurately determining whether solutions will neutralize each other during a reaction, and is the first step in the stoichiometry of the reaction.

Stoichiometry

Stoichiometry is a branch of chemistry that involves calculating the quantities of reactants and products involved in chemical reactions. It's based on the conservation of mass, where the total mass of reactants equals the total mass of products. In the context of acid-base reactions, stoichiometry allows you to predict how much acid is needed to neutralize a given amount of base, and vice versa.

Understanding stoichiometry requires familiarity with chemical equations and the ability to balance them. In a balanced chemical equation, the number of atoms for each element is the same on both sides of the reaction. This balance is achieved by adding coefficients in front of the chemical formulas. These coefficients represent the mole ratios of the reactants and products and are essential in stoichiometric calculations.

Examples in Stoichiometry

• To find the amount of product formed, you multiply the moles of the limiting reagent by the stoichiometric ratio from the balanced chemical equation.
• To determine the excess reagent, you subtract the amount reacted from the initial amount.

With these principles, it's possible to determine whether the acids and bases in a given mixture will exactly neutralize each other, and if not, to calculate the concentration of excess H+ or OH- ions remaining, as seen in the exercise.

Acid-Base Titration

Acid-base titration is a laboratory method used to determine the concentration of an acid or base in a solution by performing a neutralization reaction. This analytical technique involves adding a reactant of known concentration, called the titrant, to a solution of the substance being analyzed, known as the analyte, until the reaction reaches its endpoint. The point at which neutralization is achieved is often indicated by a color change due to an added indicator or by measuring pH changes.

The stoichiometry of the reaction is crucial for titrations, as the mole ratio between the titrant and the analyte determines the amount of titrant needed for neutralization. By recording the volume of titrant required to reach the endpoint, and knowing its molarity, you can calculate the molarity of the analyte using the neutralization equation:\[\begin{equation}M_{Acid} \times V_{Acid} = M_{Base} \times V_{Base}\end{equation}\]where:

• \(M_{Acid}\) is the molarity of the acid,
• \(V_{Acid}\) is the volume of the acid,
• \(M_{Base}\) is the molarity of the base, and
• \(V_{Base}\) is the volume of the base.

Titrations can only provide accurate results if the exact concentration of the titrant is known and the reaction is a clear-cut neutralization. In the exercise, the student combines known volumes and molarities of acids and bases, which could be viewed as performing a sort of manual titration without an indicator. The steps required to determine if neutralization occurred mirror those of an acid-base titration procedure.