Question

A solution is made by adding \(0.300 \mathrm{~g} \mathrm{Ca}(\mathrm{OH})_{2}(s), 50.0 \mathrm{~mL}\) of \(1.40 \mathrm{M} \mathrm{HNO}_{3}\), and enough water to make a final volume of \(75.0 \mathrm{~mL}\) Assuming that all of the solid dissolves, what is the \(\mathrm{pH}\) of the final solution?

Short answer

The final pH of the solution is 0, indicating a highly acidic solution.

Step-by-step solution

Determine the moles of Ca(OH)₂ and HNO₃

First, we need to find the moles of Ca(OH)₂ that were initially present in the solution. To do this, we can use the mass of Ca(OH)₂, which was given as 0.300 g, and the molar mass of Ca(OH)₂, which is approximately 74.1 g/mol. Moles of Ca(OH)₂ = \( \frac{0.300 \, g}{74.1 \, g/mol} \) Next, we need to identify the moles of HNO₃ that were initially present in the solution. We can use the volume of HNO₃, which was given as 50.0 mL, and the molarity of HNO₃, which was given as 1.40 M. Moles of HNO₃ = 1.40 M × 0.050 L = 0.070 mol

Write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction between Ca(OH)₂ and HNO₃ is: Ca(OH)₂ + 2HNO₃ → Ca(NO₃)₂ + 2H₂O

Determine the moles of each reactant and product in the final solution

In the balanced chemical equation, we see that Ca(OH)₂ reacts with HNO₃ in a 1:2 ratio. Because we calculated the moles of each reactant in Step 1, we can now determine which reactant is the limiting reactant and how much each reactant and product will remain after the reaction. Moles of Ca(OH)₂ = \( \frac{0.300 \, g}{74.1 \, g/mol} \) ≈ 0.00405 mol Since the stoichiometric ratio between Ca(OH)₂ and HNO₃ is 1:2, we need twice the amount of HNO₃ to react with all of Ca(OH)₂. Moles of HNO₃ needed: 0.00405 mol × 2 = 0.0081 mol Since the moles of HNO₃ initially present (0.070 mol) are more than the moles needed to react with Ca(OH)₂, Ca(OH)₂ is the limiting reactant, and HNO₃ is the excess reactant. Hence, after the reaction, there will be no Ca(OH)₂ left, and we will have some remaining HNO₃: Moles of remaining HNO₃ = 0.070 mol - 0.0081 mol = 0.0619 mol

Calculate the concentration of remaining HNO₃ in the final solution

To calculate the concentration of the remaining HNO₃ in the final solution, we need to know the final volume of the solution, which was given as 75.0 mL (0.075 L). With the moles of remaining HNO₃ from Step 3 (0.0619 mol), we can now calculate the concentration: Final concentration of HNO₃ = \( \frac{0.0619 \, mol}{0.075 \, L} \) ≈ 0.825 M

Determine the pH of the final solution

Since the excess reactant HNO₃ is an acidic solution, the pH can be calculated as: pH = -log(0.825) ≈ -0.08 However, pH is defined between 0 and 14, so it is impossible to have a negative pH value. In this case, we will round up to the nearest significant digit, which is 0. pH = 0 This result indicates that the final solution is a highly acidic solution with a pH of 0.

Key concepts

Stoichiometry

Stoichiometry is akin to a recipe for chemistry, specifying the exact amounts of reactants needed to create desired products in a chemical reaction. When calculating pH in chemistry, stoichiometry comes into play by determining the proportion of acid to base, which ultimately influences the hydrogen ion concentration and pH of the solution.

In the given exercise, we encounter a reaction between calcium hydroxide, a strong base, and nitric acid, a strong acid. Stoichiometry demands a balance between the reactant and product amounts. For instance, calcium hydroxide reacts in a 1:2 molar ratio with nitric acid resulting in calcium nitrate and water. Understanding this ratio is vital to identify the limiting and excess reactants, and to predict the outcome of acid-base interactions in the solution, directing us toward the pH calculation.

Acid-Base Reactions

Acid-base reactions are fundamental in chemistry as they determine the pH levels of various substances. These reactions involve the transfer of protons (H+) from the acid to the base. In the context of our textbook problem, we have calcium hydroxide (base) reacting with nitric acid (acid) in a neutralization reaction.

The essence of this process is to create water and an ionic salt, in this case, calcium nitrate. Upon complete reaction, the stoichiometry dictates that all calcium hydroxide will be consumed, altering the concentration of H+ ions in the solution. This change is pivotal to finding the pH. If the acid is in excess, as in our case, the pH will reflect the acidic nature of the remaining solution.

Chemical Equilibrium

Chemical equilibrium pertains to the state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, leading to no net change in the concentrations of reactants and products over time. In our exercise, when calcium hydroxide and nitric acid are mixed, they swiftly react to reach such an equilibrium and form products.

However, since strong acids and bases are involved, the reaction goes to completion, meaning there's no reversible interplay. Thus, equilibrium concepts don't directly apply here. Yet, understanding equilibrium is crucial when dealing with weak acids or bases, which partially ionize and establish a dynamic equilibrium in solution impacting the pH.

Molarity Calculations

Molarity, the measure of concentration, is the number of moles of a solute per liter of solution. It's fundamental in quantifying the strength of an acid or base, and therefore, in figuring the pH. The exercise demonstrates the use of molarity in determining the remaining concentration of nitric acid after it has reacted with calcium hydroxide.

By calculating the leftover moles of nitric acid and considering the final volume of the solution, we can discern the molarity and directly determine the hydronium ion concentration, since the nitric acid is strong and fully dissociates. This concentration is pivotal in calculating the pH level, representing the acidity of the solution. Without proper molarity calculations, the estimation of pH cannot be accurate.