Question

What volume of \(0.0200 \mathrm{M}\) calcium hydroxide is required to neutralize \(35.00 \mathrm{mL}\) of \(0.0500 \mathrm{M}\) nitric acid?

Short answer

The volume of \(0.0200 \ \mathrm{M}\) calcium hydroxide solution required to neutralize \(35.00 \ \mathrm{mL}\) of \(0.0500 \ \mathrm{M}\) nitric acid is \(43.75 \ \mathrm{mL}\).

Step-by-step solution

Write down the given information.

We are given the following information: - Concentration of calcium hydroxide (\(C_{Ca(OH)_2}\)): \(0.0200 \ \mathrm{M}\) - Volume of nitric acid (\(V_{HNO_3}\)): \(35.00 \ \mathrm{mL}\) - Concentration of nitric acid (\(C_{HNO_3}\)): \(0.0500 \ \mathrm{M}\)

Find the moles of nitric acid.

We can find the moles of nitric acid in the given volume using the formula: Moles of nitric acid = \(C_{HNO_3} \times V_{HNO_3}\) Moles of nitric acid = \((0.0500 \ \mathrm{M}) \times (35.00 \ \mathrm{mL})\) Moles of nitric acid = \(1.75 \ \mathrm{mmol}\)

Use stoichiometry to find the moles of calcium hydroxide.

Now, we can use the stoichiometry of the balanced chemical equation to find the moles of calcium hydroxide required for the neutralization: \(Ca(OH)_2 : HNO_3 = 1 : 2\) This means that for every 2 moles of nitric acid, we require 1 mole of calcium hydroxide. Moles of calcium hydroxide required = \(1.75 \ \mathrm{mmol} (HNO_3) \times \frac{1}{2}\) Moles of calcium hydroxide required = \(0.875 \ \mathrm{mmol}\)

Calculate the volume of calcium hydroxide solution.

Finally, we can use the moles of calcium hydroxide, along with its given concentration, to find the volume of calcium hydroxide solution required for neutralization: Volume of calcium hydroxide solution (\(V_{Ca(OH)_2}\)) = \(\frac{\text{moles of }Ca(OH)_2}{C_{Ca(OH)_2}}\) \(V_{Ca(OH)_2}\) = \(\frac{0.875 \ \mathrm{mmol}}{0.0200 \ \mathrm{M}}\) \(V_{Ca(OH)_2}\) = \(43.75 \ \mathrm{mL}\) Therefore, the volume of \(0.0200 \ \mathrm{M}\) calcium hydroxide solution required to neutralize \(35.00 \ \mathrm{mL}\) of \(0.0500 \ \mathrm{M}\) nitric acid is \(43.75 \ \mathrm{mL}\).

Key concepts

Neutralization Reaction

A neutralization reaction is a type of chemical reaction where an acid and a base react to form water and a salt. This process is crucial in various fields, including environmental science, medicine, and industrial processes. In our exercise, we encounter a classic case of neutralization where calcium hydroxide, a base, reacts with nitric acid, an acid.

The reaction between calcium hydroxide and nitric acid can be represented by this balanced chemical equation:
\[ Ca(OH)_2 + 2HNO_3 \rightarrow Ca(NO_3)_2 + 2H_2O \]
This equation tells us that for every one molecule of calcium hydroxide that reacts, two molecules of nitric acid are required. The products of this reaction are water, which is the neutral substance, and calcium nitrate, the resultant salt. Understanding this ratio is vital to solving our stoichiometry problem as it guides us on how much of each reactant is needed to reach neutralization.

Molarity

Molarity is a measure of the concentration of a solution, representing the number of moles of a solute per liter of solution. Molarity is commonly expressed in units of moles per liter (M). For instance, a molarity of \(0.0200 \mathrm{M}\) suggests that there are 0.0200 moles of the solute in each liter of solution.

To calculate the molarity of a solution, we use the formula: \[ Molarity (M) = \frac{moles\text{ of solute}}{volume\text{ of solution in liters}} \]
In our exercise, we used molarity to find out how many moles of nitric acid were present in a given volume of solution. This calculation was pivotal to determine the amount of calcium hydroxide needed to neutralize the nitric acid, showcasing the real-world utility of understanding molarity in chemical reactions.

Chemical Equations

Chemical equations succinctly represent chemical reactions using the chemical symbols and formulas of the reactants and products involved. These equations are balanced to ensure the law of conservation of mass is adhered to, meaning that the same number of each type of atom appears on both sides of the equation.

In the step-by-step solution we discussed, the balanced chemical equation enabled us to understand the stoichiometry of the reaction — the ratio of reactants to products. To balance a chemical equation, we adjust coefficients, which are numbers placed in front of chemical formulas to indicate the number of units of each substance. For example, in our neutralization reaction: \[ Ca(OH)_2 + 2HNO_3 \rightarrow Ca(NO_3)_2 + 2H_2O \]
The coefficient 2 in front of \(HNO_3\) indicates that two moles of nitric acid react with one mole of calcium hydroxide. Being able to correctly interpret and balance chemical equations is essential for solving stoichiometry problems in chemistry.