Question

Challenge How much calcium hydroxide (Ca(OH) \(_{2},\) in grams, is needed to produce 1.5 \(\mathrm{L}\) of a 0.25 \(\mathrm{M}\) solution?

Short answer

To produce 1.5 L of a 0.25 M calcium hydroxide (Ca(OH)₂) solution, we need 27.79 grams of calcium hydroxide.

Step-by-step solution

Calculate moles of calcium hydroxide

Using the formula, we can find the moles of calcium hydroxide required to produce a 1.5 L, 0.25 M solution: moles = molarity × volume moles = 0.25 mol/L × 1.5 L moles = 0.375 mol We need 0.375 moles of Ca(OH)₂.

Calculate molar mass of calcium hydroxide

Next, we need to find the molar mass of calcium hydroxide (Ca(OH)₂) by adding the molar masses of each element in the compound: Molar mass of Ca = 40.08 g/mol Molar mass of O = 16.00 g/mol Molar mass of H = 1.01 g/mol Molar mass of Ca(OH)₂ = 40.08 g/mol (Ca) + 2 × (16.00 g/mol (O) + 1.01 g/mol (H)) Molar mass of Ca(OH)₂ = 40.08 + 2 × (16.00 + 1.01) Molar mass of Ca(OH)₂ = 40.08 + 2 × 17.01 Molar mass of Ca(OH)₂ = 40.08 + 34.02 Molar mass of Ca(OH)₂ = 74.10 g/mol The molar mass of Ca(OH)₂ is 74.10 g/mol.

Calculate the mass of calcium hydroxide in grams

Now, we will multiply the moles of calcium hydroxide by its molar mass to obtain the mass in grams: mass = moles × molar mass mass = 0.375 mol × 74.10 g/mol mass = 27.79 g So, we need 27.79 grams of calcium hydroxide (Ca(OH)₂) to produce 1.5 L of a 0.25 M solution.

Key concepts

Calcium Hydroxide

Calcium hydroxide, commonly known as slaked lime, is a chemical compound with the formula Ca(OH)₂. It consists of calcium, oxygen, and hydrogen. Each molecule contains one calcium atom and two hydroxide groups.

Calcium hydroxide is significant in various applications, including the preparation of lime mortar and limewater, as well as in the treatment of acidic soils.

• It appears as a white powder or crystalline solid.
• It has a moderately basic pH, making it useful for neutralizing acids.

Understanding the properties of calcium hydroxide is crucial for tasks like calculating how much is needed for a specific molarity in a solution.

Molar Mass

The molar mass of a compound is the mass of one mole of its molecules measured in grams per mole (g/mol). To determine it, you can sum up the atomic masses of all the atoms within the compound.

For calcium hydroxide, Ca(OH)₂, calculate as follows:

• Calcium (Ca) has a molar mass of 40.08 g/mol.
• Oxygen (O) has a molar mass of 16.00 g/mol.
• Hydrogen (H) has a molar mass of 1.01 g/mol.

The molar mass of Ca(OH)₂ thus becomes:
40.08 + 2 imes (16.00 + 1.01) = 74.10 ext{ g/mol}

This calculated molar mass is a fundamental component used in mass and solution calculations.

Mass Calculation

Mass calculation involves determining the mass of a substance for a solution with known molarity and volume. Here’s a simple way to do this:

Using the formula: ext{mass} = ext{moles} imes ext{molar mass}

• First, calculate the moles of the compound using the equation: ext{moles} = ext{molarity} imes ext{volume}.
• Then multiply the moles by the molar mass of the compound to find the mass.

In our example for calcium hydroxide, we calculated 0.375 moles (0.25 mol/L imes 1.5 L) and multiplied by its molar mass (74.10 g/mol) to find the mass required:
0.375 ext{ mol} imes 74.10 ext{ g/mol} = 27.79 ext{ g}

This provides the exact amount of substance needed for the desired solution.

Solution Preparation

Solution preparation involves dissolving a solute, like calcium hydroxide, in a solvent, typically water, to form a homogeneous solution. The goal is often to achieve a specific concentration, measured in molarity (M), which is moles of solute per liter of solution.

• Calculate the moles needed based on the desired molarity and volume.
• Determine the mass of solute needed by multiplying the moles by the molar mass.
• Dissolve the calculated mass in the appropriate volume of solvent.

For instance, to prepare 1.5 liters of a 0.25 M solution of calcium hydroxide, you would dissolve 27.79 grams of the compound in 1.5 liters of water.

This systematic approach ensures that each solution has the correct chemical composition for experiments or applications.