Question

A sample of solid \(\mathrm{Ca}(\mathrm{OH})_{2}\) is stirred in water at \(30^{\circ} \mathrm{C}\) until the solution contains as much dissolved \(\mathrm{Ca}(\mathrm{OH})_{2}\) as it can hold. A \(100-\mathrm{mL}\) sample of this solution is withdrawn and titrated with \(5.00 \times 10^{-2} \mathrm{M} \mathrm{HBr}\). It requires \(48.8 \mathrm{~mL}\) of the acid solution for neutralization. What is the molarity of the \(\mathrm{Ca}(\mathrm{OH})_{2}\) solution? What is the solubility of \(\mathrm{Ca}(\mathrm{OH})_{2}\) in water, at \(30^{\circ} \mathrm{C}\), in grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) per \(100 \mathrm{~mL}\) of solution?

Short answer

The molarity of the Ca(OH)_2 solution is \(1.22 \times 10^{-2}\) M, and its solubility in water at \(30^{\circ} \mathrm{C}\) is 0.0905 g/100 mL.

Step-by-step solution

Moles of HBr

To find the moles of HBr used to neutralize the Ca(OH)_2 solution, we can use the formula: moles of HBr = Molarity × Volume The molarity of HBr is given as 5.00 × 10^-2 M, and the volume used is 48.8 mL. However, we need to convert the volume from mL to L: Volume of HBr in L = 48.8 mL × (1 L / 1000 mL) = 0.0488 L Now, we can calculate the moles of HBr: moles of HBr = (5.00 × 10^-2 M) × (0.0488 L) = 2.44 × 10^-3 mol #Step 2: Determine the moles of Ca(OH)_2 in the 100 mL sample#

Moles of Ca(OH)_2

Since the ratio of HBr to Ca(OH)_2 in the neutralization reaction is 2:1 (i.e., for every 1 mole of Ca(OH)_2, 2 moles of HBr are required), we can determine the moles of Ca(OH)_2 by dividing the moles of HBr by 2: moles of Ca(OH)_2 = (2.44 × 10^-3 mol) / 2 = 1.22 × 10^-3 mol #Step 3: Calculate the molarity of the Ca(OH)_2 solution#

Molarity of Ca(OH)_2

In order to determine the molarity of the Ca(OH)_2 solution, we can use the formula: Molarity = moles of solute / volume of solution (L) The volume of the solution is given as 100 mL, and we convert it to L: Volume of Ca(OH)_2 in L = 100 mL × (1 L / 1000 mL) = 0.1 L Now, we can calculate the molarity of the Ca(OH)_2 solution: Molarity = (1.22 × 10^-3 mol) / 0.1 L = 1.22 × 10^-2 M #Step 4: Determine the solubility of Ca(OH)_2 in water at 30°C#

Solubility in grams per 100 mL

To find the solubility of Ca(OH)_2 in water at 30°C, we first need to convert the moles of Ca(OH)_2 to grams. The molar mass of Ca(OH)_2 is: Molar mass of Ca(OH)_2 = 40.08 g/mol (Ca) + (2 × (1.01 g/mol (H)) + 15.999 g/mol (O)) = 74.10 g/mol Now, we can find the mass of Ca(OH)_2 in the 100 mL sample: mass of Ca(OH)_2 = moles of Ca(OH)_2 × molar mass = (1.22 × 10^-3 mol) × (74.10 g/mol) = 0.0905 g Finally, to find the solubility of Ca(OH)_2 in grams per 100 mL of solution: Solubility = (0.0905 g Ca(OH)_2) / (100 mL solution) = 0.0905 g/100 mL The molarity of the Ca(OH)_2 solution is 1.22 × 10^-2 M, and its solubility in water at 30°C is 0.0905 g/100 mL.

Key concepts

Titration

Titration is a very useful laboratory method that helps us determine the concentration of a solute in a solution. You perform a titration by slowly adding a solution of known concentration, called the "titrant," to a solution of unknown concentration, until the reaction between the two solutions is complete. The titrant used here is the hydrobromic acid (HBr) solution of known molarity. We can determine when the reaction is complete by observing a change in color due to an indicator or by using a pH meter.
Essentially, titration allows us to find out how concentrated a particular solution is, based on how much of another substance reacts with it.

• Known concentration: Called a "standard solution."
• Endpoint: The point where the reaction is complete.
• Indicator: A chemical that changes color at the endpoint.

Neutralization Reaction

When you hear "neutralization," think of an acid meeting a base. That’s exactly what's happening in these types of reactions. In this exercise, calcium hydroxide (Ca(OH⎰)⎰) acts as the base, while hydrobromic acid (HBr) acts as the acid. When combined in the right proportions, they form water and calcium bromide, which is a salt.
This process can be symbolized as:\[ ext{Ca(OH)_2 + 2HBr} ightarrow ext{CaBr}_2 + 2 ext{H}_2 ext{O}\]This reaction demonstrates the classic acid-base reaction, where two moles of hydrobromic acid neutralize one mole of calcium hydroxide. Understanding the stoichiometry is important since it tells us how much of each reactant is needed.

• Two reactants: An acid and a base.
• Products: Water and a salt.
• Stoichiometry: Ratio of reacting species.

Molarity Calculation

Molarity is a measure of concentration, expressed in moles of solute per liter of solution. Calculating molarity requires two main components: moles of solute and volume of solution. To find it, we use the equation:\[ ext{Molarity (M)} = \frac{ ext{moles of solute}}{ ext{volume of solution in L}}\]In this task, the molarity of Ca(OH⎰)⎰) is determined after finding out how much HBr reacted with it during the titration. Once you know the number of moles of Ca(OH⎰)⎰), and the volume of the solution it was in, molarity becomes straightforward to calculate.

• Moles: Amount of substance in moles.
• Volume: Space the solution occupies, in liters.
• Concentration: Described using molarity.

Calcium Hydroxide

Calcium hydroxide, often called slaked lime, is a solid powder that dissolves sparingly in water. This exercise is about understanding its solubility properties. At a certain temperature, you can only dissolve a specific amount before the solution becomes saturated. In our context, we want to know just how much calcium hydroxide can dissolve in water at 30°C. This is described by its solubility: Solubility is the mass of solute per volume of solvent, typically expressed in grams per 100 mL of solution. Knowing its molar mass (74.10 g/mol) lets us convert from moles to grams for a clearer picture of solubility in practical terms.

• Commonly used in construction and various chemical processes.
• Sparingly soluble at room temperature.
• Applications: pH adjustment, water treatment.