Barium Hydroxide Solution: Calculating Mass Fraction
Let's dive into a chemistry problem involving barium hydroxide, barium oxide, and calculating the mass fraction of the resulting solution. This is a classic problem that combines stoichiometry with solution chemistry, and it's super important to grasp these concepts for a solid foundation in chemistry. So, grab your calculators, and let’s get started, guys!
Understanding the Problem
Before we jump into the calculations, let's break down the problem. We're starting with a 30 g aqueous solution of barium hydroxide (Ba(OH)₂) that has a 20% mass fraction of the alkali. To this solution, we're adding 15 g of barium oxide (BaO). The goal here is to figure out the mass fraction of the alkali in the final solution. Basically, we need to know how much barium hydroxide is in the end solution compared to the total mass of the solution. Remember, mass fraction is a way of expressing the concentration of a substance in a mixture, defined as the mass of the substance divided by the total mass of the mixture, often expressed as a percentage.
Initial Mass of Barium Hydroxide
First, let's find out how much barium hydroxide we initially have in the 30 g solution. Since it's a 20% solution, we can calculate the mass of Ba(OH)₂ as follows:
Mass of Ba(OH)₂ = (Mass fraction) × (Total mass of solution)
Mass of Ba(OH)₂ = 0.20 × 30 g = 6 g
So, we start with 6 grams of barium hydroxide. This is our baseline, and we'll see how this amount changes after we add the barium oxide. This initial calculation is crucial because it sets the stage for understanding how the added barium oxide will affect the overall concentration. Without knowing this initial amount, we'd be starting our calculations with a significant missing piece.
Reaction of Barium Oxide with Water
Now, let's consider what happens when we add barium oxide (BaO) to water. Barium oxide reacts with water to form barium hydroxide:
BaO + H₂O → Ba(OH)₂
This is a key reaction because the added BaO will convert into Ba(OH)₂ in the solution, increasing the total amount of barium hydroxide. To figure out how much Ba(OH)₂ is formed, we need to use stoichiometry. Stoichiometry, guys, is just a fancy word for using the balanced chemical equation to calculate the amounts of reactants and products involved in a chemical reaction. It's like a recipe for chemical reactions!
Molar Masses
We need the molar masses of BaO and Ba(OH)₂ to perform our stoichiometric calculations. The molar mass is the mass of one mole of a substance, and we can calculate it by adding up the atomic masses of all the atoms in the compound. Let's look them up:
- Molar mass of BaO ≈ 137.33 g/mol (Ba) + 16.00 g/mol (O) = 153.33 g/mol
- Molar mass of Ba(OH)₂ ≈ 137.33 g/mol (Ba) + 2 × (16.00 g/mol (O) + 1.01 g/mol (H)) = 171.35 g/mol
These molar masses are the conversion factors we'll use to switch between grams and moles, which is essential for stoichiometric calculations. Think of molar mass as the bridge that connects the macroscopic world (grams) to the microscopic world (moles).
Moles of BaO Added
Next, we'll convert the mass of BaO added (15 g) to moles:
Moles of BaO = Mass of BaO / Molar mass of BaO
Moles of BaO = 15 g / 153.33 g/mol ≈ 0.0978 mol
So, we've added approximately 0.0978 moles of BaO to the solution. This value tells us how many BaO molecules we've introduced, and since each BaO molecule will react to form one Ba(OH)₂ molecule, this number is crucial for determining the additional amount of barium hydroxide formed.
Moles of Ba(OH)₂ Formed
From the balanced equation, 1 mole of BaO reacts to form 1 mole of Ba(OH)₂. Therefore, the moles of Ba(OH)₂ formed will be the same as the moles of BaO reacted:
Moles of Ba(OH)₂ formed = Moles of BaO = 0.0978 mol
This is a direct application of stoichiometry. The mole ratio between BaO and Ba(OH)₂ is 1:1, making the calculation straightforward. Knowing this, we can now find out the mass of Ba(OH)₂ formed.
Mass of Ba(OH)₂ Formed
Now, we'll convert the moles of Ba(OH)₂ formed back to grams:
Mass of Ba(OH)₂ formed = Moles of Ba(OH)₂ × Molar mass of Ba(OH)₂
Mass of Ba(OH)₂ formed = 0.0978 mol × 171.35 g/mol ≈ 16.76 g
So, about 16.76 grams of Ba(OH)₂ are formed from the reaction of BaO with water. This is a significant increase in the amount of barium hydroxide in the solution, and it will directly impact the final mass fraction.
Total Mass of Ba(OH)₂ in the Solution
To find the total mass of Ba(OH)₂ in the solution, we'll add the initial mass of Ba(OH)₂ to the mass of Ba(OH)₂ formed from the reaction:
Total mass of Ba(OH)₂ = Initial mass of Ba(OH)₂ + Mass of Ba(OH)₂ formed
Total mass of Ba(OH)₂ = 6 g + 16.76 g ≈ 22.76 g
This is the total amount of barium hydroxide present in the final solution. It's the sum of what we started with and what was produced from the reaction of barium oxide with water. This value is crucial for calculating the final mass fraction.
Total Mass of the Solution
We need to calculate the total mass of the solution after adding BaO. The total mass will be the sum of the initial solution mass and the mass of BaO added:
Total mass of solution = Initial mass of solution + Mass of BaO added
Total mass of solution = 30 g + 15 g = 45 g
This calculation assumes that the volume change upon dissolving BaO is negligible, which is a common approximation in these types of problems. The total mass of the solution is the denominator in our mass fraction calculation.
Calculating the Final Mass Fraction
Finally, we can calculate the mass fraction of Ba(OH)₂ in the final solution:
Mass fraction of Ba(OH)₂ = (Total mass of Ba(OH)₂) / (Total mass of solution)
Mass fraction of Ba(OH)₂ = 22.76 g / 45 g ≈ 0.5058
To express this as a percentage, we multiply by 100:
Mass fraction of Ba(OH)₂ (%) = 0.5058 × 100 ≈ 50.58%
Conclusion
So, the mass fraction of barium hydroxide in the resulting solution is approximately 50.58%. This means that about half of the final solution's mass is made up of barium hydroxide. Remember, guys, these types of problems might seem tricky at first, but breaking them down into smaller steps makes them much more manageable. It’s all about understanding the chemistry and applying the right formulas and concepts. Keep practicing, and you'll become a pro at these calculations in no time!