Barium Hydroxide Molar Calculation How Many Moles Needed
Hey everyone! Let's dive into a fun chemistry problem involving barium hydroxide dissolution. This is a classic example that helps us understand stoichiometry and molarity, crucial concepts in chemistry. We're going to figure out how many moles of $Ba(OH)_2$ are needed to create a solution with a specific hydroxide ion concentration. So, grab your calculators, and let's get started!
Barium Hydroxide Dissolution: A Step-by-Step Guide
The Dissolution Equation
First, let's take a close look at the chemical equation representing the dissolution of barium hydroxide in water:
This equation tells us a lot. It shows that solid barium hydroxide ($Ba(OH)_2(s)$) dissociates in water to form one barium ion ($Ba^{2+}(aq)$) and two hydroxide ions ($2OH^-(aq)$). This 1:2 stoichiometric relationship between barium hydroxide and hydroxide ions is key to solving our problem. This means that for every one mole of $Ba(OH)_2$ that dissolves, we get two moles of $OH^-$. Keep this in mind; it's super important for our calculations!
Understanding the dissolution process is fundamental. When barium hydroxide, an ionic compound, is introduced into water, the polar water molecules surround the ions and pull them apart from the crystal lattice. This process, known as solvation or hydration, leads to the ions dispersing throughout the solution. The equilibrium established during dissolution is governed by the solubility product constant ($K_{sp}$), which indicates the extent to which a compound dissolves in water. While we aren't directly using the $K_{sp}$ in this specific problem, it’s worth noting that barium hydroxide is considered a strong base, meaning it dissociates almost completely in water, leading to a high concentration of hydroxide ions.
The significance of this dissolution process extends beyond simple calculations. In practical applications, controlling the concentration of hydroxide ions is crucial in various chemical reactions and industrial processes. For instance, in titrations, a solution of barium hydroxide can be used to neutralize acids, and the precise concentration of the hydroxide ions is essential for accurate results. Similarly, in environmental chemistry, understanding the solubility and dissociation of hydroxides is important for water treatment and pollution control. Moreover, in the synthesis of certain materials, the pH of the solution, which is directly influenced by the hydroxide ion concentration, can affect the properties of the resulting product.
Molarity and Moles: The Connection
Now, let's clarify the concepts of molarity and moles. Molarity (M) is a measure of the concentration of a solution, defined as the number of moles of solute per liter of solution. Mathematically, it's expressed as:
Moles, on the other hand, represent the amount of a substance. One mole is defined as 6.022 x 10^23 entities (atoms, molecules, ions, etc.), a number known as Avogadro's constant. The relationship between moles, mass, and molar mass is given by:
In our problem, we're given the desired hydroxide ion concentration (0.050 M) and the volume of the solution (1.0 L), and we need to find the moles of barium hydroxide required. We'll use the molarity formula and the stoichiometric relationship from the dissolution equation to bridge this gap. Remember, molarity provides a direct link between concentration and the number of moles present in a given volume, making it an indispensable tool in quantitative chemistry.
The importance of understanding molarity extends to various laboratory procedures. When preparing solutions, chemists need to accurately calculate the mass of the solute required to achieve a specific molar concentration. For instance, in pharmaceutical chemistry, the precise concentration of a drug in a solution is critical for its efficacy and safety. Similarly, in analytical chemistry, molarity calculations are essential for determining the concentration of unknown substances using techniques like titrations and spectrophotometry. In research settings, controlling the molarity of reactants is often necessary to optimize reaction rates and yields. By mastering the concept of molarity, chemists can ensure the reproducibility and reliability of their experiments.
Solving the Problem: A Practical Approach
Alright, guys, let's get our hands dirty and solve the problem! We're asked to determine the moles of $Ba(OH)_2$ needed to prepare 1.0 L of a solution where the concentration of $OH^-$ is 0.050 M. This is where our understanding of stoichiometry and molarity comes into play. We'll break down the problem into manageable steps to ensure we nail the solution.
First, we need to determine the moles of $OH^-$ required. We know the desired molarity (0.050 M) and the volume of the solution (1.0 L). Using the molarity formula, we can rearrange it to solve for moles:
Plugging in the given values:
So, we need 0.050 moles of $OH^-$ in our solution. Now, let's use the stoichiometric relationship from the dissolution equation. We know that 1 mole of $Ba(OH)_2$ produces 2 moles of $OH^-$ when it dissolves. We can use this ratio to convert moles of $OH^-$ to moles of $Ba(OH)_2$:
Moles ext{ of } Ba(OH)_2 = rac{Moles ext{ of } OH^-}{2}
Substituting the moles of $OH^-$ we calculated:
Moles ext{ of } Ba(OH)_2 = rac{0.050 ext{ moles}}{2} = 0.025 ext{ moles}
Therefore, we need 0.025 moles of $Ba(OH)_2$ to make 1.0 L of solution with an $OH^-$ concentration of 0.050 M. See? It wasn't so bad after all! By carefully applying the principles of stoichiometry and molarity, we were able to solve the problem systematically.
This problem-solving approach can be applied to a wide range of stoichiometry calculations. For instance, in acid-base titrations, you might need to determine the molarity of an unknown acid or base using a standardized solution. The same principles of stoichiometry and molarity are crucial in these calculations. Additionally, in reaction yield calculations, you might need to determine the limiting reactant and the theoretical yield of a product, which also involves applying stoichiometric ratios and molar mass conversions. By practicing these types of problems, you can build a strong foundation in quantitative chemistry.
Final Answer: The Required Moles of Barium Hydroxide
In conclusion, to prepare 1.0 L of a solution with a hydroxide ion concentration of 0.050 M, you would need 0.025 moles of barium hydroxide ($Ba(OH)_2$). This calculation highlights the importance of understanding stoichiometry and molarity in chemistry. Remember, the balanced chemical equation provides the crucial stoichiometric ratios, and molarity links the concentration of a solution to the number of moles of solute present.
Key Takeaways
- The dissolution of barium hydroxide follows the equation: $Ba(OH)_2(s) \rightarrow Ba^{2+}(aq) + 2OH^-(aq)$
- Molarity (M) is defined as moles of solute per liter of solution.
- Stoichiometric ratios are essential for converting between moles of different species in a reaction.
- Careful attention to units and significant figures is crucial for accurate calculations.
Keep practicing these types of problems, and you'll become a pro at solution chemistry in no time! Chemistry is all about understanding the relationships between different substances and how they interact, and problems like this one help solidify those fundamental concepts. Keep up the great work, and happy calculating!
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