Ca(OH)₂ Concentration Calculation In Acid-Base Titration

Unlocking the secrets of chemical analysis often involves understanding the nuances of titrations. In this comprehensive guide, we will delve into the intricacies of a specific chemistry problem, Q.11, which revolves around determining the concentration of Calcium Hydroxide, Ca(OH)₂, in a sample using acid-base titration. This exploration is not just about solving a numerical problem; it’s about grasping the underlying principles of stoichiometry, molarity, and the equivalence point in chemical reactions. By dissecting the question, we aim to provide a clear, step-by-step explanation that will not only answer the question but also enhance your understanding of titration concepts. Whether you're a student grappling with chemistry problems or a professional seeking a refresher, this guide will serve as a valuable resource. So, let's embark on this journey of chemical discovery and unravel the complexities of Q.11 together.

The Problem: A Deep Dive

The question at hand presents a scenario where a 50 ml sample containing Ca(OH)₂ is titrated against 0.02 N H₂SO₄ (Sulfuric Acid). The key piece of information is that 10 ml of the H₂SO₄ solution is required to reach the equivalence point. Our ultimate goal is to determine the concentration of Ca(OH)₂ in the sample, expressed in milligrams per liter (mg/l) as an integer. To solve this problem effectively, we must first break it down into manageable parts. Understanding the chemical reaction between Ca(OH)₂ and H₂SO₄ is crucial. This reaction is a classic acid-base neutralization, where the hydroxide ions from Ca(OH)₂ react with the hydrogen ions from H₂SO₄ to form water and a salt. The stoichiometry of this reaction – the ratio in which the reactants combine – is the foundation of our calculations. Furthermore, we need to understand the concept of normality (N), which is a measure of concentration related to the number of reactive species in a solution. By converting normality to molarity if necessary, we can then use the volumes of the solutions and the stoichiometry of the reaction to find the number of moles of Ca(OH)₂ in the sample. Finally, we can convert this amount to milligrams per liter, giving us the desired concentration. This methodical approach, combining chemical principles with careful calculations, is the key to successfully tackling titration problems.

Unpacking the Chemistry: Reaction and Stoichiometry

At the heart of this problem lies the chemical reaction between Ca(OH)₂ (Calcium Hydroxide) and H₂SO₄ (Sulfuric Acid). This reaction is a prime example of an acid-base neutralization, a fundamental concept in chemistry. To fully grasp the solution, we must first dissect the balanced chemical equation for this reaction. Calcium Hydroxide, Ca(OH)₂, is a strong base, meaning it readily dissociates in water to release hydroxide ions (OH⁻). Sulfuric Acid, H₂SO₄, is a strong acid, donating hydrogen ions (H⁺) in solution. When these two compounds mix, the H⁺ ions from H₂SO₄ react with the OH⁻ ions from Ca(OH)₂ to form water (H₂O). The remaining ions, Ca²⁺ from Ca(OH)₂ and SO₄²⁻ from H₂SO₄, combine to form Calcium Sulfate (CaSO₄), a salt. The balanced chemical equation for this reaction is:

Ca(OH)₂ (aq) + H₂SO₄ (aq) → CaSO₄ (s) + 2 H₂O (l)

This equation is the key to unlocking the stoichiometry of the reaction. Stoichiometry, in simple terms, is the quantitative relationship between reactants and products in a chemical reaction. The balanced equation tells us the exact molar ratios in which the substances react. In this case, the equation reveals that one mole of Ca(OH)₂ reacts with one mole of H₂SO₄. This 1:1 stoichiometric ratio is crucial for our calculations. It allows us to directly relate the amount of H₂SO₄ used in the titration to the amount of Ca(OH)₂ present in the sample. Understanding this relationship is the cornerstone of solving the problem, as it allows us to use the known volume and concentration of H₂SO₄ to determine the unknown amount of Ca(OH)₂ in the solution.

Normality vs. Molarity: The Concentration Connection

In the given problem, the concentration of H₂SO₄ is provided in terms of normality (N), which is a measure of the reactive capacity of a solution. However, for many stoichiometric calculations, molarity (M), which represents the number of moles of solute per liter of solution, is often more convenient. Therefore, understanding the relationship between normality and molarity is crucial for solving this problem. Normality is defined as the number of gram equivalent weights of solute per liter of solution. The equivalent weight depends on the reaction taking place. In acid-base reactions, the equivalent weight is the molar mass divided by the number of replaceable hydrogen ions (for acids) or hydroxide ions (for bases). For H₂SO₄, which has two replaceable hydrogen ions, the relationship between normality and molarity is:

Normality (N) = Molarity (M) × Number of replaceable H⁺ ions

Since H₂SO₄ has two replaceable hydrogen ions, its normality is twice its molarity. Therefore, a 0.02 N H₂SO₄ solution is equivalent to a 0.01 M solution. This conversion is essential because molarity directly relates to the number of moles, which is what we need for stoichiometric calculations. Once we have the molarity of the H₂SO₄ solution, we can use the volume used in the titration to calculate the number of moles of H₂SO₄ that reacted. This, in turn, will allow us to determine the number of moles of Ca(OH)₂ in the sample, thanks to the 1:1 stoichiometric ratio we established earlier. This conversion from normality to molarity is a critical step in bridging the information provided in the problem statement to the calculations needed to find the solution.

Titration Calculations: Finding the Equivalence Point

With a firm grasp of the stoichiometry and the concentration units, we can now proceed to the core of the problem: the titration calculations. The key to solving this problem lies in understanding the equivalence point. The equivalence point in a titration is the point at which the acid and base have completely neutralized each other. In other words, it's the point where the number of moles of acid is stoichiometrically equivalent to the number of moles of base, as dictated by the balanced chemical equation. In our case, the equivalence point is reached when the moles of H₂SO₄ added are equal to the moles of Ca(OH)₂ in the sample. We know that 10 ml (0.010 L) of 0.02 N H₂SO₄ is required to reach this equivalence point. As we determined earlier, 0.02 N H₂SO₄ is equivalent to 0.01 M H₂SO₄. Therefore, we can calculate the number of moles of H₂SO₄ used in the titration using the formula:

Moles = Molarity × Volume

Plugging in the values, we get:

Moles of H₂SO₄ = 0.01 M × 0.010 L = 0.0001 moles

Since the stoichiometric ratio between Ca(OH)₂ and H₂SO₄ is 1:1, the number of moles of Ca(OH)₂ in the 50 ml sample is also 0.0001 moles. This is a crucial finding, as it directly links the experimental data (volume and concentration of titrant) to the amount of the substance we're trying to quantify (Ca(OH)₂). With this value in hand, we're just one step away from determining the concentration of Ca(OH)₂ in the desired units of milligrams per liter.

Concentration Conversion: From Moles to mg/L

Having determined the number of moles of Ca(OH)₂ in the sample, our final task is to convert this quantity into the desired units of milligrams per liter (mg/L). This conversion involves two steps: first, converting moles to mass (in milligrams), and second, adjusting for the volume of the sample (50 ml) to express the concentration per liter. To convert moles to mass, we need the molar mass of Ca(OH)₂. The molar mass is the mass of one mole of a substance, which can be calculated by summing the atomic masses of all the atoms in the chemical formula. For Ca(OH)₂, the molar mass is:

Ca: 40.08 g/mol O: 16.00 g/mol (× 2 = 32.00 g/mol) H: 1.01 g/mol (× 2 = 2.02 g/mol)

Total molar mass = 40.08 + 32.00 + 2.02 = 74.10 g/mol

Now we can convert the moles of Ca(OH)₂ to grams:

Mass (g) = Moles × Molar mass

Mass of Ca(OH)₂ = 0.0001 moles × 74.10 g/mol = 0.00741 g

To express this in milligrams, we multiply by 1000 (since 1 g = 1000 mg):

Mass of Ca(OH)₂ = 0.00741 g × 1000 mg/g = 7.41 mg

This is the mass of Ca(OH)₂ in the 50 ml sample. To find the concentration in mg/L, we need to scale this up to a liter. Since 1 liter is 1000 ml, we can set up a proportion:

Concentration (mg/L) = (Mass in mg / Sample volume in ml) × 1000 ml/L

Concentration of Ca(OH)₂ = (7.41 mg / 50 ml) × 1000 ml/L = 148.2 mg/L

The problem asks for the answer as an integer, so we round 148.2 mg/L to the nearest whole number, which is 148 mg/L. Therefore, the concentration of Ca(OH)₂ in the sample is 148 mg/L.

Conclusion: Mastering Titration Calculations

In this comprehensive guide, we have meticulously dissected the problem of determining the concentration of Ca(OH)₂ in a sample using acid-base titration. We started by understanding the chemical reaction between Ca(OH)₂ and H₂SO₄, emphasizing the importance of the balanced chemical equation and the stoichiometric relationships it reveals. We then navigated the concept of normality and its relationship to molarity, a crucial step in converting the given concentration of H₂SO₄ into a usable form for calculations. The heart of the problem lay in the titration calculations, where we identified the equivalence point and used it to link the amount of H₂SO₄ used to the amount of Ca(OH)₂ present in the sample. Finally, we performed the necessary conversions to express the concentration in the desired units of milligrams per liter, arriving at the answer of 148 mg/L. This journey through the problem highlights the interconnectedness of various chemical concepts. Stoichiometry, concentration units, and the principles of acid-base chemistry all play vital roles in solving titration problems. By mastering these concepts and practicing problem-solving techniques, you can confidently tackle a wide range of analytical chemistry challenges. Titration, as we have seen, is not just a laboratory technique; it's a window into the quantitative world of chemistry, where careful measurements and calculations unlock the composition of matter.

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Original Question: A 50 ml sample containing Ca(OH)₂ requires 10 ml of 0.02 N H₂SO₄ to reach the equivalence point in an acid-base titration. The concentration (in mg/l, in integer) of Ca(OH)₂ in the sample is?

Rewritten Question: In an acid-base titration, a 50 ml sample containing Calcium Hydroxide (Ca(OH)₂) requires 10 ml of 0.02 N Sulfuric Acid (H₂SO₄) to reach the equivalence point. What is the concentration of Ca(OH)₂ in the sample, expressed in milligrams per liter (mg/L) as an integer?