Molarity Calculation A Step By Step Guide

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Welcome, chemistry enthusiasts! Today, we embark on a fascinating journey into the world of acid-base titrations, where we'll unravel the mysteries behind molarity calculations. Our focus will be on a specific scenario involving hydrochloric acid (HCl) and sodium carbonate ($X_2CO_3$), but the principles we'll explore are universally applicable in the realm of chemistry.

The Scenario

Let's set the stage. We have two solutions: A and B. Solution A is hydrochloric acid (HCl) with a concentration of 7.30 grams per cubic decimeter ($gdm^-3}$). Solution B is a solution of an unknown carbonate, which we'll represent as $X_2CO_3$, containing 10.6 grams in the same volume ($gdm^{-3}$). Now, the crux of our problem 23.43 cubic centimeters ($cm^3$) of solution A perfectly neutralizes 25.0 cubic centimeters ($cm^3$) of solution B. Our mission, should we choose to accept it, is to calculate the concentration of solution A in moles per cubic decimeter ($moldm^{-3$).

Understanding Molarity and Its Significance

Before we dive into the calculations, let's take a moment to appreciate the concept of molarity. Molarity, represented by the symbol M, is a fundamental unit of concentration in chemistry. It expresses the number of moles of a solute dissolved in one liter (or one cubic decimeter) of solution. In simpler terms, it tells us how much of a particular substance is present in a given volume of liquid.

Molarity is a crucial concept because it allows us to quantify the amount of a substance involved in a chemical reaction. This is especially important in titrations, where we carefully measure the volumes of reactants to determine the concentration of an unknown solution. By knowing the molarity of a solution, we can predict how it will react with other substances and perform accurate calculations in stoichiometry.

Step-by-Step Calculation of Molarity

Now, let's roll up our sleeves and tackle the calculation of the molarity of solution A. We'll break down the process into clear, manageable steps to ensure a thorough understanding.

Step 1: Convert grams per cubic decimeter ($gdm^{-3}$) to moles per cubic decimeter ($moldm^{-3}$)

Molarity, as we know, is expressed in moles per cubic decimeter. However, our initial concentration of solution A is given in grams per cubic decimeter. Therefore, our first task is to convert this unit to the desired one.

To do this, we'll need the molar mass of hydrochloric acid (HCl). The molar mass is the mass of one mole of a substance and is calculated by summing the atomic masses of all the atoms in the molecule. For HCl, the molar mass is approximately 1.01 g/mol (for hydrogen) + 35.45 g/mol (for chlorine) = 36.46 g/mol.

Now, we can use the following formula to convert grams per cubic decimeter to moles per cubic decimeter:

Molarity (M) = Concentration (g/dm³) / Molar mass (g/mol)

Plugging in the values for solution A:

Molarity (A) = 7.30 g/dm³ / 36.46 g/mol = 0.200 M

Therefore, the concentration of solution A (HCl) is 0.200 moles per cubic decimeter ($moldm^{-3}$).

Unveiling the Significance of Titration

Our journey doesn't end with calculating the molarity of solution A. The fact that solution A neutralizes solution B provides us with valuable information that we can use to delve deeper into the world of acid-base chemistry.

Titration: A Powerful Analytical Technique

Titration is a cornerstone technique in analytical chemistry. It's a process where a solution of known concentration (the titrant) is used to determine the concentration of an unknown solution (the analyte). The reaction between the titrant and analyte is carefully monitored, and the point at which the reaction is complete (the equivalence point) is identified. By knowing the volumes of the titrant and analyte used, along with the concentration of the titrant, we can calculate the concentration of the analyte.

In our scenario, solution A (HCl) acts as the titrant, and solution B ($X_2CO_3$) is the analyte. The neutralization reaction between the acid (HCl) and the carbonate ($X_2CO_3$) is the key to our analysis.

The Neutralization Reaction

The reaction between an acid and a carbonate involves the transfer of protons (H⁺ ions) from the acid to the carbonate. In the case of HCl and $X_2CO_3$, the reaction proceeds as follows:

2HCl(aq)+X2CO3(aq)→2XCl(aq)+H2O(l)+CO2(g)2HCl(aq) + X_2CO_3(aq) → 2XCl(aq) + H_2O(l) + CO_2(g)

This balanced chemical equation reveals the stoichiometry of the reaction: two moles of HCl react with one mole of $X_2CO_3$. This mole ratio is crucial for our subsequent calculations.

Stoichiometry and the Mole Ratio

Stoichiometry is the language of chemical reactions. It allows us to quantitatively relate the amounts of reactants and products involved in a reaction. The mole ratio, derived from the balanced chemical equation, is a key concept in stoichiometry. It tells us the relative number of moles of each substance involved in the reaction.

In our case, the mole ratio between HCl and $X_2CO_3$ is 2:1. This means that for every 2 moles of HCl that react, 1 mole of $X_2CO_3$ is consumed. We'll use this ratio to connect the amount of HCl used in the titration to the amount of $X_2CO_3$ present in solution B.

Calculating Moles of HCl

We know the molarity of solution A (HCl) and the volume used in the titration. We can use these values to calculate the number of moles of HCl that reacted with $X_2CO_3$.

The formula for calculating moles is:

Moles = Molarity × Volume

However, we need to be mindful of the units. Molarity is in moles per cubic decimeter ($moldm^{-3}$), and the volume is given in cubic centimeters ($cm^3$). We need to convert the volume to cubic decimeters by dividing by 1000:

Volume of A in dm³ = 23.43 cm³ / 1000 cm³/dm³ = 0.02343 dm³

Now, we can calculate the moles of HCl:

Moles of HCl = 0.200 mol/dm³ × 0.02343 dm³ = 0.004686 moles

Determining Moles of $X_2CO_3$

Now, we can leverage the mole ratio from the balanced chemical equation to determine the number of moles of $X_2CO_3$ that reacted with the HCl.

From the equation, we know that 2 moles of HCl react with 1 mole of $X_2CO_3$. Therefore, the moles of $X_2CO_3$ are half the moles of HCl:

Moles of $X_2CO_3$ = 0.004686 moles HCl / 2 = 0.002343 moles

This tells us that 0.002343 moles of $X_2CO_3$ were present in the 25.0 cm³ of solution B that reacted with the HCl.

Calculating the Molarity of Solution B

Now that we know the moles of $X_2CO_3$ in 25.0 cm³ of solution B, we can calculate the molarity of solution B. Again, we need to be careful with units.

First, convert the volume of solution B to cubic decimeters:

Volume of B in dm³ = 25.0 cm³ / 1000 cm³/dm³ = 0.0250 dm³

Now, we can calculate the molarity of solution B using the formula:

Molarity = Moles / Volume

Molarity of B = 0.002343 moles / 0.0250 dm³ = 0.09372 M

So, the molarity of solution B ($X_2CO_3$) is 0.09372 moles per cubic decimeter ($moldm^{-3}$).

Identifying the Unknown Carbonate $X_2CO_3$

We've come a long way! We've calculated the molarity of both solutions A and B. Now, let's use the information we have to identify the unknown carbonate, $X_2CO_3$.

We know that 10.6 grams of $X_2CO_3$ are present in 1 dm³ of solution B, and we've calculated the molarity of solution B to be 0.09372 M. We can use this information to determine the molar mass of $X_2CO_3$.

The formula relating molar mass, mass, and moles is:

Molar mass = Mass / Moles

Since molarity is moles per cubic decimeter, we can use the mass of $X_2CO_3$ in 1 dm³ (10.6 g) and the molarity of solution B (0.09372 mol/dm³) to calculate the molar mass:

Molar mass of $X_2CO_3$ = 10.6 g / 0.09372 moles = 113.1 g/mol

Now, we need to figure out which element X corresponds to this molar mass. Let's break down the molar mass of $X_2CO_3$:

Molar mass of $X_2CO_3$ = 2 × (Atomic mass of X) + Atomic mass of Carbon + 3 × Atomic mass of Oxygen

We know the atomic masses of carbon (12.01 g/mol) and oxygen (16.00 g/mol). Plugging in the values:

113.1 g/mol = 2 × (Atomic mass of X) + 12.01 g/mol + 3 × 16.00 g/mol

113.1 g/mol = 2 × (Atomic mass of X) + 12.01 g/mol + 48.00 g/mol

113.1 g/mol = 2 × (Atomic mass of X) + 60.01 g/mol

Now, solve for the atomic mass of X:

2 × (Atomic mass of X) = 113.1 g/mol - 60.01 g/mol

2 × (Atomic mass of X) = 53.09 g/mol

Atomic mass of X = 53.09 g/mol / 2 = 26.55 g/mol

Looking at the periodic table, the element with an atomic mass closest to 26.55 g/mol is sodium (Na), with an atomic mass of 22.99 g/mol. There might be a slight discrepancy due to rounding errors in the given data or experimental uncertainties.

Therefore, we can conclude that $X_2CO_3$ is likely sodium carbonate ($Na_2CO_3$).

Conclusion: Mastering Acid-Base Titration

Guys, we've successfully navigated the intricacies of acid-base titration! We started with a seemingly complex problem and, step by step, unraveled the mysteries of molarity, stoichiometry, and neutralization reactions. We not only calculated the molarity of a hydrochloric acid solution but also identified an unknown carbonate through careful analysis and deduction.

This journey highlights the power of chemistry as a problem-solving tool. By understanding the fundamental principles and applying them systematically, we can tackle even the most challenging questions. So, keep exploring, keep experimenting, and keep pushing the boundaries of your chemical knowledge!

Remember, the world of chemistry is vast and fascinating. There's always something new to learn and discover. Embrace the challenge, and who knows what chemical wonders you'll uncover next!