Problem 40 - Chemistry: Principles and Reactions

Question

What is the volume of \(0.885 M\) hydrochloric acid required to react with (a) \(25.00 \mathrm{~mL}\) of \(0.288 \mathrm{M}\) aqueous ammonia? (b) \(10.00 \mathrm{~g}\) of sodium hydroxide? (c) \(25.0 \mathrm{~mL}\) of a solution \(\left(d=0.928 \mathrm{~g} / \mathrm{cm}^{3}\right)\) containing \(10.0 \%\) by mass of methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right) ?\)

Step-by-Step Solution

Answer

Answer: The volumes of 0.885 M hydrochloric acid required to react with the given solutions are (a) 8.14 mL, (b) 282.5 mL, and (c) 84.4 mL.

Step 1: (a) Given information

We are given the following information: - The volume of the aqueous ammonia solution: 25.00 mL - The concentration of the aqueous ammonia: 0.288 M - The concentration of the hydrochloric acid: 0.885 M First, we need to convert the given volume and concentration into moles of ammonia.

Step 2: (a) Convert volume and concentration to moles

Moles of ammonia can be calculated using the molarity formula: moles = Molarity * Volume moles of NH3 = 0.288 mol/L * 25.00 mL * (1 L/1000 mL) moles of NH3 = 0.0072 moles Now, let's write the balanced chemical equation for the reaction between ammonia and hydrochloric acid: NH3(aq) + HCl(aq) -> NH4Cl(aq) This tells us that one mole of aqueous ammonia reacts with one mole of hydrochloric acid to produce one mole of ammonium chloride.

Step 3: (a) Determine the volume of HCl required

Since the stoichiometry is 1:1, we need 0.0072 moles of HCl to react with the ammonia. Now we can use the molarity of HCl to find the required volume: Volume = moles / Molarity Volume of HCl = 0.0072 moles / 0.885 M Volume of HCl = 0.00814 L Convert the volume to milliliters: Volume of HCl = 0.00814 L * (1000 mL/1 L) = 8.14 mL The volume of 0.885 M hydrochloric acid required to react with 25.00 mL of 0.288 M aqueous ammonia is 8.14 mL.

Step 4: (b) Given information

We are given the following information: - The mass of sodium hydroxide: 10.00 g - The concentration of the hydrochloric acid: 0.885 M - Sodium hydroxide molar mass: 40.00 g/mol First, we need to convert the given mass into moles of sodium hydroxide.

Step 5: (b) Convert mass to moles

Moles of sodium hydroxide can be calculated using the mass and molar mass: moles of NaOH = mass of NaOH / molar mass of NaOH = 10.00 g / 40.00 g/mol moles of NaOH = 0.25 moles Now, let's write the balanced chemical equation for the reaction between sodium hydroxide and hydrochloric acid: NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l) This tells us that one mole of sodium hydroxide reacts with one mole of hydrochloric acid.

Step 6: (b) Determine the volume of HCl required

Since the stoichiometry is 1:1, we need 0.25 moles of HCl to react with the sodium hydroxide. Now we can use the molarity of HCl to find the required volume: Volume of HCl = moles of HCl / Molarity of HCl = 0.25 moles / 0.885 M Volume of HCl = 0.2825 L Convert the volume to milliliters: Volume of HCl = 0.2825 L * (1000 mL/1 L) = 282.5 mL The volume of 0.885 M hydrochloric acid required to react with 10.00 g of sodium hydroxide is 282.5 mL.

Step 7: (c) Given information

We are given the following information: - The volume of the methylamine solution: 25.0 mL - The density of the solution: 0.928 g/mL - The mass percentage of methylamine: 10.0% - The concentration of the hydrochloric acid: 0.885 M - Methylamine molar mass: 31.06 g/mol First, we need to convert the given volume, density, and mass percentage into moles of methylamine.

Step 8: (c) Convert volume and density to moles

Mass of the solution can be calculated using the volume and density: mass of the solution = volume * density = 25.0 mL * 0.928 g/mL mass of the solution = 23.2 g Since the mass percentage of methylamine is 10.0%, we can calculate the mass of methylamine in the solution: mass of CH3NH2 = 0.1 * 23.2 g = 2.32 g Now, convert the mass of methylamine to moles using its molar mass: moles of CH3NH2 = mass of CH3NH2 / molar mass of CH3NH2 = 2.32 g / 31.06 g/mol moles of CH3NH2 = 0.0747 moles Now, let's write the balanced chemical equation for the reaction between methylamine and hydrochloric acid: CH3NH2(aq) + HCl(aq) -> CH3NH3Cl(aq) This tells us that one mole of methylamine reacts with one mole of hydrochloric acid.

Step 9: (c) Determine the volume of HCl required

Since the stoichiometry is 1:1, we need 0.0747 moles of HCl to react with the methylamine. Now we can use the molarity of HCl to find the required volume: Volume of HCl = moles of HCl / Molarity of HCl = 0.0747 moles / 0.885 M Volume of HCl = 0.0844 L Convert the volume to milliliters: Volume of HCl = 0.0844 L * (1000 mL/1 L) = 84.4 mL The volume of 0.885 M hydrochloric acid required to react with 25.0 mL of a solution containing 10.0% by mass of methylamine is 84.4 mL.

Key Concepts

Molarity and Volume Relationship

Molarity, often denoted as 'M', is a unit of concentration that measures the number of moles of a substance per liter of solution. The relationship between molarity and volume is central to various calculations in chemistry, particularly when dealing with solutions and their reactions. Understanding this relationship is crucial when you need to determine how much of a reactant is required to completely react with another substance in a chemical reaction.

For instance, in the given exercise, we are asked to calculate the volume of hydrochloric acid (HCl) needed to react with other chemicals. The formula connecting molarity and volume is:
\[\text{Moles} = \text{Molarity} \times \text{Volume}\]
This formula helps in calculating the number of moles when the molarity and volume of the solution are known, as seen in the first step of the solution provided for part (a). It's essential to carefully convert volumes into liters when using the molarity formula, as molarity is always given in moles per liter.

To improve comprehension of this concept, visualize molarity as the number of 'solute particles' in each liter of solution. By knowing the molarity and the volume, we find out the total number of these particles, hence the amount of the substance present. In analytical and preparative chemistry, this is a fundamental concept that helps in preparing desired solutions and determining reactant quantities in reactions.

Chemical Reaction Stoichiometry

Stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction. It's based on the balanced chemical equation and the principles of the conservation of mass and the conservation of atoms. By employing stoichiometry, we can determine how much of each reactant is needed or how much of a product will be produced in a chemical reaction.

Each reactant and product in a balanced equation is represented by coefficients that indicate their mole ratios. For example, in the reaction provided for the exercise:
\[\text{NH}_3(aq) + \text{HCl}(aq) \rightarrow \text{NH}_4\text{Cl}(aq)\]
The coefficients imply a 1:1 mole ratio between ammonia (NH3) and hydrochloric acid (HCl). Understanding these ratios is indispensable in predicting the outcome of chemical reactions.

Importance of a Balanced Equation

Every stoichiometry calculation relies on a balanced chemical equation because it provides the mole ratio needed to find out how much of each substance is involved. For instance, in step 6 of the exercise (b), we apply the 1:1 mole ratio to determine the volume of HCl needed to react with a given amount of sodium hydroxide (NaOH).

From Stoichiometry to Practical Applications

Precise stoichiometry calculations are not only vital for theoretical practices but also for practical applications such as pharmaceuticals formulation, chemical manufacturing, and in any scenario where the exact amounts of substances are critical. Stoichiometry is indeed the backbone of chemical reactions and a gateway to mastering chemistry.

Mass to Mole Conversion

Converting mass to moles is a fundamental skill in stoichiometry that allows chemists to transition from the macroscopic world of grams and kilograms to the molecular world of moles and molecules. This conversion is pivotal when dealing with chemical reactions, as seen in the exercise's problem (b) and (c).

The mass to mole conversion relies on the formula:
\[\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}\]
Here, the molar mass is the mass of one mole of a given substance and can be found on the periodic table as the atomic or molecular weight in grams per mole.

In the context of our exercise, to find the volume of HCl required to react with sodium hydroxide (NaOH), we first needed to convert the mass of NaOH to moles using its molar mass (40.00 g/mol). Similarly, for the solution containing methylamine (CH3NH2), the mass of CH3NH2 was converted to moles using the molar mass of 31.06 g/mol (step 8).

Visualizing the Mole Concept

Think of a mole as a 'chemist’s dozen'. Just as a dozen represents 12 items, a mole represents approximately 6.022 x 10^23 particles of a substance. This number is known as Avogadro's number. The mole links the microscopic particles with macroscopic amounts we can measure and work with in the lab.

Real-World Application

Outside of academic exercises, mass to mole conversions are critically important in fields such as medicine production, where precise amounts of substances must be transformed and reacted to create pharmacological compounds. This skill is essential in ensuring that chemical reactions have the right quantities of reactants for the desired products.