Question

What volume of each of the following bases will react completely with $25.00\mathrm{mL}$ of $0.200\mathrm{M}\mathrm{HCl}$ ? a. $0.100\mathrm{M}\mathrm{NaOH}$ b. $0.0500\mathrm{M}\mathrm{Ba}(\mathrm{OH}{)}_2$ c. $0.250\mathrm{M}\mathrm{KOH}$

Short answer

a. $50.00\mathrm{mL}$ of $0.100\mathrm{M}\mathrm{NaOH}$ will react completely with $25.00\mathrm{mL}$ of $0.200\mathrm{M}\mathrm{HCl}$. b. $50.00\mathrm{mL}$ of $0.0500\mathrm{M}\mathrm{Ba}(\mathrm{OH}{)}_2$ will react completely with $25.00\mathrm{mL}$ of $0.200\mathrm{M}\mathrm{HCl}$. c. $20.00\mathrm{mL}$ of $0.250\mathrm{M}\mathrm{KOH}$ will react completely with $25.00\mathrm{mL}$ of $0.200\mathrm{M}\mathrm{HCl}$.

Step-by-step solution

Calculate moles of HCl

Let's find the moles of HCl using the given volume and concentration: moles = Molarity × Volume moles of HCl = 0.200 M × 0.025 L = 0.005 mol

Step 2a: Find moles of NaOH required

The balanced reaction between HCl and NaOH is: HCl + NaOH → NaCl + H₂O From the stoichiometry, the ratio between HCl and NaOH is 1:1. Thus, the moles of NaOH required will be equal to the moles of HCl: moles of NaOH = 0.005 mol

Step 3a: Calculate volume of NaOH

Now, find the volume of the 0.100 M NaOH solution required for the reaction: Volume = moles / Molarity Volume of NaOH = 0.005 mol / 0.100 M = 0.050 L (or 50.00 mL)

Step 2b: Find moles of Ba(OH)₂ required

The balanced reaction between HCl and Ba(OH)₂ is: 2 HCl + Ba(OH)₂ → BaCl₂ + 2 H₂O From the stoichiometry, the ratio between HCl and Ba(OH)₂ is 2:1. Thus, the moles of Ba(OH)₂ required will be half of the moles of HCl: moles of Ba(OH)₂ = 0.005 mol / 2 = 0.0025 mol

Step 3b: Calculate volume of Ba(OH)₂

Now, find the volume of the 0.0500 M Ba(OH)₂ solution required for the reaction: Volume = moles / Molarity Volume of Ba(OH)₂ = 0.0025 mol / 0.0500 M = 0.050 L (or 50.00 mL)

Step 2c: Find moles of KOH required

The balanced reaction between HCl and KOH is: HCl + KOH → KCl + H₂O From the stoichiometry, the ratio between HCl and KOH is 1:1. Thus, the moles of KOH required will be equal to the moles of HCl: moles of KOH = 0.005 mol

Step 3c: Calculate volume of KOH

Now, find the volume of the 0.250 M KOH solution required for the reaction: Volume = moles / Molarity Volume of KOH = 0.005 mol / 0.250 M = 0.020 L (or 20.00 mL) In summary: a. 50.00 mL of 0.100 M NaOH will react completely with 25.00 mL of 0.200 M HCl. b. 50.00 mL of 0.0500 M Ba(OH)₂ will react completely with 25.00 mL of 0.200 M HCl. c. 20.00 mL of 0.250 M KOH will react completely with 25.00 mL of 0.200 M HCl.

Key concepts

Stoichiometry

Stoichiometry is the mathematical relationship between the quantities of reactants and products in a chemical reaction. It is anchored in the law of conservation of mass, which states that mass is neither created nor destroyed in chemical reactions. Thus, stoichiometry allows us to predict the amount of product that will form during a reaction, given the quantities of reactants.

When analyzing reactions like the neutralization between hydrochloric acid (HCl) and various bases, stoichiometry comes into play by examining the balanced chemical equations. For instance, a 1:1 molar ratio of HCl to NaOH means that one mole of HCl will react with one mole of NaOH to produce one mole of water and one mole of sodium chloride as per the reaction equation. Similarly, in the case of HCl reacting with barium hydroxide, Ba(OH)₂, the stoichiometry of the reaction shows a 2:1 molar ratio, indicating that it takes two moles of HCl to react with one mole of Ba(OH)₂.

Molarity

Molarity is a measure of the concentration of a solution. It’s defined as the number of moles of a solute divided by the volume of the solution in liters. The unit for molarity is moles per liter (M). This concept is fundamental in chemistry, especially when preparing solutions for reactions and calculating the results of titrations.

Understanding molarity is essential for acid-base titrations because it allows chemists to determine how much acid or base solutions are required to neutralize each other. For example, knowing the molarity of an HCl solution and the required volume lets us calculate the moles of HCl present, which then helps determine the volume of a base of known molarity needed to neutralize the acid.

Neutralization Reaction

A neutralization reaction is a type of chemical reaction in which an acid and a base react with each other to form water and a salt. This is a subset of double replacement reactions and is essential in understanding acid-base chemistry. In general, the hydrogen ions (H⁺) from the acid combine with the hydroxide ions (OH⁻) from the base to form water (H₂O), and the remaining ions form the salt.

For the exercise in question, HCl is a strong acid, and the bases (NaOH, Ba(OH)₂, KOH) are strong bases. The reactions are straightforward: for every mole of HCl, we need an equivalent mole of NaOH or KOH, but only half a mole of Ba(OH)₂. This mole relationship is due to the stoichiometry of each balanced chemical equation, which governs the proportion in which the reactants will combine.

Mole Calculation

In chemistry, mole calculations are used to convert between mass, particles, and volume in relation to a chemical substance. The mole is the base unit in the International System of Units (SI) for amount of substance and represents approximately 6.022 x 10²³ particles - be they atoms, molecules, ions, or others - and this number is known as Avogadro's number.

In the context of our problem, the exercise begins with mole calculations for the acid HCl. We use the molarity and volume to find the moles of HCl. These moles are then key to finding out how much of each base is needed to completely neutralize the existing amount of acid. The process demonstrates the direct application of mole calculations in determining reactant volumes, which is integral to solving titration problems.