Question

What volume of each of the following bases will react completely with 25.00 mL of 0.200 M HCl? a. 0.100\(M \mathrm{NaOH}\) b. 0.0500\(M \mathrm{Sr}(\mathrm{OH})_{2}\) c. 0.250 \(\mathrm{M} \mathrm{KOH}\)

Short answer

The volume of each base required to react completely with 25.00 mL of 0.200 M HCl is: a. 50.00 mL of 0.100 M NaOH b. 50.00 mL of 0.0500 M Sr(OH)₂ c. 20.00 mL of 0.250 M KOH

Step-by-step solution

Calculate moles of HCl

We know the volume (25 mL) and molarity (0.200M) of the given HCl solution, so we can calculate the moles of HCl using the following formula: Moles of HCl = Volume (L) × Molarity (M) First, convert 25 mL to liters: \(25\,\mathrm{mL}\times \frac{1\,\mathrm{L}}{1000\,\mathrm{mL}}=0.025\,\mathrm{L}\) Now, calculate the moles of HCl: Moles of HCl = 0.025 L × 0.200 M = 0.005 mol

Using stoichiometry calculate moles of bases

For each base, we will determine the number of moles required to completely react with the given moles of HCl. (a) NaOH: HCl + NaOH → NaCl + H₂O (1 mol of HCl reacts with 1 mol of NaOH) Moles of NaOH = 0.005 mol (b) Sr(OH)₂: 2HCl + Sr(OH)₂ → SrCl₂ + 2H₂O (2 mol of HCl react with 1 mol of Sr(OH)₂) Moles of Sr(OH)₂ = 0.005 mol HCl × (1 mol Sr(OH)₂ / 2 mol HCl) = 0.0025 mol (c) KOH: HCl + KOH → KCl + H₂O (1 mol of HCl reacts with 1 mol of KOH) Moles of KOH = 0.005 mol

Calculate volume of bases using molarity and moles

We will use the moles of each base and their given molarity to calculate the volume required for each base. (a) Volume of 0.100 M NaOH: Volume (L) = Moles of NaOH / Molarity of NaOH = 0.005 mol / 0.100 M = 0.050 L Convert to mL: 0.050 L × 1000 = 50.00 mL (b) Volume of 0.0500 M Sr(OH)₂: Volume (L) = Moles of Sr(OH)₂ / Molarity of Sr(OH)₂ = 0.0025 mol / 0.0500 M = 0.050 L Convert to mL: 0.050 L × 1000 = 50.00 mL (c) Volume of 0.250 M KOH: Volume (L) = Moles of KOH / Molarity of KOH = 0.005 mol / 0.250 M = 0.020 L Convert to mL: 0.020 L × 1000 = 20.00 mL The volume of each base that will react completely with 25 mL of 0.200 M HCl is: a. 50.00 mL of 0.100 M NaOH b. 50.00 mL of 0.0500 M Sr(OH)₂ c. 20.00 mL of 0.250 M KOH

Key concepts

Molarity Calculations

Molarity is a way to express the concentration of a solution. It's defined as the number of moles of solute divided by the volume of the solution in liters. This concept is crucial when you need to determine how much of a solution is needed to react with another.
For instance, in an acid-base titration problem, knowing the molarity lets you calculate how many moles of the reactant are present in the solution. Once you have the number of moles, you can easily find out how much of another substance is required for a complete reaction.
Here’s how you calculate it: using the formula \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]In our exercise, we calculated the moles of HCl first, because we knew both its molarity (0.200 M) and volume (25 mL or 0.025 L). Multiplying these values together gave us the moles of HCl: 0.005 mol.

Stoichiometry

Stoichiometry is like the recipe for chemical reactions, where you measure out the exact amounts of reactants and products involved. It's essential to understand stoichiometry to predict how much of each substance is needed or produced.
This concept uses mole ratios, derived from the balanced chemical equation of the reaction, to relate quantities of reactants to products. In our problem, we used these relationships to decide how much of each base is needed to react with a given amount of HCl.
For instance, in the reaction with NaOH, each mole of HCl reacts with 1 mole of NaOH. For Sr(OH)₂, the ratio is different: 2 moles of HCl react with 1 mole of Sr(OH)₂. This means you need to adjust the amount of Sr(OH)₂ based on this ratio.
By ensuring you have the correct stoichiometric proportions, you can determine the precise quantity of each base required for a complete reaction.

Chemical Reactions

Chemical reactions involve substances called reactants being transformed into different substances known as products. In the case of acid-base titration, the focus is on the neutralization reaction, where an acid reacts with a base to form water and salt.

• **Acid-base neutralization:** This reaction type is crucial in our exercise, involving HCl (an acid) reacting with various bases (NaOH, Sr(OH)₂, KOH).
• **Balanced chemical equations:** These are essential for stoichiometry. They show the correct ratio of reactants to products.
• **Example:** For NaOH and HCl, the reaction is \( \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \). For every mole of HCl, you need one mole of NaOH.

Each reaction adheres to the principle of conservation of mass, meaning the mass of reactants equals the mass of products. This principle allows you to predict the outcomes of chemical reactions and ensure the necessary quantities of each substance are present to react completely, just as we did when determining needed base volumes in the original exercise.