Problem 81 - Chemistry The Central Science

Question

(a) What volume of \(0.115 \mathrm{M} \mathrm{HClO}_{4}\) solution is needed to neutralize \(50.00 \mathrm{~mL}\) of \(0.0875 \mathrm{M} \mathrm{NaOH}\) ? (b) What volume of \(0.128 \mathrm{M} \mathrm{HCl}\) is needed to neutralize \(2.87 \mathrm{~g}\) of \(\mathrm{Mg}(\mathrm{OH})_{2} ?\) (c) If \(25.8 \mathrm{~mL}\) of \(\mathrm{AgNO}_{3}\) is needed to precipitate all the \(\mathrm{Cl}^{-}\) ions in a \(785-\mathrm{mg}\) sample of \(\mathrm{KCl}\) (forming \(\mathrm{AgCl}\) ), what is the molarity of the \(\mathrm{AgNO}_{3}\) solution? (d) If \(45.3 \mathrm{~mL}\) of \(0.108 \mathrm{M} \mathrm{HCl}\) solution is needed to neutralize a solution of KOH, how many grams of KOH must be present in the solution?

Step-by-Step Solution

Answer

The answers to the given problems are as follows: (a) To neutralize 50.00 mL of 0.0875 M NaOH, approximately 38.04 mL of 0.115 M HClO₄ solution is needed. (b) To neutralize 2.87 g of Mg(OH)₂, 553 mL of 0.128 M HCl solution is needed. (c) The molarity of the AgNO₃ solution needed to precipitate all the Cl⁻ ions in a 785-mg sample of KCl is 0.4004 M. (d) 0.269 g of KOH must be present in the solution if 45.3 mL of 0.108 M HCl is needed to neutralize it.

Step 1: Identify the neutralization reaction

The neutralization reaction between HClO₄ and NaOH is: \( HClO_4 + NaOH \rightarrow NaClO_4 + H_2O \)

Step 2: Calculate the moles of NaOH

Moles of NaOH = Molarity × Volume in liters Moles of NaOH = 0.0875 × (50.00/1000) = 0.004375 moles

Step 3: Calculate the moles of HClO₄

From the balanced equation, the stoichiometry is 1:1, so moles of HClO₄ = moles of NaOH = 0.004375 moles

Step 4: Calculate the volume of HClO₄

Volume of HClO₄ (L) = Moles / Molarity Volume of HClO₄ = 0.004375 / 0.115 = 0.038044 L = 38.04 mL (Approximately) Problem (b)

Step 1: Calculate the moles of Mg(OH)₂

Moles of Mg(OH)₂ = Mass / Molar mass Moles of Mg(OH)₂ = 2.87 / (24.305 + 34.02) = 0.0354 moles

Step 2: Identify the neutralization reaction

The neutralization reaction between HCl and Mg(OH)₂ is: \( Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O \)

Step 3: Calculate the moles of HCl

From the balanced equation, the stoichiometry is 2:1, so moles of HCl = 2 × moles of Mg(OH)₂ = 2 × 0.0354 = 0.0708 moles

Step 4: Calculate the volume of HCl

Volume of HCl (L) = Moles / Molarity Volume of HCl = 0.0708 / 0.128 = 0.553 L = 553 mL (Approximately) Problem (c)

Step 1: Calculate moles of KCl

Moles of KCl = Mass / Molar mass Moles of KCl = 0.785 / (39.098 + 35.453) = 0.01033 moles

Step 2: Calculate moles of Cl⁻ ions

Moles of Cl⁻ ions = moles of KCl = 0.01033 moles

Step 3: Calculate the molarity of AgNO₃

Molarity of AgNO₃ = Moles of Cl⁻ ions / Volume of AgNO₃ (L) Molarity of AgNO₃ = 0.01033 / (25.8 / 1000) = 0.4004 M Problem (d)

Step 1: Calculate the moles of HCl

Moles of HCl = Molarity × Volume in liters Moles of HCl = 0.108 × (45.3 / 1000) = 0.004886 moles

Step 2: Identify the neutralization reaction

The neutralization reaction between HCl and KOH is: \( HCl + KOH \rightarrow KCl + H_2O \)

Step 3: Calculate the moles of KOH

From the balanced equation, the stoichiometry is 1:1, so moles of KOH = moles of HCl = 0.004886 moles

Step 4: Calculate the mass of KOH

Mass of KOH = Moles × Molar mass Mass of KOH = 0.004886 × (39.098 + 15.999) = 0.269 g (Approximately)

Key Concepts

Acid-Base Titration

Acid-base titration is a lab technique used to find the concentration of an unknown acid or base solution by adding a titrant of known concentration until the reaction reaches its end point. This technique is vital for determining how much of an acid or base is needed to neutralize its opposite. For instance, if you want to know how much hydrochloric acid (HCl) to add to a sodium hydroxide (NaOH) solution for them to completely react, you'll use a titration process.

The end point of titration is often identified by a color change in an indicator added to the solution, signaling that neutralization is complete. A balanced chemical equation helps in predicting how much titrant is necessary by using the stoichiometric relationships between reactants and products.

Stoichiometry

Stoichiometry involves calculating the exact quantities of reactants and products involved in chemical reactions. It is based on balanced chemical equations that reflect the conservation of mass and charge. Each molecule or formula unit in a reaction represents a specific number of moles, and these relationships are vital for calculations.

For example, consider the reaction of perchloric acid (\(HClO_4\)) with sodium hydroxide (\(NaOH\)): \[ HClO_4 + NaOH \rightarrow NaClO_4 + H_2O\]. The 1:1 stoichiometry means that one mole of \(HClO_4\) neutralizes one mole of \(NaOH\). By using stoichiometric coefficients, you can determine the amounts of other substances involved in the reaction.

Solution Molarity

Molarity is a measure of the concentration of a solute in a solution, defined as the number of moles of solute per liter of solution (\(M = \frac{\text{moles of solute}}{\text{liters of solution}}\)). Understanding molarity is essential for preparing solutions with precise concentrations necessary for reactions like titrations.

When calculating how much reactant is needed, like \(0.115 M \ HClO_4\) to neutralize \(50.00 \ mL\) of \(0.0875 M \ NaOH\), you'll rely on the molarity formula. By rearranging the formula, you can find the volume of \(HClO_4\) needed for a complete reaction, yielding precise solution concentrations for lab work.

Chemical Equations

Chemical equations represent reactions using symbols for the substances involved. A balanced equation shows equal numbers of each type of atom on both sides, respecting the law of conservation of mass. This balance is crucial for understanding how substances interact.

When considering reactions like the neutralization of \(Mg(OH)_2\) with \(HCl\):\[ Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O\], the equation's coefficients indicate that two moles of \(HCl\) are required for every mole of \(Mg(OH)_2\) to ensure complete reaction, producing consistent results in experiments. Balancing equations allows chemists to predict the outcomes and proportions in any chemical process.