Problem 81 - Chemistry: The Central Science in SI Units, Global Edition

Question

(a) What volume of \(0.115 \mathrm{MHClO}_{4}\) solution is needed to neutralize \(50.00 \mathrm{~mL}\) of \(0.0875 \mathrm{MNaOH}\) ? (b) What volume of \(0.128 \mathrm{MHCl}\) is needed to neutralize \(2.87 \mathrm{~g}\) of \(\mathrm{Mg}(\mathrm{OH})_{2} ?\) (c) If \(25.8 \mathrm{~mL}\) of an \(\mathrm{AgNO}_{3}\) solution 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 a 0.108 \(M\) HCl solution is needed to neutralize a solution of \(\mathrm{KOH}\), how many grams of KOH must be present in the solution?

Step-by-Step Solution

Answer

(a) 38.043 mL of 0.115 M HClO4 solution is needed to neutralize 50.00 mL of 0.0875 M NaOH. (b) 387.34 mL of 0.128 M HCl is needed to neutralize 2.87 g of Mg(OH)2. (c) The molarity of the AgNO3 solution is 0.4065 M. (d) There are 0.2742 g of KOH present in the solution.

Step 1: (a) Find the necessary volume of HClO4 to neutralize NaOH

To find the volume of the HClO4 solution needed, we will use the balanced chemical equation and the concept of moles: \[HClO_{4} + NaOH \rightarrow NaClO_{4} + H_2O\] Since the balanced chemical equation shows a 1:1 mole ratio between HClO4 and NaOH, we can use the given molarity of NaOH to find the moles of NaOH: moles of NaOH = Molarity of NaOH × volume of NaOH = 0.0875 M × 50.00 mL = 4.375 mmol Now, using the moles of NaOH, we can find the volume of HClO4 needed: volume of HClO4 = (moles of NaOH / Molarity of HClO4) = (4.375 mmol / 0.115 M) = 38.043 mL Therefore, 38.043 mL of 0.115 M HClO4 solution is needed to neutralize 50.00 mL of 0.0875 M NaOH.

Step 2: (b) Find the necessary volume of HCl to neutralize Mg(OH)2

First, let's write the balanced chemical equation for the reaction between HCl and Mg(OH)2: \[2HCl + Mg(OH)_{2} \rightarrow MgCl_{2} + 2H_2O\] Now, let's convert the mass of Mg(OH)2 given (2.87 g) into moles, using its molar mass (58.32 g/mol for Mg and 34.02 g/mol for 2 OH): moles of Mg(OH)2 = (2.87 g) / (58.32 g/mol + 2 × 34.02 g/mol) = 0.02479 mol Using the mole ratio from the balanced chemical equation (2:1), we can find the moles of HCl needed: moles of HCl = 2 × moles of Mg(OH)2 = 2 × 0.02479 mol = 0.04958 mol Now, using the given molarity of HCl (0.128 M), we can find the volume of HCl needed: volume of HCl = (moles of HCl / Molarity of HCl) = (0.04958 mol / 0.128 M) = 387.34 mL Therefore, 387.34 mL of 0.128 M HCl is needed to neutralize 2.87 g of Mg(OH)2.

Step 3: (c) Calculate the molarity of AgNO3 solution

We are given the information that 25.8 mL of AgNO3 solution is needed to precipitate all Cl- ions from a 785-mg sample of KCl, forming AgCl. First, let's find the moles of Cl- ions in the 785-mg sample of KCl using its molar mass (39.10 g/mol for K and 35.45 g/mol for Cl): moles of Cl- = (0.785 g) / (39.10 g/mol + 35.45 g/mol) = 0.01049 mol Now, we can write the balanced chemical equation for the reaction between AgNO3 and KCl: \[AgNO_{3} + KCl \rightarrow AgCl + KNO_{3}\] Since the balanced chemical equation shows a 1:1 mole ratio between AgNO3 and Cl-, the moles of AgNO3 required for the reaction is equal to the moles of Cl- ions. moles of AgNO3 = moles of Cl- = 0.01049 mol Now, using the given volume of the AgNO3 solution (25.8 mL), we can find the molarity of AgNO3: Molarity of AgNO3 = (moles of AgNO3 / volume of AgNO3) = (0.01049 mol / 25.8 mL) = 0.4065 M Therefore, the molarity of the AgNO3 solution is 0.4065 M.

Step 4: (d) Calculate the grams of KOH present in the solution

We are given the information that 45.3 mL of a 0.108 M HCl solution is needed to neutralize a solution of KOH. First, let's find the moles of HCl using its molarity and volume: moles of HCl = Molarity of HCl × volume of HCl = 0.108 M × 45.3 mL = 4.8876 mmol Now, let's write the balanced chemical equation for the reaction between HCl and KOH: \[HCl + KOH \rightarrow KCl + H_2O\] Since the balanced chemical equation shows a 1:1 mole ratio between HCl and KOH, the moles of KOH required for the reaction are equal to the moles of HCl: moles of KOH = moles of HCl = 4.8876 mmol Finally, using the molar mass of KOH (56.11 g/mol), we can convert the moles of KOH to grams: mass of KOH = moles of KOH × molar mass of KOH = 4.8876 mmol × 56.11 g/mol = 0.2742 g Therefore, there are 0.2742 g of KOH present in the solution.

Key Concepts

Molarity Calculation

When dealing with chemical solutions, **molarity** is a crucial concept. Molarity (M) is a measure of the concentration of a solution, defined as the number of moles of a solute divided by the volume of the solution in liters. It provides a powerful way to determine how much solute is present in a given volume.
To calculate molarity, use the formula:

• \( M = \frac{moles\ of\ solute}{liters\ of\ solution} \)

For example, if a solution contains 1 mole of a substance in 1 liter of water, it has a molarity of 1 M. Understanding this allows you to effectively prepare and react solutions in the laboratory.
In the exercises mentioned, the volume of HCl or AgNO₃ needed for a reaction was calculated by acknowledging the relationship between the moles present in a given solution and its volume.

Neutralization Reactions

Neutralization reactions play a significant role in chemistry, involving the reaction of an acid and a base to produce water and a salt. Generally, the equation for a simple neutralization reaction can be exemplified with:

• \( HCl + NaOH \rightarrow NaCl + H_2O \)

These reactions are fundamental in analytical chemistry for determining substances' concentrations.
In the step-by-step solution, the given chemical reactions of HCl neutralizing KOH or Mg(OH)₂ are examples of such reactions. Each one involves equimolar amounts of hydrogen ions (from the acid) and hydroxide ions (from the base) combining to form water.
Neutralization is completed when the amounts of acid and base are equivalent, which may involve using stoichiometry to calculate the required amount of each reactant.

Balanced Chemical Equations

Chemical equations need to be balanced to obey the Law of Conservation of Mass, which asserts that matter cannot be created or destroyed. This means each side of a chemical equation must have equal numbers of each type of atom.
For example, in the reaction:

• \( 2HCl + Mg(OH)_2 \rightarrow MgCl_2 + 2H_2O \)

Both sides of the equation have 2 chlorine atoms, 2 hydrogen molecules, and 1 magnesium atom, ensuring the equation balances. This balancing allows chemists to determine the ratios of reactants and products required for a reaction.
Properly balanced equations ensure correct stoichiometry calculations for determining the amount of products and reactants involved in a reaction, as seen in the problem set where equivalent volumes of solutions are needed to reach complete neutralization.

Molar Mass Conversion

Molar mass is essential when converting between grams and moles of a substance. It acts as a bridge between mass (grams) and the amount of substance (moles).
The molar mass, expressed in g/mol, is the mass of one mole of a substance, allowing you to translate mass into moles and vice versa. Use the formula:

• \( \, moles = \frac{mass\ (g)}{molar\ mass\ (g/mol)} \)

For instance, converting 2.87 g of \( Mg(OH)_2 \) into moles involves using its molar mass derived from adding the atomic masses of magnesium and hydroxide ions.
This concept is foundational in calculating how much of a reactant is needed or produced in a reaction, such as determining the moles of Mg(OH)₂ required in a neutralization reaction with HCl.