Problem 81 - Chemistry The Central Science In Si Units

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 \(\mathrm{Mg}(\mathrm{OH})_{2} ?(\mathbf{c})\) If \(25.8 \mathrm{~mL}\) of an \(\mathrm{AgNO}_{3}\) solution is needed to precipitate all the Cl \(^{-}\) ions in a \(785-m g\) sample of \(K C l\) (forming \(\left.A g C l\right),\) 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) 38.04 mL, (b) 768.75 mL, (c) 408.14 M, (d) 0.2745 g

Step 1: Determine the moles of NaOH

To find the volume of HClO₄ needed, first calculate the moles of NaOH using the formula: \[ \text{moles} = \text{molarity} \times \text{volume (L)} \]Given the molarity of NaOH is 0.0875 M and the volume is 50.00 mL (or 0.05000 L), we have: \[ \text{moles of NaOH} = 0.0875 \times 0.05000 = 0.004375 \text{ mol} \]

Step 2: Use stoichiometry for neutralization

The neutralization reaction between NaOH and HClO₄ is a 1:1 molar ratio:\[ \text{HClO}_4 + \text{NaOH} \rightarrow \text{NaClO}_4 + \text{H}_2\text{O} \]Thus, moles of HClO₄ needed = moles of NaOH = 0.004375 mol.

Step 3: Find the volume of HClO₄

Use the formula to calculate volume:\[ \text{volume (L)} = \frac{\text{moles}}{\text{molarity}} \]Given moles of HClO₄ = 0.004375 mol and molarity = 0.115 M:\[ \text{volume of HClO}_4 = \frac{0.004375}{0.115} = 0.03804 \text{ L} = 38.04 \text{ mL} \]

Step 4: Calculate moles of Mg(OH)₂

The molar mass of Mg(OH)₂ is approximately 58.32 g/mol. We find its moles:\[ \text{moles of Mg(OH)}_2 = \frac{2.87}{58.32} = 0.0492 \text{ mol} \]

Step 5: Determine moles of HCl needed

The balanced equation for the neutralization is:\[ \text{Mg(OH)}_2 + 2 \text{HCl} \rightarrow \text{MgCl}_2 + 2 \text{H}_2\text{O} \]Thus, moles of HCl = 2 \times 0.0492 = 0.0984 mol.

Step 6: Calculate volume of HCl

Apply the volume formula:\[ \text{volume (L)} = \frac{\text{moles}}{\text{molarity}} \]Here, moles of HCl = 0.0984 mol and molarity = 0.128 M:\[ \text{volume of HCl} = \frac{0.0984}{0.128} = 0.76875 \text{ L} = 768.75 \text{ mL} \]

Step 7: Calculate moles of KCl in the sample

The molar mass of KCl is approximately 74.55 g/mol. Thus, \[ \text{moles of KCl} = \frac{785}{74.55} = 10.53 \text{ mol} \]

Step 8: Determine moles of AgNO₃ needed

The stoichiometry of the reaction between KCl and AgNO₃ is 1:1:\[ \text{KCl} + \text{AgNO}_3 \rightarrow \text{AgCl} + \text{KNO}_3 \]Therefore, moles of AgNO₃ = moles of KCl = 10.53 mol.

Step 9: Calculate molarity of AgNO₃

Apply the molarity formula:\[ \text{molarity} = \frac{\text{moles}}{\text{volume (L)}} \]Here, volume = 25.8 mL = 0.0258 L:\[ \text{molarity of AgNO}_3 = \frac{10.53}{0.0258} = 408.14 \text{ M} \]

Step 10: Calculate moles of HCl used

Calculate the moles of HCl:\[ \text{moles of HCl} = \text{molarity} \times \text{volume (L)} = 0.108 \times 0.0453 = 0.0048924 \text{ mol} \]

Step 11: Find moles of KOH neutralized

Since the reaction between KOH and HCl is 1:1,\[ \text{moles of KOH} = \text{moles of HCl} = 0.0048924 \text{ mol} \]

Step 12: Calculate mass of KOH

Using the molar mass of KOH (approximately 56.11 g/mol):\[ \text{mass of KOH} = moles \times \text{molar mass} = 0.0048924 \times 56.11 = 0.2745 \text{ g} \]

Key Concepts

Neutralization reactions

Neutralization reactions are fascinating chemical processes where an acid and a base react to form water and a salt. This kind of reaction is fundamental in chemistry because it involves the transfer of protons from the acid to the base. In many cases, the reaction results in a solution that is neither acidic nor basic, hence the name "neutralization."

When you look at the exercise, it highlights how neutralization works using hydrochloric acid (HCl) and sodium hydroxide (NaOH). In this instance, the net ionic equation is simple:

• HClO₄ + NaOH → NaClO₄ + H₂O

The concept is straightforward—the hydrogen ions ( H^+ ) from the acid combine with the hydroxide ions ( OH^- ) from the base, resulting in water. Neutralization reactions often have a 1:1 stoichiometric ratio, as seen here, but this ratio can vary depending on the specific reactants involved.

Understanding these reactions is essential to mastering stoichiometry, as they directly relate to the calculation of reactant-product relationships during chemical reactions.

Molarity calculations

Molarity is a way of expressing concentration, specified as moles of solute per liter of solution ( M ). In stoichiometry and neutralization reactions, calculating molarity is a critical skill, as it determines how much of one solution is needed to react completely with a given amount of another.

In the provided exercise, one task is to determine the amount of HClO₄ needed to neutralize NaOH . The formula for molarity is:

• ext{Molarity (M)} = rac{ ext{moles of solute}}{ ext{volume of solution (L)}}

To determine how much HClO₄ to use, you first calculate the moles of NaOH . Then, use stoichiometry to find how many moles of HClO₄ are needed, and finally, calculate the volume using molarity.

The exercise involves converting milliliters to liters (by dividing by 1000), determining the mole count, and using these calculations to find the needed volumes of acid or base solutions. This showcases how molarity links to practical applications in titration and other chemical industries.

Chemical equations

Chemical equations are foundational to chemistry, providing a concise way to describe chemical reactions. They indicate the reactants, products, and the molar relationships between them. The balanced chemical equation is key, as it obeys the law of conservation of mass, meaning the number of atoms for each element is the same on both sides of the equation.

In the problems provided, we see balanced equations such as the one for neutralization:

• ext{Mg(OH)} ₂ + 2 ext{HCl} → ext{MgCl} ₂ + 2 ext{H} ₂O

Such equations illustrate stoichiometry, helping us determine the needed reactant or expected product quantities through the relationships established by coefficients.

Chemical equations also highlight the importance of the stoichiometric coefficients, which tell us how many molecules or moles of each substance are involved, directly impacting calculations and predictions in chemical reactions. Understanding and being able to balance chemical equations is a crucial skill for anyone studying chemistry, as it serves as the foundation for answering diverse chemical questions and problems.