Problem 42 - Chemistry The Central Science In Si Units

Question

How many milliliters of \(0.105 \mathrm{MHCl}\) are needed to titrate each of the following solutions to the equivalence point: (a) 45.0 \(\mathrm{mL}\), of \(0.0950 \mathrm{MNaOH}\) (b) \(22.5 \mathrm{~mL}\) of \(0.118 \mathrm{MNH}_{3}\), (c) 125.0 \(\mathrm{mL}\). of a solution that contains \(1.35 \mathrm{~g}\) of \(\mathrm{NaOH}\) per liter?

Step-by-Step Solution

Answer

(a) 40.71 mL, (b) 25.29 mL, (c) 40.18 mL of HCl are needed.

Step 1: Use the Neutralization Reaction

In each titration, hydrochloric acid (HCl) reacts with a base (either NaOH or NH3), producing water and a salt. For strong acids and strong bases, the neutralization equation is given as: \[ ext{HCl} + ext{NaOH} o ext{NaCl} + ext{H}_2 ext{O}\] For weak bases like NH3, consider: \[ ext{HCl} + ext{NH}_3 o ext{NH}_4 ext{Cl}\] At the equivalence point, moles of HCl will equal moles of the base.

Step 2: Calculate Moles of NaOH (Part a)

First, calculate the number of moles of NaOH using the given volume and molarity. \[ ext{moles of NaOH} = ext{volume (L)} imes ext{molarity (M)} = 0.0450 imes 0.0950 = 0.004275 ext{ moles}\]

Step 3: Determine Volume of HCl for NaOH (Part a)

Apply the relation that moles of HCl equals moles of NaOH at the equivalence point. Rearrange the relation to solve for the volume of HCl needed: \[ ext{volume of HCl (L)} = \frac{ ext{moles of HCl}}{ ext{molarity of HCl}} = \frac{0.004275}{0.105} = 0.04071 ext{ L}\] Convert this volume to milliliters: \[ ext{volume of HCl (mL)} = 0.04071 imes 1000 = 40.71 ext{ mL}\]

Step 4: Calculate Moles of NH3 (Part b)

Calculate the number of moles of NH3 using the given volume and molarity. \[ ext{moles of NH}_3 = ext{volume (L)} imes ext{molarity (M)} = 0.0225 imes 0.118 = 0.002655 ext{ moles}\]

Step 5: Determine Volume of HCl for NH3 (Part b)

Similar to part a, at equivalence point, moles of HCl equals moles of NH3. Calculate the volume of HCl required:\[ ext{volume of HCl (L)} = \frac{0.002655}{0.105} = 0.02529 ext{ L}\] Convert to milliliters:\[ ext{volume of HCl (mL)} = 0.02529 imes 1000 = 25.29 ext{ mL}\]

Step 6: Calculate Moles of NaOH (Part c)

The given concentration (1.35 ext{ g/L}) is used to find the number of moles of NaOH.First, convert the mass to moles using the molar mass of NaOH (approximately 40.0 g/mol):\[ ext{moles of NaOH per L} = \frac{1.35}{40.0} = 0.03375 ext{ moles/L}\]Find the total moles in the given volume (125.0 mL = 0.125 L):\[ ext{moles of NaOH} = 0.03375 imes 0.125 = 0.00421875 ext{ moles}\]

Step 7: Determine Volume of HCl for NaOH (Part c)

Calculate the required volume of HCl for the moles of NaOH:\[ ext{volume of HCl (L)} = \frac{0.00421875}{0.105} = 0.04018 ext{ L}\] Convert to milliliters:\[ ext{volume of HCl (mL)} = 0.04018 imes 1000 = 40.18 ext{ mL}\]

Key Concepts

Neutralization Reaction

A neutralization reaction is a chemical process where an acid and a base react to form water and a salt. This occurs during titration, a method used to determine the concentration of an unknown solution. In a typical titration involving hydrochloric acid (HCl) and bases such as sodium hydroxide (NaOH) or ammonia (NH₃), the reaction leads to the formation of a neutral solution at the equivalence point.
For a strong acid like HCl and a strong base such as NaOH, the reaction formula is given by:

• \( \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \)

If HCl reacts with a weak base like NH₃, the reaction results in the formation of ammonium chloride (NH₄Cl):

• \( \text{HCl} + \text{NH}_3 \rightarrow \text{NH}_4\text{Cl} \)

At the equivalence point of these reactions, the quantity of acid is stoichiometrically equivalent to the quantity of base, meaning they have reacted in equal moles to form a neutral solution.

Molarity Calculation

Molarity is a measure of the concentration of a solution, defined as the number of moles of solute present in one liter of solution. This is an essential concept when performing titrations, as it allows for the calculation of the volume of titrant needed to reach the equivalence point.
When given the volume and molarity of a solution, the number of moles of solute can be calculated using the formula:

• \[ \text{Moles of solute} = \text{Volume of solution (L)} \times \text{Molarity (M)} \]

For example, to find the moles of NaOH in a 45.0 mL solution with a molarity of 0.0950 M, convert the volume to liters (0.0450 L) and use the formula:

• \[ 0.0450 \text{ L} \times 0.0950 \text{ M} = 0.004275 \text{ moles of NaOH} \]

This calculation is crucial for determining how much titrant (such as HCl) is required to reach the equivalence point during a titration.

Equivalence Point

The equivalence point in a titration is the stage at which the amount of titrant added is just enough to completely neutralize the analyte solution. At this point, the number of moles of acid is equal to the number of moles of base present in the reaction.
In practical terms, reaching the equivalence point means that there is no excess acid or base left in the solution, and the reaction between them is complete.
Using a balanced chemical equation is vital to accurately identify the equivalence point. From the stoichiometry of the reaction, the relationship between the moles of reactants can be determined. For example, with the reaction \( \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \), one mole of HCl will react with one mole of NaOH.
This concept is also applied when a colored indicator or a pH meter is used to visually or instrumentally detect the completion of the reaction, marking the exact moment to stop adding titrant.

Acid-Base Reactions

Acid-base reactions are a type of chemical reaction that occur between an acid and a base, often resulting in the formation of water and a salt. These reactions are fundamental in many aspects of chemistry, especially in titrations used to determine the concentration of a given solution.
The reaction can be characterized by the transfer of protons from the acid to the base. For example, in the process where HCl and NaOH both participate, the proton from HCl (the acid) reacts with the hydroxide ion OH⁻ from NaOH (the base) to form water.

• \( \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \)

In cases involving weak bases like NH₃, even though the base is not fully ionized, the reaction still proceeds successfully. The reactions depend on the extent of dissociation of the acid and base in water, determining how readily they react. Understanding acid-base reactions allows chemists to predict the outcome of a reaction and to manipulate chemical processes for desired results, such as achieving the equivalence point in a titration.