Problem 42 - Chemistry: The Central Science in SI Units, Global Edition

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 mL of a solution that contains \(1.35 \mathrm{~g}\) of \(\mathrm{NaOH}\) per liter?

Step-by-Step Solution

Answer

To titrate each solution mentioned to the equivalence point, the following volumes of 0.105 M HCl are needed: (a) 40.71 mL for 45.0 mL of 0.0950 M NaOH. (b) 25.29 mL for 22.5 mL of 0.118 M NH3. (c) 40.18 mL for 125.0 mL of a solution that contains 1.35 g of NaOH per liter.

Step 1: Determine moles of NaOH

To find the moles of NaOH, we will multiply its molarity (0.0950 M) by its volume in liters (0.0450 L): Moles of NaOH = (0.0950 mol/L) * (0.0450 L) = 0.004275 mol

Step 2: Find moles of HCl reactant

Since the molar ratio of NaOH to HCl is 1:1 in their reaction, we need an equal amount of moles of HCl to titrate the NaOH: Moles of HCl = 0.004275 mol

Step 3: Calculate the volume of 0.105 M HCl

Now, we'll divide the moles of HCl by its molarity to find the required volume to titrate the NaOH: Volume of HCl = (0.004275 mol) / (0.105 mol/L) = 0.040714 L = 40.71 mL Answer for problem (a): 40.71 mL of 0.105 M HCl. ### Problem (b): 22.5 mL of 0.118 M NH3 ###

Step 1: Determine moles of NH3

Moles of NH3 = (0.118 mol/L) * (0.0225 L) = 0.002655 mol

Step 2: Find moles of HCl reactant

The molar ratio of NH3 to HCl is 1:1, therefore: Moles of HCl = 0.002655 mol

Step 3: Calculate the volume of 0.105 M HCl

Volume of HCl = (0.002655 mol) / (0.105 mol/L) = 0.025286 L = 25.29 mL Answer for problem (b): 25.29 mL of 0.105 M HCl. ### Problem (c): 125 mL of a solution containing 1.35 g of NaOH per liter ###

Step 1: Calculate moles of NaOH in the solution

First, find the molarity of the NaOH solution: 1.35 g/L * (1 mol/39.997 g) = 0.033749 M (we multiply by the inverse of NaOH's molar mass to convert grams to moles). Now, determine moles of NaOH in our given volume: Moles of NaOH = (0.033749 mol/L) * (0.125 L) = 0.0042186 mol

Step 2: Find moles of HCl reactant

Moles of HCl = 0.0042186 mol (1:1 molar ratio of NaOH to HCl)

Step 3: Calculate the volume of 0.105 M HCl

Volume of HCl = (0.0042186 mol) / (0.105 mol/L) = 0.0401771 L = 40.18 mL Answer for problem (c): 40.18 mL of 0.105 M HCl.

Key Concepts

Molarity

Molarity is a way to measure the concentration of a solution. It refers to the number of moles of solute, like HCl or NaOH, dissolved in one liter of solution. Molarity is expressed as M, such as 0.105 M HCl, meaning 0.105 moles of HCl per liter of solution.
To calculate molarity, you divide the number of moles of solute by the volume of solution in liters. For example, in part c of the exercise, the NaOH solution's molarity is found by dividing the moles of NaOH (calculated from the given mass and molar mass) by the volume of solution.

• Determine moles of solute: use the molar mass for conversion from grams to moles.
• Molarity calculation: moles of solute/volume of solution in liters.

Understanding molarity helps us know the exact concentration of reactive agents participating in titration.

Equivalence Point

In titration, the equivalence point is a critical stage where the quantity of titrant added is enough to completely neutralize the analyte solution. At this point, the number of moles of the titrant equals the number of moles of the substance being titrated, following the reaction stoichiometry.
For example, when you titrate 45 mL of 0.0950 M NaOH with 0.105 M HCl, the equivalence point is where the moles of NaOH equal the moles of HCl. Since their molar ratio is 1:1, the numbers of moles of NaOH and HCl are equivalent, hence the equivalence point is reached when they are equal.

• Equivalence point marks complete reaction between acid and base.
• Hint: it's not always the same as the end point, which might be indicated by a color change.

Recognizing the equivalence point is crucial for calculating how much titrant is needed to exactly react with all the analyte.

Stoichiometry

Stoichiometry involves using balanced chemical equations to determine the relationships between reactants and products. In titration, stoichiometry helps us understand and calculate the amounts of each substance involved in the reaction. It tells us how much of a titrant is needed to react with a specific amount of reactant.
When titrating, stoichiometry gives us the ratio of moles in the balanced chemical equation. For instance, in the titration between NaOH and HCl, the equation: \[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \] indicates a 1:1 mole ratio. This helps in calculating the exact volume of titrant needed.
Key steps include:

• Write down the balanced equation.
• Use mole ratios to find relationships.
• Convert given units to moles and apply the ratio.

By understanding stoichiometry, you can predict the outcomes of reactions and solve various chemistry problems with precision.

Acid-Base Reaction

An acid-base reaction is a chemical process where an acid reacts with a base, resulting in the formation of a salt and usually water. It is a prominent class of reactions in chemistry, especially noticeable in titration processes.
In the exercise, reactions involve hydrochloric acid (HCl) as the titrant neutralizing bases such as NaOH and NH3. Acid-base reactions are central to titration, as they reach completion at the equivalence point. These reactions follow the general formula: \[ \text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water} \]
A few key points about acid-base reactions:

• Frequently involve H+ ions from the acid and OH- ions from the base.
• Equivalence point occurs when all acid and base molecules in solution have reacted.
• Used extensively in analytical chemistry to determine concentrations.

Balancing these reactions using stoichiometry ensures stoichiometric calculations are correct and reflects the conservation of mass principle.