Problem 41 - Chemistry The Central Science In Si Units

Question

How many milliliters of \(0.0750 \mathrm{M} \mathrm{KOH}\) are required to titrate each of the following solutions to the equivalence point: (a) \(30.0 \mathrm{~mL}\) of \(0.0900 \mathrm{M} \mathrm{HCOOH}\), (b) \(45.0 \mathrm{~mL}\) of \(0.0750 \mathrm{M} \mathrm{HNO}_{3},(\mathbf{c}) 50.0 \mathrm{~mL}\) of a solution that contains \(3.00 \mathrm{~g}\) of HBr per liter?

Step-by-Step Solution

Answer

(a) 36.0 mL KOH, (b) 45.0 mL KOH, (c) 24.67 mL KOH.

Step 1: Determine moles of HCOOH

First, calculate the moles of \mathrm{HCOOH} in the solution. Use the formula \[ \text{moles} = \text{concentration (M)} \times \text{volume (L)} \]For \(a\), the concentration is \(0.0900 \, \mathrm{M}\) and the volume is \(30.0 \, \mathrm{mL} = 0.0300 \, \mathrm{L}\).\[ \text{moles of } \mathrm{HCOOH} = 0.0900 \, \mathrm{M} \times 0.0300 \, \mathrm{L} = 0.00270 \, \text{moles} \]

Step 2: Calculate KOH required for HCOOH

The reaction between \mathrm{KOH} and \mathrm{HCOOH} is a 1:1 molar reaction, meaning one mole of \mathrm{KOH} reacts with one mole of \mathrm{HCOOH}.Thus, \textbf{the moles of \mathrm{KOH}} required is also \(0.00270\, \text{moles}\).Using the concentration of \(0.0750 \, \mathrm{M} \, \mathrm{KOH}\), we calculate the volume required:\[ \text{Volume (L)} = \frac{\text{moles}}{\text{concentration}} = \frac{0.00270 \, \text{moles}}{0.0750 \, \mathrm{M}} = 0.0360 \, \mathrm{L} \]Convert to milliliters:\[ 0.0360 \, \mathrm{L} \times 1000 = 36.0 \, \mathrm{mL} \]

Step 3: Determine moles of HNO3

Calculate the moles of \mathrm{HNO_3} in the solution. For \(b\), the concentration is \(0.0750 \, \mathrm{M}\) and the volume is \(45.0 \, \mathrm{mL} = 0.0450 \, \mathrm{L}\).\[ \text{moles of } \mathrm{HNO_3} = 0.0750 \, \mathrm{M} \times 0.0450 \, \mathrm{L} = 0.00338 \, \text{moles} \]

Step 4: Calculate KOH required for HNO3

Since \mathrm{KOH} and \mathrm{HNO_3} also react in a 1:1 molar fashion, the moles of \mathrm{KOH} required is \(0.003375 \, \,\text{moles}\).The volume of \mathrm{KOH} needed is:\[ \text{Volume (L)} = \frac{0.00338 \, \text{moles}}{0.0750 \, \mathrm{M}} = 0.0450 \, \mathrm{L} \]Convert to milliliters:\[ 0.0450 \, \mathrm{L} \times 1000 = 45.0 \, \mathrm{mL} \]

Step 5: Determine moles of HBr

For \(c\), calculate the concentration of \mathrm{HBr} from the given mass per liter. Molar mass of \mathrm{HBr} = 1 + 80 = 81 \, \text{g/mol}.Concentration of \mathrm{HBr}:\[ = \frac{3.00 \, \text{g}}{81 \, \text{g/mol}}= 0.0370 \, \text{M} \]Calculate the moles of \mathrm{HBr} using this concentration:\[ \text{moles of } \mathrm{HBr} = 0.0370 \, \text{M} \times 0.0500 \, \mathrm{L} = 0.00185 \, \text{moles} \]

Step 6: Calculate KOH required for HBr

The moles of \mathrm{KOH} needed for \mathrm{HBr} is equivalent to its moles: \(0.00185 \, \text{moles}\).Volume of \mathrm{KOH}:\[ \text{Volume (L)} = \frac{0.00185 \, \text{moles}}{0.0750 \, \mathrm{M}} = 0.02467 \, \mathrm{L} \]Convert to milliliters:\[ 0.02467 \, \mathrm{L} \times 1000 = 24.67 \, \mathrm{mL} \]

Key Concepts

Acid-Base Reactions

In a titration involving acid-base reactions, the goal is to find the exact amount of an acid that can neutralize a base, or vice versa. This process helps us determine the concentration of one solution using a solution of known concentration. The key reaction in the given exercise involves potassium hydroxide (KOH) and different acids like HCOOH (formic acid), HNO3 (nitric acid), and HBr (hydrobromic acid).

A titration reaches its endpoint at the equivalence point, where the moles of acid equal the moles of base. The chemical reaction involves the hydrogen ions from the acid reacting with the hydroxide ions from the base to form water and a salt. For example:

• KOH + HCOOH → H2O + KCOOH
• KOH + HNO3 → H2O + KNO3
• KOH + HBr → H2O + KBr

Understanding the stoichiometry of the reactions is essential as they guide the calculations required to find the amounts needed during titration.

Molarity Calculations

Molarity is a measure of the concentration of a solution, given in moles of solute per liter of solution (mol/L). Accurate molarity calculations are crucial in titrations as the solution’s concentration directly affects the volume measurement when reacting with another substance.

In this exercise, you calculate the amount of potassium hydroxide (KOH) needed by first determining how many moles of acid are present in the solution. Use the formula:

• \[\text{moles of solute} = \text{molarity} \times \text{volume in liters}\]

Then, calculate the volume of KOH needed to reach the equivalence point using:

• \[\text{Volume of solution} = \frac{\text{moles required}}{\text{molarity of } KOH}\]

For example, if you have 0.0900 M of HCOOH in 30.0 mL, convert the volume to liters, calculate the moles, and then find how much 0.0750 M KOH is needed to neutralize the acid.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. The stoichiometric coefficients of a balanced chemical equation tell us the proportions in which chemicals react. In acid-base titrations, it's typically a 1:1 ratio unless specified differently by the reaction equation.

For titrating HCOOH, HNO3, and HBr with KOH, the reactions happen at a 1:1 ratio. This means that one mole of KOH neutralizes one mole of acid. Here's a step-by-step on using stoichiometry in titration:

• Calculate the moles of the acid present using molarity and volume.
• Assuming the reaction is a 1:1 ratio, the moles of KOH needed will be equal to the acid's moles.
• Finally, determine the moles of KOH in the required volume of solution through its molarity.

Once stoichiometry is understood, you can confidently calculate how much titrant is needed for any given reaction, assuring the balance and completion of a chemical reaction.