Question

If \(55.60 \mathrm{~mL}\) of \(0.2221 \mathrm{M}\) HCl was needed to titrate a sample of \(\mathrm{NaOH}\) to its equivalence point, what mass of \(\mathrm{NaOH}\) was present?

Short answer

The mass of NaOH is 0.4942 g.

Step-by-step solution

Understand the Given Data and What is Required

We are given the volume and molarity of HCl, a strong acid, used to titrate NaOH, a strong base. We need to find the mass of NaOH present.

Write the Balanced Chemical Equation

The balanced chemical equation for the reaction is: \[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \] This equation tells us that one mole of HCl reacts with one mole of NaOH.

Calculate Moles of HCl

We use the molarity equation, \( \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in L}} \), to calculate the moles of HCl.\[ \text{Moles of HCl} = 0.2221 \, \text{mol/L} \times 0.05560 \, \text{L} = 0.01235516 \, \text{moles} \]

Apply the Mole Ratio to Find Moles of NaOH

From the balanced equation, the mole ratio of HCl to NaOH is 1:1. Thus, the moles of NaOH are the same as the moles of HCl.\[ \text{Moles of NaOH} = 0.01235516 \, \text{moles} \]

Calculate the Mass of NaOH

The mass of a substance can be calculated using the formula \( \text{mass} = \text{moles} \times \text{molar mass} \). The molar mass of NaOH is 40.00 g/mol.\[ \text{Mass of NaOH} = 0.01235516 \, \text{moles} \times 40.00 \, \text{g/mol} = 0.4942064 \, \text{g} \]Rounding to appropriate significant figures: 0.4942 g.

Key concepts

Mole Concept

The mole concept is a fundamental idea in chemistry that helps us to measure and compare amounts of chemical substances. Essentially, a "mole" is a unit—a bit like a dozen or a pair. However, instead of 12 or 2, a mole consists of Avogadro's number, which is approximately \(6.022 imes 10^{23}\) particles (atoms, molecules, ions, etc.).
This concept allows chemists to work with the submicroscopic world on a macroscopic scale, making it easier to perform calculations. For instance, in a lab setting, we handle moles of substances rather than individual atoms or molecules.

• To convert between moles and number of particles, we use Avogadro's number.
• To convert between moles and grams, we use the substance's molar mass, which is the mass of one mole of that substance in grams.

Understanding the mole concept is crucial for solving problems in titration, as it allows you to relate quantities of reactants and products.

Stoichiometry

Stoichiometry is the aspect of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. When dealing with a problem like titration, stoichiometry becomes essential, particularly through balanced chemical equations. These equations show us the exact proportions in which different substances react.
In our exercise, the equation \( \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \) indicates a 1:1 molar relationship between hydrochloric acid (HCl) and sodium hydroxide (NaOH). This relationship tells us that one mole of HCl will react with one mole of NaOH. If we know how many moles of HCl are used in the titration, we can directly find the moles of NaOH using this stoichiometric ratio.

• It simplifies complex chemical calculations.
• It is vital for figuring out the correct amounts of reactive chemicals.
• Helps in maximizing efficiency and reduces waste in chemical processes.

Stoichiometry guides us in determining the right amount of substances to use and predict end results correctly.

Molarity

Molarity is a way of expressing the concentration of a solution. It is defined as the number of moles of solute (the substance being dissolved) per liter of solution. The unit of molarity is mol/L or simply M. In titration problems, knowing the molarity of one reactant can help identify the amount needed to fully react with another.
In our titration exercise, we are given a molarity of the HCl solution at \(0.2221 \text{ M}\). Using this information, we can calculate the moles of HCl present in the given volume. The formula we often use is: \[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in L}} \]This formula is rearranged to find the moles if molarity and volume are known:
\[ \text{Moles of solute} = \text{Molarity} \times \text{Volume (L)} \]Understanding molarity helps in determining how "strong" a solution is and in calculating the precise amounts needed for reactions.

Equivalence Point

The equivalence point in a titration is a significant concept where the amount of titrant added is exactly enough to completely neutralize the analyte solution. At this point, the number of moles of acid equals the number of moles of base, often indicated by a noticeable change in the indicator's color.
In our exercise with NaOH and HCl, reaching the equivalence point means that the acids and bases have reacted perfectly according to their stoichiometric ratio. With NaOH and HCl having a 1:1 stoichiometry, the equivalence point tells us that 0.01235516 moles of HCl means the same number of moles for NaOH.

• The equivalence point is where the titration calculation centers around.
• It's vital for determining precise concentrations and amounts.
• Helps in understanding the complete reaction of the titrant and analyte.

Achieving the equivalence point allows for accurate calculation of unknown concentrations or amounts in titration exercises.