Problem 53 - General Chemistry: Principles and Modern Applications

Question

An \(\mathrm{NaOH}(\mathrm{aq})\) solution cannot be made up to an exact concentration simply by weighing out the required mass of NaOH, because the NaOH is not pure. Also, water vapor condenses on the solid as it is being weighed. The solution must be standardized by titration. For this purpose, a 25.00 \(\mathrm{mL}\) sample of an NaOH(aq) solution requires 28.34 \(\mathrm{mL}\) of 0.1085 \(\mathrm{M}\) HCl. What is the molarity of the NaOH(aq)? \(\mathrm{HCl}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq}) \longrightarrow \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)\)

Step-by-Step Solution

Answer

The molarity of the NaOH(aq) is approximately 0.123 M

Step 1: Calculate moles of HCl

Moles of HCl can be calculated from its volume and molarity. Since volume is given in mL, convert it to L: \( Volume_{HCl} = 28.34 mL = 28.34/1000 = 0.02834 L \). Now, calculate moles using the formula \( Moles = Molarity * Volume \): \( Moles_{HCl} = 0.1085 * 0.02834 = 0.00307469 moles \). It's significant to remember that at the equivalence point, moles of HCl = moles of NaOH.

Step 2: Calculate molarity of NaOH

Now as per the information, a 25.00 ml sample of an NaOH solution is equivalent to the calculated moles of HCl. Convert this volume also to liters \( Volume_{NaOH} = 25 mL = 25/1000 = 0.025 L \). Hence the molarity of NaOH is calculated by rearranging the formula for moles to \( Molarity = Moles/Volume \): \( Molarity_{NaOH} = Moles_{NaOH}/Volume_{NaOH} = 0.00307469 / 0.025 = 0.1229876 M \) .

Key Concepts

Standardization

Standardization is a crucial process in chemistry that ensures the precise concentration of a solution. It involves using a titration method to compare a solution with a known concentration, called a standard, to another solution with an unknown concentration. In this example of NaOH solution, standardization is essential because the substance is often impure. Factors like water vapor absorption can affect its mass and, consequently, its measured concentration.
To standardize, a known concentration of \( ext{HCl}\) is used in the titration as a reference to determine the exact concentration of the \( ext{NaOH}\) solution. By reacting \( ext{NaOH}\) with \( ext{HCl}\) until they neutralize, we accurately determine \( ext{NaOH}\)'s molarity, overcoming initial uncertainties in its mass or purity. This step ensures accurate measurements in experiments and industrial applications.

Molarity Calculation

Molarity is a measure of the concentration of a solution. It represents moles of solute per liter of solvent. Calculating molarity requires knowing both the volume and the moles of the solution. Let's break down the steps involved in calculating the molarity of \( ext{NaOH}\) in this context.
First, you need to find the moles of the solution being titrated (here, \( ext{HCl}\)). This is done using the relation: \[\text{Moles} = \text{Molarity} \times \text{Volume}.\] After determining the moles of \( ext{HCl}\), we realize that at equivalence point, moles of \( ext{HCl}\) equal the moles of \( ext{NaOH}\) due to a 1:1 reaction ratio.
With the mole value in hand, calculate \( ext{NaOH}\) molarity using \[\text{Molarity} = \frac{\text{Moles}}{\text{Volume (in L)}}.\] Adjust the units carefully to ensure accurate calculations, as seen when converting milliliters to liters before applying the formula. This calculation ensures the solution’s proper chemical identity and reaction balance.

Stoichiometry

Stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction. It's vital for determining the exact amounts needed or produced in reactions. In titration, stoichiometry helps us understand the interaction between acid and base, ensuring the molar balance is achieved.
In the given exercise, the stoichiometric equation is \[\text{HCl (aq) + NaOH (aq) } \rightarrow \text{ NaCl (aq) + } H_2O(l)\], showcasing a simple acid-base neutralization. Each \( ext{HCl}\) molecule reacts with one \( ext{NaOH}\) molecule, indicating a 1:1 molar ratio. Understanding this ratio is essential, as it allows direct conversion of moles from one substance (\( ext{HCl}\)) to another (\( ext{NaOH}\)) under the conditions of the reaction.
This stoichiometric relationship guides the calculations for determining concentrations and ensures predictability in chemical processes, making it crucial in laboratory and industrial settings.