Problem 56 - General Chemistry: Principles and Modern Applications

Question

For use in titrations, we want to prepare \(20 \mathrm{L}\) of \(\mathrm{HCl}(\mathrm{aq})\) with a concentration known to four significant figures. This is a two-step procedure beginning with the preparation of a solution of about \(0.10 \mathrm{M}\) HCl. A sample of this dilute HCl(aq) is titrated with a NaOH(aq) solution of known concentration. (a) How many milliliters of concentrated \(\mathrm{HCl}(\mathrm{aq})\) \((d=1.19 \mathrm{g} / \mathrm{mL} ; 38 \% \mathrm{HCl}, \text { by mass })\) must be diluted with water to 20.0 L to prepare \(0.10 \mathrm{M} \mathrm{HCl}\) ? (b) \(\mathrm{A} 25.00\) \(\mathrm{mL}\) sample of the approximately \(0.10\) \(\mathrm{M}\) HCl prepared in part (a) requires \(20.93\) \(\mathrm{mL}\) of \(0.1186\) \(\mathrm{M}\) NaOH for its titration. What is the molarity of the \(\mathrm{HCl}(\mathrm{aq}) ?\) (c) Why is a titration necessary? That is, why not prepare a standard solution of \(0.1000\) \(\mathrm{M} \mathrm{HCl}\) simply by an appropriate dilution of the concentrated HCl(aq)?

Step-by-Step Solution

Answer

(a) 161 mL concentrated HCl is needed. (b) The molarity of the HCl solution is 0.0991 M. (c) Titration is necessary for accurate determination of the solution’s actual concentration, especially when it needs to be known to several significant figures.

Step 1: Calculate the Volume of Concentrated HCl for Preparation

To prepare a 20.0 L of 0.10 M HCl solution, we will use the formula \( M_{1}V_{1}=M_{2}V_{2} \) for dilution of solutions where\n\( M_{1} \) = Molarity of concentrated HCl (to be found),\n\( V_{1} \) = Volume of concentrated HCl to be added,\n\( M_{2} \) = Molarity of the diluted HCl (0.10 M),\n\( V_{2} \) = Volume of the diluted HCl (20.0 L).\nFrom the given information that the 38% by mass of concentrated HCl solution and \( \text{density (d) is } 1.19 \text{g/mL} \), we calculate \( M_{1} \) as follows:\nThe mass of 1 mL of solution is 1.19 g (from the density). Thus, the mass of HCl in 1 mL of solution will be \( 38\% \) of 1.19 g = 0.4522 g. Now, calculating the moles of HCl in 1 mL of solution using the molar mass of HCl (36.5 g/mol), we find \( M_{1} \) = moles/volume = \( \frac{0.4522 \text{ g}}{36.5 \text{ g/mol}} \) / (1 mL / 1000) = 12.40 mol/L.

Step 2: Substitute and Solve for Unknown

The only unknown value in the dilution equation is \( V_{1} \), the volume of concentrated HCl. Substituting \( M_{1} = 12.4 \text{ M}, M_{2} = 0.1 \text{ M}, V_{2} = 20.0 \text{ L} \) into the dilution formula, we solve for \( V_{1} \): \( V_{1} = \frac{M_{2}V_{2}}{M_{1}} = \frac{(0.1)(20.0)}{12.4} = 0.161 \text{ L} \) which is 161 mL.

Step 3: Determining the Molarity of Dilute HCl after Titration

Following a titration process, 25.00 mL sample of the 0.10 M HCl solution was neutralized by 20.93 mL of 0.1186 M NaOH. Given that HCl and NaOH react in a 1:1 ratio, we can determine the actual molarity of the HCl solution by setting up a mole ratio and solving for unknown molarity. Because the reaction is 1:1, the molarity of HCl will be \( \frac{0.1186 \text{ M NaOH} \times 20.93 \text{ mL NaOH}}{25.00 \text{ mL HCl}} = 0.0991 \text{ M HCl} \)

Step 4: Explanation for the Necessity of Titration

The titration is necessary because the molarity of a solution prepared by dilution isn't always exactly known due to possible experimental errors. Performing titration allows for accurate determination of the solution’s actual concentration, especially when it needs to be known to several significant figures, as in this case. It ensures the precision and reliability of the results.

Key Concepts

Molarity Calculation

Molarity, often represented by the symbol \( M \), is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. In the context of the problem, we needed to prepare 20 liters of a 0.10 M HCl solution. The initial step in determining the volume of concentrated hydrochloric acid (HCl) required involves using the dilution formula:

• \( M_1V_1 = M_2V_2 \)

Where \( M_1 \) is the molarity of the concentrated solution, \( V_1 \) is the volume of the concentrated solution, \( M_2 \) is the molarity of the diluted solution (0.10 M), and \( V_2 \) is the volume of the diluted solution (20.0 L). First, we calculated \( M_1 \) using the information that the density of the concentrated HCl (38% by mass) is 1.19 g/mL. This gives us the moles of HCl in one milliliter of solution, allowing us to find \( M_1 = 12.40 \text{ mol/L} \). By substituting known values into the dilution equation, we solve for \( V_1 \), which represents how many milliliters of concentrated HCl are required.

Dilution Process

In chemistry, dilution is the process of decreasing the concentration of a solute in a solution, usually by adding more solvent. The main purpose of dilution is to make a solution less concentrated to achieve a desired molarity. This is relevant when using chemicals in titration experiments, as precise concentrations are crucial.

• To dilute the concentrated HCl to a specific molarity, you must know both the initial concentration \( M_1 \) and amount of solvent to add, to achieve the desired concentration \( M_2 \)
• When you perform a dilution, the amount of solute remains the same; only the solvent increases.

Using the dilution formula \( M_1V_1 = M_2V_2 \), the number of milliliters of concentrated HCl needed was accurately calculated, ensuring the 20 L of 0.10 M HCl solution is precisely prepared for further titration procedures.

Neutralization Reaction

A neutralization reaction occurs when an acid and a base react to form water and a salt. This type of reaction is crucial in titrations, which are used to determine the unknown concentration of an acid or base in a solution. In our specific exercise, this reaction takes place between HCl (an acid) and NaOH (a base).

• The balanced chemical equation for this reaction is \( \text{HCl (aq) + NaOH (aq) } \rightarrow \text{ NaCl (aq) + } \frac{1}{2} \text{H}_2 \text{O} \).
• Such reactions follow a 1:1 molar ratio, meaning one mole of HCl reacts with one mole of NaOH.

Through the titration, we used a known concentration of NaOH to determine the molarity of the HCl solution by measuring how much NaOH is required to completely neutralize the HCl sample. This is crucial because it helps in finding the actual concentration of HCl, which might differ from the expected molarity due to potential experimental errors during preparation. This procedure yields a more precise and reliable measurement of molarity, fully essential in quantitative chemical analysis.