Calculating The Volume Of Distilled Water For A PH 2 HCl Solution

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Understanding pH and its relationship to the concentration of acids and bases is fundamental in chemistry. This article addresses the question of how much distilled water is needed to dissolve 0.01 mol of hydrochloric acid (HCl) to create a solution with a pH of 2. This is a common type of problem encountered in chemistry, particularly in the areas of acid-base chemistry and solution preparation. We'll delve into the concepts, calculations, and steps required to solve this problem, making it clear and understandable for anyone studying chemistry.

Understanding pH and Acid Concentration

To tackle this problem effectively, it's crucial to grasp the fundamental concepts of pH and how it relates to the concentration of strong acids like HCl. pH is a measure of the acidity or alkalinity of a solution. It is defined as the negative base-10 logarithm of the hydrogen ion (H+) concentration:

pH=−log⁡10[H+]\text{pH} = -\log_{10}[H^+]

A solution with a pH of 2 indicates a relatively acidic environment. Specifically, a pH of 2 means that the hydrogen ion concentration, [H+][H^+], is 10−210^{-2} moles per liter (M). Mathematically, this can be represented as:

[H+]=10−pH=10−2 M[H^+] = 10^{-\text{pH}} = 10^{-2} \text{ M}

Hydrochloric acid (HCl) is a strong acid, which means it completely dissociates in water to form hydrogen ions (H+) and chloride ions (Cl-). This complete dissociation is key to calculating the required volume. For every mole of HCl dissolved in water, one mole of H+ ions is produced. Therefore, the concentration of H+ ions in the solution directly corresponds to the concentration of HCl.

In our problem, we want to prepare a solution with a pH of 2, meaning the [H+][H^+] should be 10−210^{-2} M. Since HCl completely dissociates, the concentration of HCl in the solution also needs to be 10−210^{-2} M. This understanding bridges the pH requirement with the concentration of the acid, setting the stage for calculating the required volume of water.

Calculating the Required Volume of Water

Now that we understand the desired HCl concentration, we can calculate the volume of distilled water needed. We have 0.01 moles of HCl, and we want a solution with a concentration of 10−210^{-2} M (or 0.01 M). Molarity (M) is defined as the number of moles of solute per liter of solution:

Molarity (M)=Moles of soluteVolume of solution in liters\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}

We can rearrange this formula to solve for the volume:

Volume of solution in liters=Moles of soluteMolarity (M)\text{Volume of solution in liters} = \frac{\text{Moles of solute}}{\text{Molarity (M)}}

In our case, the moles of solute (HCl) are 0.01 mol, and the desired molarity is 0.01 M. Plugging these values into the formula, we get:

Volume of solution in liters=0.01 mol0.01 M=1 liter\text{Volume of solution in liters} = \frac{0.01 \text{ mol}}{0.01 \text{ M}} = 1 \text{ liter}

This calculation shows that we need 1 liter of solution to achieve the desired concentration. Since the question asks for the volume of distilled water, and we're assuming the volume of HCl is negligible compared to the final volume, we need to dissolve the 0.01 mol of HCl in 1 liter of water.

However, the options are given in dm3dm^3 (cubic decimeters). Since 1 liter is equal to 1 dm3dm^3, the required volume is 1 dm3dm^3. Looking at the options provided, none of them match our calculated volume directly. This suggests there might be a misunderstanding or a subtle trick in the question. Let's re-evaluate the question and our approach to ensure accuracy.

Re-evaluating and Refining the Calculation

Upon re-evaluating the problem, we realize that we have calculated the final volume of the solution, which is 1 liter or 1 dm3dm^3. However, the question asks for the volume of distilled water to be added. Our initial calculation assumed the volume of HCl added is negligible, but we should verify this assumption.

We have 0.01 mol of HCl. The molar mass of HCl is approximately 36.46 g/mol. Therefore, the mass of HCl we are using is:

0.01 mol×36.46 g/mol=0.3646 g0.01 \text{ mol} \times 36.46 \text{ g/mol} = 0.3646 \text{ g}

The density of concentrated HCl (37%) is about 1.19 g/mL. However, we're dealing with pure HCl gas dissolved in water, so we can't directly use this density. Since we're aiming for a pH of 2, the solution will be very dilute, and the volume change upon dissolving 0.3646 g of HCl in water will be minimal.

Therefore, our initial assumption that the volume of HCl is negligible is valid in this context. We still need 1 liter (1 dm3dm^3) of solution. However, the options provided do not include 1 dm3dm^3. This discrepancy indicates a potential error in the question or the provided options. Let's examine the options more closely and see if we can deduce the correct answer based on the given choices.

Given the options: A. 0.4dm30.4 dm^3 B. 0.5dm30.5 dm^3 C. 0.6dm30.6 dm^3 D. 0.8dm30.8 dm^3

None of these options result in a final volume of 1 dm3dm^3 when considering the 0.01 mol of HCl added. This suggests a possible mistake in the options or the question itself. However, if we were to choose the closest answer based on a reasonable scenario, we would need to think about what volume would give us approximately 0.01 M concentration.

To achieve 0.01 M concentration with 0.01 mol of HCl, we need 1 L (1dm31 dm^3) of solution. If the question intended to ask for an approximate volume, then adding 0.01 mol of HCl to any of the volumes listed would not result in a pH of 2. The closest option that would yield a concentration nearing 0.01 M would be the largest volume provided, which is 0.8 dm3dm^3. However, even this would result in a slightly higher concentration than desired.

Determining the Correct Answer

Given our calculations and analysis, it appears there might be an issue with the provided options. The correct volume of distilled water required to make a solution of pH 2 by dissolving 0.01 mol HCl should result in a final solution volume of 1 dm3dm^3 (1 liter). However, none of the options (0.4, 0.5, 0.6, or 0.8 dm3dm^3) lead to this result.

If we were forced to choose the closest answer from the given options, we would need to consider which volume, when used to dissolve 0.01 mol HCl, would result in a pH closest to 2. A smaller volume would lead to a higher concentration and thus a lower pH (more acidic), while a larger volume would lead to a lower concentration and a higher pH (less acidic).

Let's calculate the HCl concentration for each option:

A. 0.4dm30.4 dm^3: Concentration = 0.01 mol0.4 dm3\frac{0.01 \text{ mol}}{0.4 \text{ }dm^3} = 0.025 M B. 0.5dm30.5 dm^3: Concentration = 0.01 mol0.5 dm3\frac{0.01 \text{ mol}}{0.5 \text{ }dm^3} = 0.02 M C. 0.6dm30.6 dm^3: Concentration = 0.01 mol0.6 dm3\frac{0.01 \text{ mol}}{0.6 \text{ }dm^3} ≈ 0.0167 M D. 0.8dm30.8 dm^3: Concentration = 0.01 mol0.8 dm3\frac{0.01 \text{ mol}}{0.8 \text{ }dm^3} = 0.0125 M

We want a concentration of 0.01 M, which corresponds to a pH of 2. Option D (0.8 dm3dm^3) gives us the closest concentration to 0.01 M. Therefore, if we must choose from the given options, D. 0.8 dm3dm^3 is the most reasonable answer, although it's not perfectly accurate.

Conclusion

The problem of determining the volume of distilled water to dissolve 0.01 mol HCl to achieve a pH of 2 highlights the importance of understanding pH, concentration, and the behavior of strong acids in solution. While our calculations show that 1 dm3dm^3 of solution is required, the closest option provided is 0.8 dm3dm^3. This discrepancy underscores the need for careful problem analysis and critical evaluation of given options. In real-world scenarios, it is crucial to double-check experimental setups and calculations to ensure accurate results. This exercise not only reinforces key chemical concepts but also enhances problem-solving skills essential in chemistry and related fields.