Problem 88 - General Chemistry: Principles and Modern Applications

Question

You are asked to prepare a 100.0 mL sample of a solution with a pH of 5.50 by dissolving the appropriate amount of a solute in water with \(\mathrm{pH}=7.00 .\) Which of these solutes would you use, and in what quantity? Explain your choice. (a) \(15 \mathrm{M} \mathrm{NH}_{3}(\mathrm{aq}) ;\) (b) \(12 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) (c) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) ;\) (d) glacial (pure) acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\)

Step-by-Step Solution

Answer

To prepare a 100.0 mL solution with a pH of 5.50, use about 0.003 g of \(\mathrm{NH}_{4}\mathrm{Cl}\).

Step 1: Identify the nature of the solutes

Identify the nature of the solutes. Among them, only ammonium chloride would result in a solution with a pH less than 7, because it behaves as a weak acid when dissolved in water.

Step 2: Determine the amount of \( \mathrm{NH}_{4} \mathrm{Cl} )

Use the formula: pH = pKa + log([A-]/[HA]), where A- is NH3 and HA is NH4+. Given that pKa of NH4+ is 9.25 and pH is 5.50, we can solve for the ratio of [A-]/[HA] using the above formula. This gives [A-]/[HA] = 10^(pH-pKa) = 10^(5.5 - 9.25) = 0.000562. The ratio is less than 1, which means that we need more HA, or NH4Cl, than A-, or NH3, to make the solution.

Step 3: Calculate the concentration of NH3 and NH4Cl

Now, since NH3 is the solvent, we can assume its initial concentration to be 1M and NH4Cl to be xM. We can use the ratio obtained in step 2 to write: x / 1 = 0.000562. Solving for x, we get x = 1 * 0.000562 = 0.000562 M.

Step 4: Calculate the amount of NH4Cl in grams

Now we can calculate the amount of NH4Cl in grams needed. First, convert the volume of solution required from milliliters to liters. 100 mL = 0.1 L. Then calculating the mass = Molarity * volume * molecular weight. For NH4Cl, Molarity = 0.000562 M, volume= 0.1 L, and molecular weight of NH4Cl = 53.5 g/mol. Solving the above equation gives the mass of NH4Cl in g = 0.000562 * 0.1 * 53.5 =~ 0.003 g.

Key Concepts

Acid-Base Chemistry

In acid-base chemistry, we classify substances based on how they interact with water. Acids donate protons (\( H^+ \)) to water, while bases accept protons from water. This interaction affects the pH of the solution, which is a measure of its acidity or basicity.
The pH scale runs from 0 to 14:

• A pH less than 7 indicates an acidic solution.
• A pH of 7 is neutral, as is pure water.
• A pH greater than 7 indicates a basic/alkaline solution.

To create a solution with a desired pH, one needs to select an appropriate solute. In the provided exercise, the goal was to achieve a pH of 5.50, which is slightly acidic. This requires choosing a solute that, upon dissolving, releases ions to slightly increase the concentration of hydrogen ions (\( H^+ \)), thereby lowering the pH from 7 to 5.50. When selecting solutes, it's important to consider whether they act as acids or bases when dissolved.

Solution Preparation

Solution preparation involves dissolving a specific solute into a solvent to achieve the desired concentration and properties, such as pH. In this exercise, the solution needs to have a pH of 5.50, thereby requiring careful selection and measurement of the solute.
Preparation steps include:

• Choosing the correct solute: For the desired pH, ammonium chloride (\( NH_4Cl \)) is suitable because it forms a solution with a pH less than 7.
• Calculating the required concentration: Use the formula for buffer solutions: \[ \text{pH} = \text{p}K_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
• Determining the amount of solute needed: Convert the calculated molarity of \( NH_4Cl \) into grams using its molar mass.
• Dissolving the calculated amount into the solvent volume: In this case, 100 mL of water.

This careful preparation ensures the final solution will have the desired chemical properties, particularly the correct pH.

Buffer Solutions

Buffer solutions are special mixtures that maintain a relatively constant pH, even when small amounts of acids or bases are added. They are crucial in many biochemical applications where maintaining pH is important for stability and reaction rates.
A buffer is typically made from a weak acid and its conjugate base or vice versa. In the exercise, we use \( NH_4Cl \) and \( NH_3 \). Here:

• \( NH_4^+ \) acts as the weak acid.
• \( NH_3 \) serves as its conjugate base.

Together, they create a buffer that stabilizes the pH around their pKa value. The Henderson-Hasselbalch equation, \[ \text{pH} = \text{p}K_a + \log\left(\frac{[A^-]}{[HA]}\right) \], is essential for calculating the pH of buffer solutions. The balance between these components determines the pH, allowing for adjustments based on the desired pH level.