Question

Use information from Appendix \(D\) to calculate the pH of (a) a solution that is \(0.250 \mathrm{M}\) in sodium formate \((\mathrm{HCOONa})\) and \(0.100 M\) in formic acid \((\mathrm{HCOOH}) ;\) (b) a solution that is \(0.510 \mathrm{M}\) in pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\) and \(0.450 \mathrm{M}\) in pyridinium chloride \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHCl}\right) ;\) (c) a solution that is made by combining \(55 \mathrm{~mL}\) of \(0.050 \mathrm{M}\) hydrofluoric acid with \(125 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) sodium fluoride.

Short answer

For solution (a), the pH is approximately 4.24. For solution (b), the pH is approximately 5.40. For solution (c), the pH is approximately 3.78.

Step-by-step solution

Identify the concentrations of the acid and its conjugate base

The concentrations of the acid (formic acid, \(\mathrm{HCOOH}\)) and its conjugate base (sodium formate, \(\mathrm{HCOONa}\)) are given as \(0.150\,\mathrm{M}\) and \(0.250\,\mathrm{M}\), respectively.

Use the Henderson-Hasselbalch equation

For a buffer solution, we can use the Henderson-Hasselbalch equation, which is given as follows: \(pH = pK_a + \log_{10} \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\) According to Appendix D, the \(pK_a\) value for formic acid is 3.75. Now we can substitute the values into the equation: \(pH = 3.75 + \log_{10} \frac{0.250}{0.150}\)

Calculate the pH

Now we can perform the calculation: \(pH = 3.75 + \log_{10} \frac{0.250}{0.150} \approx 4.24\) Thus, the pH of solution (a) is approximately 4.24. ###Solution (b)###

Identify the concentrations of the base and its conjugate acid

The concentrations of the base (pyridine, \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N}\)) and its conjugate acid (pyridinium chloride, \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{NHCl}\)) are given as \(0.490\,\mathrm{M}\) and \(0.450\,\mathrm{M}\), respectively.

Use the Henderson-Hasselbalch equation

For a buffer solution, we can use the Henderson-Hasselbalch equation. Since this is a basic buffer solution, we need to use the \(pK_b\) value for pyridine. According to Appendix D, the \(pK_b\) value for pyridine is 8.75. First, we need to find the \(pK_a\) value using the relationship: \(pK_a + pK_b = 14\) \(pK_a = 14 - pK_b = 14 - 8.75 = 5.25\) Now we can substitute the values into the equation: \(pH = 5.25 + \log_{10} \frac{0.490}{0.450}\)

Calculate the pH

Now we can perform the calculation: \(pH = 5.25 + \log_{10} \frac{0.490}{0.450} \approx 5.40\) Thus, the pH of solution (b) is approximately 5.40. ###Solution (c)###

Calculate the concentrations of the acid and its conjugate base

Since we are given the volumes and concentrations of the solutions, we can first calculate the moles of the acid (\(\mathrm{HF}\)) and its conjugate base (\(\mathrm{F}^{-}\)): moles of \(\mathrm{HF} = 0.050\,\mathrm{M} \times 0.080\,\mathrm{L} = 0.004\,\mathrm{mol}\) moles of \(\mathrm{F}^{-} = 0.10\,\mathrm{M} \times 0.100\,\mathrm{L} = 0.010\,\mathrm{mol}\) Now that we have the moles, we need to find the concentrations in the final mixed solution. To do this, we first calculate the total volume of the solution: \(V_{\text{total}} = 80\,\mathrm{mL} + 100\,\mathrm{mL} = 180\,\mathrm{mL} = 0.180\,\mathrm{L}\) Now we can calculate the concentrations: \([\mathrm{HF}] = \frac{0.004\,\mathrm{mol}}{0.180\,\mathrm{L}} = 0.0222\,\mathrm{M}\) \([\mathrm{F}^{-}] = \frac{0.010\,\mathrm{mol}}{0.180\,\mathrm{L}} = 0.0556\,\mathrm{M}\)

Use the Henderson-Hasselbalch equation

Now that we have the concentrations, we can use the Henderson-Hasselbalch equation. According to Appendix D, the \(pK_a\) value for hydrofluoric acid is 3.17: \(pH = 3.17 + \log_{10} \frac{0.0556}{0.0222}\)

Calculate the pH

Now we can perform the calculation: \(pH = 3.17 + \log_{10} \frac{0.0556}{0.0222} \approx 3.78\) Thus, the pH of solution (c) is approximately 3.78.

Key concepts

Buffer Solutions

Buffer solutions are fascinating mixtures in chemistry that help resist changes in pH. Imagine them like shock absorbers. They maintain the pH levels stable, even when acids or bases are added. This property is crucial in many applications, from biological systems to industrial processes.

A buffer solution usually comprises a weak acid and its conjugate base, or a weak base and its conjugate acid.

• For example, a solution with formic acid (\( \text{HCOOH} \)) and sodium formate (\( \text{HCOONa} \)) acts as a buffer.
• If an acid is added, the conjugate base will neutralize it.
• Conversely, if a base is added, the weak acid will help in minimizing the change in pH.

This equilibrium between the weak acid and its base ensures the solution can handle external changes efficiently, making it an essential concept in acid-base chemistry.

pH Calculations

Calculating the pH of solutions gives powerful insight into their acidity or basicity. We use the Henderson-Hasselbalch equation for this task, especially for buffer solutions.

The equation is expressed as follows:\[pH = pK_a + \log_{10} \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)\]Here is what the equation means:

• \( pK_a \) stands for the acid dissociation constant, indicating the equilibrium point.
• The \([\text{A}^-]\) represents the concentration of the conjugate base.
• The \([\text{HA}]\) is the concentration of the acid.

To find the pH, simply plug the concentrations into the equation after determining \( pK_a \) from reliable data sources like appendix tables. By solving this equation, you can effectively understand how acidic or basic a buffer solution is.

Acid-Base Chemistry

Acid-base chemistry sits at the heart of understanding various chemical processes. It revolves around the concept of acids donating protons (H\( ^+ \)) and bases accepting them.

Here's the general breakdown:

• Acids are substances like hydrochloric acid (\( \text{HCl} \)) that donate protons.
• Bases, such as sodium hydroxide (\( \text{NaOH} \)), accept protons or donate hydroxide ions \( OH^- \).
• In a buffer system, weak acids and bases are in careful balance, maintaining the pH.

Understanding the interaction between acids and bases forms the basis for manipulating chemical reactions, analyzing buffer capacities, and predicting the outcomes of reactions in which these entities are involved.