Problem 12 - Chemistry: Principles and Reactions

Question

A buffer is prepared by dissolving \(0.037\) mol of potassium fluoride in \(135 \mathrm{~mL}\) of \(0.0237 \mathrm{M}\) hydrofluoric acid. Assume no volume change after KF is dissolved. Calculate the \(\mathrm{pH}\) of this buffer.

Step-by-Step Solution

Answer

Answer: No, we cannot accurately calculate the pH of the buffer solution with the given data as the initial moles of hydrofluoric acid have been completely neutralized by the potassium fluoride, leaving only F- ions and no HF in the buffer solution. The given data is incorrect to calculate the pH accurately.

Step 1: Calculate the initial moles of hydrofluoric acid (HF)

To find the initial moles of HF, we can use the formula: moles = molarity × volume. In this case, the molarity of HF is given as \(0.0237 \mathrm{M}\), and the volume is given as \(135 \mathrm{~mL}\). Convert the volume to liters by dividing by 1000: \(135 \mathrm{~mL} = 0.135 \mathrm{~L}\). Now calculate the initial moles of HF: moles = \((0.0237 \mathrm{M}) \times (0.135 \mathrm{~L}) = 0.0031995 \mathrm{~mol}\).

Step 2: Calculate the moles of F- ions from KF

The moles of KF are given as \(0.037 \mathrm{~mol}\). Since one molecule of KF dissociates into one fluoride ion (F-) and one potassium ion (K+), the moles of F- ions in the buffer solution will also be \(0.037 \mathrm{~mol}\).

Step 3: Calculate the moles of HF after KF dissolves

When the potassium fluoride dissolves, it reacts with the hydrofluoric acid (HF) to form potassium hydrogen fluoride (KHF) and potassium ions (K+). Therefore, the moles of HF after KF dissolves will be the difference between the initial moles of HF and the moles of F- ions from KF: moles = \((0.0031995 \mathrm{~mol}) - (0.037 \mathrm{~mol}) = -0.0338005 \mathrm{~mol}\). As we cannot have a negative quantity of moles, the moles of HF after KF dissolution will be 0.

Step 4: Calculate the concentration of F- and HF in the buffer solution

Now that we know the moles of F- and HF in the buffer solution, we can calculate their concentrations. Since we assumed no volume change after KF dissolves, the total volume of the buffer solution remains constant at \(0.135 \mathrm{~L}\). Concentration of F- = \( \frac{0.037 \mathrm{~mol}}{0.135 \mathrm{~L}} = 0.274 \mathrm{M}\). Concentration of HF = \( \frac{0 \mathrm{~mol}}{0.135 \mathrm{~L}} = 0 \mathrm{M}\).

Step 5: Use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution

The Henderson-Hasselbalch equation is given as: \(\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log{\left(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\right)}\), where pKa is the dissociation constant for the acid, \([\mathrm{A}^-]\) is the concentration of the conjugate base, and \([\mathrm{HA}]\) is the concentration of the acid. For hydrofluoric acid, the value of pKa is \(3.17\). Using the concentrations calculated in step 4, we substitute the values into the equation: \(\mathrm{pH} = 3.17 + \log{\left(\frac{0.274 \mathrm{M}}{0 \mathrm{M}}\right)}\). In this case, the ratio is undefined, as we cannot divide by zero. Therefore, the hydrofluoric acid has been completely neutralized by the addition of the potassium fluoride, leaving only F- ions and no HF in the buffer solution. In this situation, we should calculate the pH based on the concentration of F- ions. To do this, we can use the formula: \(\mathrm{pOH} = -\log{[\mathrm{OH}^-]}\). Then, we can convert to pH using: \(\mathrm{pH} = 14 - \mathrm{pOH}\). However, this will not give an accurate answer since, in reality, there would still be some HF present in the solution as it is not fully neutralized by KF. We can conclude that given data is incorrect to calculate the pH of the buffer solution accurately.

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a useful tool for calculating the pH of buffer solutions. A buffer solution consists of a weak acid and its conjugate base or a weak base and its conjugate acid. This equation helps us understand how the pH is dependent on these components.

The formula is expressed as:\[\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log{\left(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]\right)}\]where:

• \(\mathrm{pK}_{\mathrm{a}}\) is the negative logarithm of the acid dissociation constant, \(\mathrm{K}_{\mathrm{a}}\), reflecting the strength of the weak acid.
• \([\mathrm{A}^-]\) represents the concentration of the conjugate base.
• \([\mathrm{HA}]\) is the concentration of the weak acid.

Using this equation requires knowing the concentrations of both the weak acid and its conjugate base. When applying this equation, if the concentration of the acid becomes zero (as seen in the problem), the solution is no longer a buffer, and other methods must be used to find the pH.

Hydrofluoric Acid

Hydrofluoric acid, often abbreviated as HF, is a weak acid known for its ability to dissolve many materials, especially oxides and glasses. This property makes HF useful in industrial and laboratory settings, yet it requires careful handling due to its corrosive nature.

As a weak acid, HF doesn't completely dissociate in water, distinguishing it from strong acids. Hence, in buffer solutions, HF often partially dissociates, contributing to the buffer's ability to maintain a stable pH.

• Its dissociation is represented by the chemical equation: \(\mathrm{HF} \rightleftharpoons \mathrm{H}^+ + \mathrm{F}^-\)
• The dissociation constant \(\mathrm{K}_a\) for HF is relatively small, which corresponds to a higher \(\mathrm{pK}_a\) value (3.17 in our case).

When calculating buffer pH using HF, consider how it behaves in the solution. As described in the solution, if a compound like potassium fluoride is added, it can neutralize HF, removing its buffering capabilities.

Potassium Fluoride

Potassium fluoride (KF) is a salt that, when dissolved in water, dissociates into potassium ions (\(\mathrm{K}^+\)) and fluoride ions (\(\mathrm{F}^-\)). It is often used in buffer solutions containing hydrofluoric acid (HF) due to its ability to shift the equilibrium of HF dissociation.

The role of KF in buffer solutions includes:

• Providing fluoride ions, \(\mathrm{F}^-\), the conjugate base that pairs with HF in the buffer system.
• Reacting with HF to form additional \(\mathrm{F}^-\) ions, which affects the overall pH calculation.

In our example, KF was added to a solution of hydrofluoric acid to form a buffer. However, it neutralized all HF, leaving F- ions predominant, thus disrupting the buffering action. It's important to understand the equilibrium between acid and base in buffers can be affected by excessive additions of either component, leading as in this case, to the limitation of the Henderson-Hasselbalch equation when concentrations fall outside expected ranges.