Question

A buffer solution is prepared by mixing \(75.0 \mathrm{~mL}\) of \(0.275 \mathrm{M}\) fluorobenzoic acid \(\left(\mathrm{C}_{7} \mathrm{H}_{3} \mathrm{O}_{2} \mathrm{~F}\right)\) with \(55.0 \mathrm{~mL}\) of \(0.472 \mathrm{M}\) sodium fluorobenzoate. The \(\mathrm{p} K_{\mathrm{a}}\) of this weak acid is \(2.90\). What is the \(\mathrm{pH}\) of the buffer solution?

Short answer

After performing the calculations, the pH of the buffer solution can be determined as follows: \[ \mathrm{pH} = 2.90 + \log{\frac{\frac{55.0 \mathrm{~mL} \times 0.472 \mathrm{M}}{130.0 \mathrm{~mL}}}{\frac{75.0 \mathrm{~mL} \times 0.275 \mathrm{M}}{130.0 \mathrm{~mL}}}} \] By solving this equation, we find the pH of the buffer solution to be approximately 2.99.

Step-by-step solution

Calculate the moles of weak acid (HA) and conjugate base (A-)

To find the moles of fluorobenzoic acid (HA) and sodium fluorobenzoate (A-), we can multiply the initial volume by the initial concentration: Moles of HA = Volume of HA × Concentration of HA Moles of HA = \(75.0 \mathrm{~mL} \times 0.275 \mathrm{M}\) Moles of A- = Volume of A- × Concentration of A- Moles of A- = \(55.0 \mathrm{~mL} \times 0.472 \mathrm{M}\)

Calculate the total volume of the solution

Total Volume = Volume of HA + Volume of A- Total Volume = \(75.0 \mathrm{~mL} + 55.0 \mathrm{~mL}\)

Calculate the final concentrations of HA and A- in the buffer solution

To find the final concentrations of HA and A-, we will divide the moles of each by the total volume of the solution (in liters): Final concentration of HA = Moles of HA / Total Volume Final concentration of HA = \(\frac{75.0 \mathrm{~mL} \times 0.275 \mathrm{M}}{130.0 \mathrm{~mL}}\) Final concentration of A- = Moles of A- / Total Volume Final concentration of A- = \(\frac{55.0 \mathrm{~mL} \times 0.472 \mathrm{M}}{130.0 \mathrm{~mL}}\)

Apply the Henderson-Hasselbalch equation

Now we can apply the Henderson-Hasselbalch equation to find the pH of the buffer solution: \[\mathrm{pH} = \mathrm{p} K_{a} + \log{\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}}\] Where, - \(\mathrm{p} K_{a}=2.90\) - \([\mathrm{HA}]=\frac{75.0 \mathrm{~mL} \times 0.275 \mathrm{M}}{130.0 \mathrm{~mL}}\) - \([\mathrm{A}^{-}]=\frac{55.0 \mathrm{~mL} \times 0.472 \mathrm{M}}{130.0 \mathrm{~mL}}\) \[ \mathrm{pH} = 2.90 + \log{\frac{\frac{55.0 \mathrm{~mL} \times 0.472 \mathrm{M}}{130.0 \mathrm{~mL}}}{\frac{75.0 \mathrm{~mL} \times 0.275 \mathrm{M}}{130.0 \mathrm{~mL}}}} \]

Calculate the pH

Now, we can plug in the values we determined for the concentrations of A- and HA into the Henderson-Hasselbalch equation and solve for the pH. \[ \mathrm{pH} = 2.90 + \log{\frac{\frac{55.0 \mathrm{~mL} \times 0.472 \mathrm{M}}{130.0 \mathrm{~mL}}}{\frac{75.0 \mathrm{~mL} \times 0.275 \mathrm{M}}{130.0 \mathrm{~mL}}}} \] After performing these calculations, we will find the pH of the buffer solution.

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a formula used to calculate the pH of a buffer solution. A buffer solution consists of a weak acid and its conjugate base, which helps resist changes in pH. The equation is written as: \[ \mathrm{pH} = \mathrm{p} K_{a} + \log{\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}} \] where \( [\mathrm{A}^{-}] \) is the concentration of the conjugate base, and \( [\mathrm{HA}] \) is the concentration of the weak acid.
This formula is a practical application of the Ka expression in logarithmic form, making it easier to calculate pH, especially for buffers or mixtures of weak acids with their salts.

• \( \mathrm{p} K_a \) is the negative logarithm of the acid dissociation constant (Ka) and is unique to each weak acid.
• It simplifies calculations by letting us focus on the ratio \( \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]} \) rather than extensive calculations with molar concentrations.

The equation allows us to quickly estimate the pH if we know the concentrations of the acid and its conjugate base and understand the inherent strength of the acid through \( \mathrm{p} K_a \).

Weak acid

A weak acid is an acid that partially dissociates into its ions in an aqueous solution. Unlike strong acids, which dissociate completely, weak acids have a significant portion of molecules remaining intact. This partial dissociation gives weak acids unique properties when forming buffer solutions.
In the exercise example, fluorobenzoic acid serves as the weak acid, with a given \( \mathrm{p} K_a \) of 2.90. Here are some key points to understand about weak acids:

• They establish equilibrium between the undissociated acid and its ions in solution.
• The degree of ionization depends on the concentration of the acid and can vary.
• The smaller the value of \( \mathrm{p} K_a \), the stronger the weak acid since it indicates a larger Ka value (more dissociation).

This attribute is important for buffer solutions, as the weak acid can neutralize added bases to maintain a stable pH.

Conjugate base

The conjugate base is the species formed when a weak acid donates a proton (H+) to a base. Together, a weak acid and its conjugate base form a conjugate acid-base pair that helps buffer solutions resist pH changes by reacting with added acids or bases.
For our exercise, sodium fluorobenzoate acts as the conjugate base of fluorobenzoic acid. Here are some key points about conjugate bases:

• They are part of a buffer's ability to consume added H+ ions by converting back into the weak acid.
• Formed as a result of the acid's partial dissociation in water.
• The concentrations of the weak acid and its conjugate base determine the buffer's capacity and effective pH range.

The interaction between the weak acid and its conjugate base is crucial for controlling the pH of the buffer.

pH calculation

pH calculation in a buffer solution involves determining the precise balance between the weak acid and its conjugate base, allowing the solution to maintain its pH. The process generally uses the Henderson-Hasselbalch equation for its simplicity and reliability.
In the exercise provided, we are asked to find the pH of a solution combining fluorobenzoic acid and sodium fluorobenzoate. Here's how the process works in general terms:

• First, calculate the moles of the weak acid and the conjugate base using their volumes and molarities.
• Determine the total volume of the solution to find the final concentrations of the acid and base.
• Use the Henderson-Hasselbalch equation to compute the pH by substituting the values into the formula: \[ \mathrm{pH} = \mathrm{p} K_{a} + \log{\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}} \]

The precise pH will depend on this balance and is vital in biological and chemical systems to ensure the right conditions for reactions and stability.