Problem 22 - Chemistry: The Central Science

Question

(a) Calculate the \(\mathrm{pH}\) of a buffer that is \(0.105 \mathrm{M}\) in \(\mathrm{NaHCO}_{3}\) and \(0.125 \mathrm{M}\) in \(\mathrm{Na}_{2} \mathrm{CO}_{3} .\) (b) Calculate the \(\mathrm{pH}\) of a solution formed by mixing \(65 \mathrm{~mL}\) of \(0.20 \mathrm{M}\) \(\mathrm{NaHCO}_{3}\) with \(75 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\)

Step-by-Step Solution

Answer

For part (a), the pH of the given buffer is approximately 10.594. For part (b), the pH of the mixed solution is approximately 10.140.

Step 1: (a) Calculate the pH of the given buffer

To calculate the pH of the buffer, we will use the Henderson-Hasselbalch equation: pH = pKa + log(\(\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\)) Here, the weak acid is HCO₃⁻ (bicarbonate) and its conjugate base is CO₃²⁻ (carbonate). The pKa value for the HCO₃⁻/CO₃²⁻ pair is 10.33. We are given the concentration of bicarbonate (0.105 M) and carbonate (0.125 M) ions in the buffer solution. Now we can plug these values into the equation: pH = 10.33 + log(\(\frac{0.125}{0.105}\))

Step 2: (a) Calculate the log term and resulting pH

First, we will calculate the value of the log term in the equation: log(\(\frac{0.125}{0.105}\)) ≈ 0.264 Now, add the pKa value: pH = 10.33 + 0.264 ≈ 10.594 So, the pH of the given buffer is approximately 10.594.

Step 3: (b) Calculate the concentrations after mixing the solutions

To determine the pH of the resulting solution after mixing, we first need to calculate the new concentrations of HCO₃⁻ (bicarbonate) and CO₃²⁻ (carbonate). The initial amounts of each ion are: \[n_{HCO_3^-} = C_{HCO_3^-} \times V_{HCO_3^-} = (0.20 \; M) \times (65 \; mL)\] and \[n_{CO_3^{2-}} = C_{CO_3^{2-}} \times V_{CO_3^{2-}} = (0.15 \; M) \times (75 \; mL)\]

Step 4: (b) Calculate the total volume and mole amounts

The total volume of the mixed solution is \(65 \; mL + 75 \; mL = 140 \; mL\). Now, we can calculate the new concentrations of HCO₃⁻ and CO₃²⁻ in the mixed solution: \[[HCO_3^-]_{new} = \frac{(0.20 \; M)(65 \; mL)}{140 \; mL} \] \[[CO_3^{2-}]_{new} = \frac{(0.15 \; M)(75 \; mL)}{140 \; mL} \]

Step 5: (b) Calculate the new concentrations and apply the Henderson-Hasselbalch equation

Now, let's calculate these concentration values: \[[HCO_3^-]_{new} \approx 0.093 \; M\] \[[CO_3^{2-}]_{new} \approx 0.080 \; M\] With these new concentrations of HCO₃⁻ and CO₃²⁻, we can use the Henderson-Hasselbalch equation to calculate the pH of the mixed solution: pH = 10.33 + log(\(\frac{0.080}{0.093}\))

Step 6: (b) Calculate the resulting pH of the mixed solution

First, we will calculate the value of the log term: log(\(\frac{0.080}{0.093}\)) ≈ -0.190 Now, add the pKa value: pH = 10.33 - 0.190 ≈ 10.140 So, the pH of the mixed solution is approximately 10.140.

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a fundamental tool in understanding the pH of buffer solutions. It's derived from the acid dissociation constant equation and relates the pH, the pKₐ (acid dissociation constant), and the ratio of the concentration of the salt form (A⁻) to the acid form (HA). The equation is:\[pH = pK_a + \log \left(\frac{[A^-]}{[HA]}\right) \]This relation is incredibly useful in buffer chemistry where you have a weak acid and its conjugate base present. The buffer system resists changes in pH when small amounts of acids or bases are added. In this particular exercise, bicarbonate (HCO₃⁻) acts as the weak acid, and carbonate (CO₃²⁻) as its conjugate base. Using the equation requires knowing the concentrations of acid and base, plus the pKₐ value related to them. Plug these into the equation, and you can easily find the pH of the buffer solution.

Bicarbonate-Carbonate Buffer System

Buffer systems are crucial for maintaining the pH levels in various chemical and biological contexts. The bicarbonate-carbonate buffer system is prominent because of its role in processes like blood's pH regulation. This particular system involves the equilibrium between bicarbonate ions (HCO₃⁻) and carbonate ions (CO₃²⁻). When there is an increase in H⁺ ions (making the solution more acidic), the bicarbonate reacts to neutralize them, resisting a pH drop. Similarly, when OH⁻ ions increase (more basic solution), carbonate ions capture them to form bicarbonate, again resisting pH changes. In the given problem, we calculate how such a buffer resists changes when mixed in different concentrations. By using the initial conditions provided, the adjustment of the bicarbonate and carbonate ion concentrations, we explored how this buffer minimized the change in pH.

pKa Calculation

Understanding pKₐ is essential when using the Henderson-Hasselbalch equation. pKₐ is the negative log of the acid dissociation constant (Kₐ), reflecting a weak acid's strength. The lower the pKₐ, the stronger the acid, as it more readily donates protons in solution. In the bicarbonate-carbonate buffer system, the specific pKₐ value is crucial in determining how the system behaves. For the bicarbonate (HCO₃⁻) to carbonate (CO₃²⁻) transition, the pKₐ value utilized is 10.33. This value was used in our exercise to calculate the buffer pH and see how well the system maintains its pH despite changing concentrations of the buffer components. Knowing and using the correct pKₐ enables precise control when predicting or adjusting the behavior of buffer solutions in different contexts.