Problem 22 - Chemistry: The Central Science in SI Units, Global Edition

Question

(a) Calculate the pH of a buffer that is \(0.125 \mathrm{M}\) in \(\mathrm{NaHCO}_{3}\) and \(0.095 \mathrm{M}\) in \(\mathrm{Na}_{2} \mathrm{CO}_{3} .\) (b) Calculate the pH of a solution formed by mixing \(25 \mathrm{~mL}\) of \(0.25 \mathrm{M} \mathrm{NaHCO}_{3}\) with \(75 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\)

Step-by-Step Solution

Answer

The pH of the buffer solution in part (a) is approximately \(6.12\). In part (b), after mixing the solutions, the pH of the resulting solution is approximately \(6.33\).

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

In this part, we have a solution that is \(0.125 \, \mathrm{M}\) in \(\mathrm{NaHCO}_{3}\) and \(0.095\, \mathrm{M}\) in \(\mathrm{Na}_{2}\mathrm{CO}_{3}\). The \(\mathrm{NaHCO}_{3}\) is the weak acid, and \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) is the conjugate base. The acid dissociation constant (\(K_a\)) for \(\mathrm{HCO}_{3}^{-}\) can be found in a reference: \(K_a = 4.3 \times 10^{-7}\) First, we will find the \(pK_a\) using the given \(K_a\): \[pK_a = -\log(K_a) = -\log(4.3 \times 10^{-7})\] Now use the Henderson-Hasselbalch equation to find the pH of the buffer: \[pH = pK_a + \log(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]})\] Plug in the values for \(pK_a\) and the concentrations: \[pH = -\log(4.3 \times 10^{-7}) + \log(\frac{0.095}{0.125})\] Calculate the pH: \[pH \approx 6.12\]

Step 2: Part (b): Calculate the pH of a solution formed by mixing two solutions.

In this part, we have to calculate the pH of the solution formed by mixing \(25 \, \mathrm{mL}\) of \(0.25\, \mathrm{M} \, \mathrm{NaHCO}_{3}\) with \(75\, \mathrm{mL}\) of \(0.15\, \mathrm{M}\, \mathrm{Na}_{2}\mathrm{CO}_{3}\). First, we will find the new concentrations of \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) in the mixed solution: Initial moles of \(\mathrm{NaHCO}_3\): moles = (volume in L) × (molarity) \[\text{moles of }\mathrm{NaHCO}_{3} =0.025\, \text{L} \times 0.25\, \mathrm{M} = 0.00625 \, \text{mol}\] Initial moles of \(\mathrm{Na}_{2}\mathrm{CO}_3\): \[\text{moles of }\mathrm{Na}_{2}\mathrm{CO}_{3} = 0.075\, \text{L} \times 0.15\, \mathrm{M} = 0.01125\, \text{mol}\] Total volume of the mixture: \[V_{total} = 0.025\, \text{L} + 0.075\, \text{L} = 0.1\, \text{L}\] New concentrations: \[[\mathrm{HCO}_{3}^{-}] =\frac{0.00625\, \text{mol}}{0.1\, \text{L}} = 0.0625 \, \mathrm{M}\] \[[\mathrm{CO}_{3}^{2-}] =\frac{0.01125\, \text{mol}}{0.1\, \text{L}} = 0.1125\, \mathrm{M}\] Now, we can use the Henderson-Hasselbalch equation again, plugging in the new concentrations: \[pH = pK_a + \log(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]})\] \[pH = -\log(4.3 \times 10^{-7}) + \log(\frac{0.1125}{0.0625})\] Calculate the pH: \[pH \approx 6.33\]

Key Concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a critical formula for calculating the pH of buffer solutions. By using this equation, we can easily find the pH of a solution containing a weak acid and its conjugate base. The formula is expressed as:\[pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right)\]- Here, \([A^-]\) represents the concentration of the conjugate base.- \([HA]\) stands for the concentration of the weak acid.- The term \(pK_a\) is the negative logarithm of the acid dissociation constant \(K_a\).This equation assumes that the concentrations of the weak acid and its conjugate base do not change significantly when the solution is diluted. One of its advantages is its simplicity, making it a favorite among students studying chemistry. The Henderson-Hasselbalch equation is especially useful because it allows for straightforward calculations when determining how the addition of acids or bases impacts the pH of a buffer system.

Acid Dissociation Constant

The acid dissociation constant, often denoted as \(K_a\), is a vital concept in understanding the behavior of weak acids in solution. It describes the extent to which a weak acid dissociates into its ions. In mathematical terms:\[K_a = \frac{[H^+][A^-]}{[HA]}\]Where:- \([H^+]\) is the concentration of hydrogen ions.- \([A^-]\) is the concentration of the conjugate base.- \([HA]\) represents the concentration of the undissociated acid.A smaller \(K_a\) value indicates a weaker acid because it signifies that less of the acid dissociates in solution. Conversely, a larger \(K_a\) value demonstrates a stronger tendency to dissociate. The magnitude of \(K_a\) can also be used to determine the acid's \(pK_a\) by taking its negative logarithm: \(pK_a = -\log(K_a)\). This \(pK_a\) value is essential for pH calculations in buffers, especially when using the Henderson-Hasselbalch equation.

Molarity Concentration Calculation

Molarity is a way to express the concentration of a solute in a solution. It's defined as the number of moles of a solute per liter of solution, represented by the equation:\[\text{Molarity} (M) = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\]This calculation is essential to understand how much solute is present in a given volume of solution and is crucial when mixing different solutions. For instance, when calculating the pH of a buffer solution resulting from mixed solutions, it's necessary to determine the new concentrations of the acid and base components. This involves:

• Calculating the initial moles of each component from their initial molarity and volume.
• Adding the volumes together to get the total volume of the solution.
• Dividing the number of moles of each component by the total volume to find their new molarity.

Understanding molarity helps in precisely calculating changes in concentrations after mixing solutions, which is fundamental for accurate pH calculations in various chemical processes.