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 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

The pH of the buffer solution in Part (a) consisting of 0.105 M NaHCO3 and 0.125 M Na2CO3 is 6.15. The pH of the mixed solution in Part (b) with 65 mL of 0.20 M NaHCO3 and 75 mL of 0.15 M Na2CO3 is 6.56.

Step 1: Find the relevant pK_a

The buffer solution consists of the bicarbonate (\(\mathrm{HCO}_{3}^{-}\)) conjugate base and carbonic acid (\(\mathrm{H}_{2}\mathrm{CO}_{3}\)) weak acid pair. Find the dissociation constant, \(K_a\), for carbonic acid and then calculate its \(pK_a\), which will be used in the Henderson-Hasselbalch equation. For carbonic acid, \(K_a = 4.3 \times 10^{-7}\), so \(pK_a = -\log(K_a) = 6.37\).

Step 2: Plug concentrations into the Henderson-Hasselbalch equation

We are given the following concentrations: \(\mathrm{[NaHCO}_{3}] = 0.105\, \mathrm{M}\) (acts as the conjugate base, \([\mathrm{A}^{-}]\), in the equation) \(\mathrm{[Na}_{2}\mathrm{CO}_{3}] = 0.125\, \mathrm{M}\) (acts as the weak acid, \([\mathrm{HA}]\), in the equation) Plug them into the Henderson-Hasselbalch equation: \[pH = 6.37 + \log \frac{0.105}{0.125}\]

Step 3: Calculate pH

Perform the calculations: \[pH = 6.37 + \log \frac{0.105}{0.125} = 6.37 - 0.221 = 6.15\] So, the pH of the buffer solution in Part (a) is 6.15. Part (b):

Step 1: Find new molar concentrations

Calculate the new molar concentrations of the two components after mixing the solutions: \[\mathrm{[NaHCO}_{3}] = \frac{65\, \mathrm{mL} \times 0.20\, \mathrm{M}}{65\, \mathrm{mL} + 75\, \mathrm{mL}} = \frac{13\, \mathrm{mmol}}{140\, \mathrm{mL}} = 0.093\, \mathrm{M}\] \[\mathrm{[Na}_{2}\mathrm{CO}_{3}] = \frac{75\, \mathrm{mL} \times 0.15\, \mathrm{M}}{65\, \mathrm{mL} + 75\, \mathrm{mL}} = \frac{11.25\, \mathrm{mmol}}{140\, \mathrm{mL}} = 0.080\, \mathrm{M}\]

Step 2: Plug concentrations into the Henderson-Hasselbalch equation

Use the same \(pK_a\)-value from Part (a) and plug the new concentrations into the Henderson-Hasselbalch equation: \[pH = 6.37 + \log \frac{0.093}{0.080}\]

Step 3: Calculate pH

Perform the calculations: \[pH = 6.37 + \log \frac{0.093}{0.080} = 6.37 + 0.189 = 6.56\] So, the pH of the solution in Part (b) is 6.56.

Key Concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a vital formula in chemistry for calculating the pH of buffer solutions. It is expressed as:\[ pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right) \]Where:

• \(pH\) is the measure of acidity or basicity of the solution.
• \(pK_a\) is the negative logarithm of the acid dissociation constant \(K_a\), representing the strength of a weak acid.
• \([A^-]\) is the concentration of the conjugate base.
• \([HA]\) is the concentration of the weak acid.

This equation originates from rearranging the acid dissociation equilibrium expressions. It's particularly useful because it incorporates both the acid and its conjugate base, allowing precise pH calculation in weak acid and base pair solutions. Such functionality makes it crucial for understanding buffer systems and maintaining specific pH levels, like in biological systems where enzymes operate within narrow pH ranges.

pH calculation

The calculation of pH is fundamental in understanding the acidity or basicity of a solution. The pH is calculated using the formula:\[ pH = -\log [H^+] \]Where \([H^+]\) is the hydrogen ion concentration in mol/L. However, when dealing with buffer solutions, we often use the Henderson-Hasselbalch equation for more complicated systems, as it directly relates the concentrations of acid and conjugate base in a practical formula.In the context of buffer solutions, calculating pH involves:

• Identifying the concentrations of both the weak acid and its conjugate base.
• Using the acid dissociation constant \(K_a\) to find \(pK_a\).
• Plugging these values into the Henderson-Hasselbalch equation.

The exercise demonstrated clear steps with numerical examples. First, the relevant \(pK_a\) value was deduced from \(K_a\) using \(pK_a = -\log(K_a)\), a critical first step for accurate pH computation.Understanding pH calculations is essential because the pH level influences many chemical reactions and processes, including those in industrial and biological applications.

Bicarbonate buffer system

The bicarbonate buffer system is one of the most significant physiological buffer systems. It maintains the pH balance in blood and other bodily fluids. It primarily consists of carbonic acid (\(H_2CO_3\)), a weak acid, and bicarbonate ions (\(HCO_3^-\)), its conjugate base.In a bicarbonate buffer system:

• Carbonic acid dissociates into hydrogen (\(H^+\)) and bicarbonate ions.
• This equilibrium helps resist drastic changes in pH.

The system works efficiently by utilizing the reversible reaction \(H_2CO_3 \rightleftharpoons H^+ + HCO_3^-\). When excess hydrogen ions \((H^+)\) are introduced, they combine with bicarbonate ions to form more carbonic acid, minimizing the pH change. Conversely, when there's an excess of hydroxide ions \((OH^-)\), carbonic acid dissociates to neutralize the base.This buffer system is not only critical in our bodies but also widely used in laboratory applications and industrial processes where maintaining a stable pH is necessary. Understanding how this system works aids in grasping more complex concepts in chemistry and biology related to acid-base homeostasis.