Problem 30 - Chemistry: The Central Science in SI Units, Global Edition

Question

A buffer, consisting of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-},\) helps control the pH of physiological fluids. Many carbonated soft drinks also use this buffer system. What is the \(\mathrm{pH}\) of a soft drink in which the major buffer ingredients are \(10.0 \mathrm{~g}\) of \(\mathrm{KH}_{2} \mathrm{PO}_{4}\) and \(10.0 \mathrm{~g}\) of \(\mathrm{K}_{2} \mathrm{HPO}_{4}\) per \(0.500 \mathrm{~L}\) of solution?

Step-by-Step Solution

Answer

The pH of the soft drink with the given buffer system and concentrations is approximately 1.53.

Step 1: Find the moles and concentrations of KH2PO4 and K2HPO4

First, we need to determine the molar masses of both compounds: For \(\mathrm{KH}_{2} \mathrm{PO}_{4}:\) \(\mathrm{K = 39.10 ~g/mol, ~H_{2} = 2.02 ~g/mol, ~P = 30.97 ~g/mol, ~O_{4} = 64.00 ~g/mol}\) So, molar mass of \(\mathrm{KH}_{2} \mathrm{PO}_{4} = 39.10 + 2.02 + 30.97 + 64.00 = 136.09 \mathrm{~g/mol}\) For \(\mathrm{K}_{2}\mathrm{HPO}_{4}:\) \(\mathrm{K_{2} = 78.20 ~g/mol, ~P = 30.97 ~g/mol, ~O_{4} = 64.00 ~g/mol}\) So, molar mass of \(\mathrm{K}_{2}\mathrm{HPO}_{4} = 78.20 + 30.97 + 64.00 = 173.17 \mathrm{~g/mol}\) Now, we can find the moles of each compound: For \(\mathrm{KH}_{2} \mathrm{PO}_{4}:\) moles = mass / molar mass = \(10.0 \mathrm{~g} / 136.09 \mathrm{~g/mol} = 0.0735 \mathrm{~mol}\) For \(\mathrm{K}_{2}\mathrm{HPO}_{4}:\) moles = mass / molar mass = \(10.0 \mathrm{~g} / 173.17 \mathrm{~g/mol} = 0.0577 \mathrm{~mol}\) Now, calculate the concentrations: For \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}:\) concentration = moles / volume = \(0.0735 \mathrm{~mol} / 0.500 \mathrm{~L} = 0.147 \mathrm{~M}\) For \(\mathrm{HPO}_{4}^{2-}:\) concentration = moles / volume = \(0.0577 \mathrm{~mol} / 0.500 \mathrm{~L} = 0.115 \mathrm{~M}\)

Step 2: Calculate the concentrations of H2PO4- and HPO42-

Since \(\mathrm{KH}_{2} \mathrm{PO}_{4}\) dissociates into \(\mathrm{K}^{+}\) and \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\), the concentration of \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\) is the same as the concentration of \(\mathrm{KH}_{2} \mathrm{PO}_{4}\). For \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}:\) concentration = \(0.147 \mathrm{~M}\) Similarly, since \(\mathrm{K}_{2}\mathrm{HPO}_{4}\) dissociates into T\(2\mathrm{K}^{+}\) and \(\mathrm{HPO}_{4}^{2-}\), the concentration of \(\mathrm{HPO}_{4}^{2-}\) is the same as the concentration of \(\mathrm{K}_{2}\mathrm{HPO}_{4}\). For \(\mathrm{HPO}_{4}^{2-}:\) concentration = \(0.115 \mathrm{~M}\)

Step 3: Determine the value of Ka

We are given that \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) are conjugate acid-base pairs, and we can find the dissociation constant \(K_{a1}\) of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) from a reference: \(K_{a1} (\mathrm{H}_{2}\mathrm{PO}_{4}^{-}) = 6.2 \times 10^{-8}\) To find the \(K_{a}\) for the reaction \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-} \rightleftharpoons \mathrm{HPO}_{4}^{2-} + \mathrm{H}^{+}\), we use the second dissociation constant, \(K_{a2}\) of \(\mathrm{H}_{3}\mathrm{PO}_{4}\): \(K_{a2} (\mathrm{H}_{3}\mathrm{PO}_{4}) = 6.2 \times 10^{-8} / 7.1 \times 10^{-13} = 7.3 \times 10^{-2}\)

Step 4: Apply the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation relates the pH of a solution to the pKa and the ratio of the concentrations of the conjugate acid \(\mathrm{A}^{-}\) and the conjugate base \(\mathrm{HA}\): \(\mathrm{pH} = \mathrm{p}K_{a} + \log \left(\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\right)\) For our buffer system, the conjugate acid is \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\), and the conjugate base is \(\mathrm{HPO}_{4}^{2-}\). The \(K_{a}\) value for \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\) is \(7.3 \times 10^{-2}\). Substitute the values into the equation: pH = \(- \log{(7.3 \times 10^{-2})} + \log \left(\frac{0.115}{0.147}\right)\)

Step 5: Calculate the pH of the soft drink

Using a calculator to perform the calculations: pH = \(1.14 - (-0.386) = 1.526\) The pH of the soft drink is approximately 1.53.

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a crucial tool in understanding buffer systems. It allows us to determine the pH of a solution by relating it to the pKa value and the ratio of the concentrations of the conjugate base and acid. This equation is vital because it simplifies the pH calculation for weak acid and conjugate base mixtures, where the acid (A^{-} ) donates a proton to become its conjugate base (HA ).
The formula is expressed as follows:
\[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^{-}]}{[\text{HA}]} \right) \]
In practice, the pKa is derived from the Ka, the acid dissociation constant, which is essential for calculating how much the acid dissociates in a solution. The Henderson-Hasselbalch equation is particularly useful when dealing with buffer solutions because it helps maintain a stable pH, as seen in applications like physiological fluids and beverages.

pH Calculation

Calculating the pH of a solution involves understanding the concentration of hydrogen ions it contains. Lower pH values indicate higher acidity, while higher pH values indicate basicity or alkalinity. The calculation typically involves using logarithmic functions because the pH is defined as the negative logarithm of the hydrogen ion concentration \([\text{H}^{+}]]\).
For buffers, using the Henderson-Hasselbalch equation simplifies this task. The equation accounts for both the pKa of the weak acid and the ratio of concentrations of the acid and its conjugate base. The following steps illustrate how to perform a pH calculation for buffers:

• Find the molar concentration of both the conjugate acid (\text{HA}) and conjugate base (\text{A}^{-}).
• Determine the pKa value of the acid using its dissociation constant, if not already known.
• Substitute these values into the Henderson-Hasselbalch equation to find the pH.

This method allows us to calculate pH more easily and is widely used in both educational and real-world chemical applications.

Conjugate Acid-Base Pairs

Conjugate acid-base pairs are integral to buffer solutions. These pairs consist of two species that transform into one another by the gain or loss of a proton (H⁺). Understanding this concept is critical for predicting the behavior of buffers in maintaining a stable pH environment.
An acid becomes a conjugate base when it loses a proton, while a base becomes a conjugate acid when it gains a proton. For example, in the given buffer system of KH₂PO₄ and K₂HPO₄:

• KH₂PO₄ (acid) donates a proton to become the conjugate base, H₂PO₄⁻.
• HPO₄²⁻ (base) accepts a proton to become the conjugate acid, H₂PO₄⁻.

These transformations are dynamic and equilibrate to form a buffer solution that can resist changes in pH when small amounts of acid or base are added. Buffer systems are incredibly useful in biological contexts where maintaining a specific pH range is crucial for proper physiological function.