Problem 30 - Chemistry The Central Science In Si Units

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 is 7.09.

Step 1: Determine Molar Mass of Compounds

First, calculate the molar mass of \( \mathrm{KH}_2\mathrm{PO}_4 \) and \( \mathrm{K}_2\mathrm{HPO}_4 \). The molar mass of \( \mathrm{KH}_2\mathrm{PO}_4 \) is calculated as follows: \( K: 39.10 \), \( H_2: 2 \times 1.01 \), \( P: 30.97 \), \( O_4: 4 \times 16.00 \), making it \( 136.09 \, \mathrm{g/mol} \). For \( \mathrm{K}_2\mathrm{HPO}_4 \), we have: \( 2K: 2 \times 39.10 \), \( H: 1.01 \), \( P: 30.97 \), \( O_4: 4 \times 16.00 \), totaling \( 174.18 \, \mathrm{g/mol} \).

Step 2: Convert Grams to Moles

Calculate the moles for \( 10.0 \, \mathrm{g} \) of each compound. For \( \mathrm{KH}_2\mathrm{PO}_4 \), \( \text{moles} = \frac{10.0 \, \mathrm{g}}{136.09 \, \mathrm{g/mol}} = 0.0735 \, \mathrm{mol} \). For \( \mathrm{K}_2\mathrm{HPO}_4 \), \( \text{moles} = \frac{10.0 \, \mathrm{g}}{174.18 \, \mathrm{g/mol}} = 0.0574 \, \mathrm{mol} \).

Step 3: Calculate Molarities

Determine the molarity for each compound using the volume of the solution, which is \( 0.500 \, \mathrm{L} \). The molarity of \( \mathrm{KH}_2\mathrm{PO}_4 \) is \( \frac{0.0735 \, \mathrm{mol}}{0.500 \, \mathrm{L}} = 0.147 \, \mathrm{M} \), while the molarity for \( \mathrm{K}_2\mathrm{HPO}_4 \) is \( \frac{0.0574 \, \mathrm{mol}}{0.500 \, \mathrm{L}} = 0.115 \, \mathrm{M} \).

Step 4: Apply Henderson-Hasselbalch Equation

Use the Henderson-Hasselbalch equation to find the \( \mathrm{pH} \). The equation is \( \mathrm{pH} = \mathrm{p}K_a + \log \left( \frac{[\mathrm{base}]}{[\mathrm{acid}]} \right) \). For the \( \mathrm{H}_2\mathrm{PO}_4^- / \mathrm{HPO}_4^{2-} \) system, \( \mathrm{p}K_a \) is approximately 7.20. Substitute the values: \( \mathrm{pH} = 7.20 + \log \left( \frac{0.115}{0.147} \right) \).

Step 5: Calculate Logarithmic Value and Final pH

Calculate \( \log \left( \frac{0.115}{0.147} \right) = -0.111 \). Therefore, the \( \mathrm{pH} = 7.20 - 0.111 = 7.089 \).

Key Concepts

Henderson-Hasselbalch equation

Buffer solutions are essential in maintaining the pH levels of various environments. The Henderson-Hasselbalch equation is a fundamental equation used to estimate the pH of a buffer solution. It provides a link between pH, pKa, and the ratio of concentrations of the acid and conjugate base. Here's the equation in its basic form:
\[\mathrm{pH} = \mathrm{p}K_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right)\]This equation assumes that the concentration of acid and its conjugate base are large compared to the concentration of hydrogen ions in the solution. It works best when the acid and base concentration ratio is between 0.1 and 10, meaning neither the acid nor the base predominates vastly over the other.
In our exercise,

• The acid is \(\mathrm{H}_2\mathrm{PO}_4^-\) and the conjugate base is \(\mathrm{HPO}_4^{2-}\).
• The ratio was derived from the given molarities, calculated after determining the moles of each compound.

The Henderson-Hasselbalch equation allowed us to calculate the pH by plugging in these values, providing us with the answer in a straightforward manner.

pH calculation

pH is a scale that measures the acidity or basicity of a solution. It is crucial in fields such as chemistry, biology, ecology, and environmental science. Calculating the pH of a solution gives insight into its chemical properties and potential reactivity.
For a buffer solution, which contains a weak acid and its conjugate base, the Henderson-Hasselbalch equation becomes a valuable tool for easy pH calculation. The pH is impacted by the concentrations of the acid and its conjugate base, along with the acid's dissociation constant, represented as \( \mathrm{p}K_a \).
In our example:

• The molarities were calculated for the acid \(\mathrm{H}_2\mathrm{PO}_4^-\) and the base \(\mathrm{HPO}_4^{2-}\).
• Using the formula \( \mathrm{pH} = \mathrm{p}K_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \), the calculated pH helps understand the solution's buffering capacity.

Our calculated pH of 7.089 demonstrates the buffer's ability to maintain a stable acidity level despite the presence of added acids or bases. This characteristic is central to many practical applications of buffer solutions.

Acid-base chemistry

Acid-base chemistry is a core concept in chemistry that revolves around the behavior of acids and bases in solution. It involves understanding how these compounds interact, their equilibrium states, and how they influence the pH of their environment.
Acids are substances that can donate a proton (\(\mathrm{H}^+\)) to another substance, while bases are substances that can accept a proton. The strength of an acid or base is determined by its tendency to donate or accept protons, which is reflected in the \(\mathrm{K}_a\) or \(\mathrm{K}_b\) values, the equilibrium constants for acid and base dissociation respectively.
In our exercise:

• We considered the phosphate buffer system, which is a common system used due to its effective buffering capacity at a physiological pH level.
• The equilibrium between \(\mathrm{H}_2\mathrm{PO}_4^-\) and \(\mathrm{HPO}_4^{2-}\) helps resist changes in pH upon addition of small amounts of acids or bases.

Understanding the principles of acid-base chemistry and buffer systems like this is critical for applications in many fields, such as medicine, where maintaining pH balance is crucial for proper biological function.