Problem 27 - Chemistry The Science in Context

Question

What masses of bromoacetic acid and sodium bromoacetate are needed to prepare \(1.00 \mathrm{L}\) of \(\mathrm{pH}=3.00\) buffer if the total concentration of the two components is \(0.200 M ?\)

Step-by-Step Solution

Answer

Answer: To prepare 1.00 L of pH 3.00 buffer with a total concentration of 0.200 M, you will need 3.375 grams of bromoacetic acid and 31.95 grams of sodium bromoacetate.

Step 1: Write down the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation relates the pH, pKa, and the ratio of the conjugate base [A-] to the acid [HA] as follows: pH = pKa + log(\frac{[A-]}{[HA]})

Step 2: Calculate the pKa of bromoacetic acid

Bromoacetic acid (C2H2BrO2) is a monoprotic weak acid. Its Ka value is \(7.2 \times 10^{-3}\). To find the pKa, use the formula: pKa = -log(Ka) pKa = -log(\(7.2\times10^{-3}\)) = 2.14

Step 3: Find the ratio of [A-] to [HA] using the Henderson-Hasselbalch equation

Now we know the pH and pKa, we can find the ratio of sodium bromoacetate ([A-]) to bromoacetic acid ([HA]): 3.00 = 2.14 + log(\frac{[A-]}{[HA]}) log(\frac{[A-]}{[HA]}) = 3.00 - 2.14 = 0.86 \frac{[A-]}{[HA]} = 10^{0.86} ≈ 7.24

Step 4: Calculate the moles of [A-] and [HA] in the buffer

The total concentration of the two components in the buffer is 0.200M and the volume is 1.00L. Therefore, the total amount of moles in the buffer is: moles = concentration × volume moles = 0.200 mol/L × 1.00 L = 0.200 mol Now we can use the ratio from Step 3 to calculate the moles of [A-] and [HA]: [A-] + [HA] = 0.200 mol [A-] = 7.24 × [HA] Substitute [A-] in the equation: 7.24 × [HA] + [HA] = 0.200 mol [HA] = \(\frac{0.200\ mol}{8.24}\) ≈ 0.0243 mol [A-] = 7.24 × 0.0243 mol ≈ 0.1757 mol

Step 5: Calculate the masses of bromoacetic acid and sodium bromoacetate

We now have the moles of each component. We can find the mass of each by multiplying the moles by their molar masses. Mass of bromoacetic acid (C2H2BrO2): Molar mass of bromoacetic acid = 138.94 g/mol mass = moles × molar mass mass = 0.0243 mol × 138.94 g/mol = 3.375 g Mass of sodium bromoacetate (C2H2BrO2Na): Molar mass of sodium bromoacetate = 181.94 g/mol (138.94 g/mol for bromoacetic acid + 23 g/mol for sodium) mass = moles × molar mass mass = 0.1757 mol × 181.94 g/mol = 31.95 g Therefore, 3.375 grams of bromoacetic acid and 31.95 grams of sodium bromoacetate are needed to prepare 1.00 L of pH 3.00 buffer with a total concentration of 0.200 M.

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation plays a crucial role in understanding buffer solutions, especially in acid-base chemistry. This equation provides a clear relationship between the pH of a solution and the concentrations of the acid and its conjugate base. It is expressed as: \[ \text{pH} = \text{pKa} + \log \left(\frac{[A^-]}{[HA]}\right) \]Here,

• \([A^-]\) represents the concentration of the conjugate base,
• \([HA]\) represents the concentration of the acid,
• and pKa is the negative logarithm of the acid dissociation constant (Ka) of the acid.

The equation gives us a useful approximation for predicting and adjusting the pH of a buffer solution. By knowing the pKa and the desired pH, you can determine the correct ratio of the acid to its conjugate base needed to prepare a buffer. This is especially valuable for creating solutions in biological systems and chemical laboratories, ensuring the pH remains stable even when small amounts of acids or bases are added.

Acid-Base Chemistry

Acid-base chemistry is the study of acids, bases, and their reactions. It is foundational for understanding how buffer solutions work. Acids can donate protons (H\(^+\)) whereas bases accept protons. When an acid donates a proton, it forms its conjugate base, completing the acid-base equilibrium:\[ \text{HA} \leftrightarrow \text{H}^+ + \text{A}^- \]In buffer solutions:

• An acid and its conjugate base pair are present together.
• This pairing resists changes in pH when small amounts of acid or base are introduced.

For instance, in the case of bromoacetic acid (C\(_2\)H\(_2\)BrO\(_2\)) and its sodium salt form a buffer solution, where bromoacetic acid acts as the weak acid and sodium bromoacetate as its conjugate base. Understanding these reactions helps in anticipating the behavior of such buffers in different environments.

pKa Calculation

The calculation of pKa is integral to working with acids and bases, particularly when using the Henderson-Hasselbalch equation. The pKa value gives insight into the strength of an acid. It can be calculated using the formula:\[ \text{pKa} = -\log(\text{Ka}) \]Where Ka is the acid dissociation constant representing the degree to which an acid dissociates in solution:\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]A low pKa value indicates a stronger acid, which means it dissociates more in water, while a higher pKa points to a weaker acid. For example, bromoacetic acid has a given Ka of \(7.2 \times 10^{-3}\), and its calculated pKa is 2.14. This value helps in adjusting and predicting the acid's position in buffer systems, determining how easily it donates a proton compared to other acids. Understanding and calculating pKa allows chemists to tailor buffer systems according to specific pH requirements in many applications.