Problem 31 - Chemistry: The Central Science in SI Units, Global Edition

Question

You have to prepare a \(\mathrm{pH}=2.50\) buffer, and you have the following \(0.20 \mathrm{M}\) solutions available: $\mathrm{HCOOH}, \mathrm{CH}_{3} \mathrm{COOH},\( \)\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{KCH}_{3} \mathrm{COO}, \mathrm{KHCOO},\( and \)\mathrm{KH}_{2} \mathrm{PO}_{4} .$ Which solutions would you use? How many liters of each solution would you use to make approximately 2 L of the buffer?

Step-by-Step Solution

Answer

To prepare a pH=2.50 buffer with the given solutions, use 0.6175 L of 0.20 M H3PO4 and 1.3825 L of 0.20 M KH2PO4. The sum of these volumes is approximately 2 L, which is the desired buffer volume.

Step 1: Identify the acid and its conjugate base

To prepare a buffer, we need an acid and its conjugate base. Let's go through the available substances and see which can form a buffer with a pH of 2.50. 1. HCOOH (Formic acid) with pKa = 3.75 2. CH3COOH (Acetic acid) with pKa = 4.75 3. H3PO4 (Phosphoric acid) with pKa1 = 2.15, pKa2 = 7.20, pKa3 = 12.35 4. KCH3COO (Potassium acetate) is salt of: CH3COOH (Acetic acid) with pKa = 4.75 5. KHCOO (Potassium formate) is salt of: HCOOH (Formic acid) with pKa = 3.75 6. KH2PO4 (Potassium dihydrogen phosphate) is salt of: H3PO4 (Phosphoric acid) with pKa1 = 2.15

Step 2: Use the Henderson-Hasselbalch equation to find the pH

The Henderson-Hasselbalch equation helps us to determine the pH of a buffer: \( pH = pKa + log \frac{[A^-]}{[HA]} \) where A- is the conjugate base and HA is the weak acid. Since we want to prepare a buffer with pH=2.50, we'll go through the available substances and check which combination of weak acid and its salt will be close to pH=2.50. 1. For HCOOH (pKa = 3.75) and KHCOO: \(2.50 = 3.75 - log \frac{[HCOO^-]}{[HCOOH]}\) 2. For CH3COOH (pKa = 4.75) and KCH3COO: \(2.50 = 4.75 - log \frac{[CH3COO^-]}{[CH3COOH]}\) 3. For H3PO4 (pKa1 = 2.15) and KH2PO4: \(2.50 = 2.15 - log \frac{[H2PO4^-]}{[H3PO4]}\)

Step 3: Determine the suitable buffer

By looking at the three cases above, we can see that H3PO4 (pKa1 = 2.15) and KH2PO4 are the most suitable as a buffering pair for the desired pH of 2.50: \(2.50 = 2.15 - log \frac{[H2PO4^-]}{[H3PO4]}\) Now, we need to determine the concentration of acid and salt required to make the buffer. Rearrange the Henderson-Hasselbalch equation to solve for the ratio of [A-]/[HA]: \(log \frac{[A^-]}{[HA]} = pH - pKa\) \(\frac{[A^-]}{[HA]} = 10^{pH - pKa}\) \(\frac{[H2PO4^-]}{[H3PO4]} = 10^{2.50 - 2.15}\)

Step 4: Calculate the required concentrations and amounts of the buffer components

Calculate the ratio of the concentrations of conjugate base to the acid: \(\frac{[H2PO4^-]}{[H3PO4]} = 10^{2.50 - 2.15} = 2.2387\) Now, let's say that we have x moles of H3PO4 and y moles of H2PO4- in 2 L of the buffer solution, and the total moles should be equal to the sum of moles of the acid and base. So, x + y = total moles We know that: \(\frac{y}{x} = 2.2387\) So, y = 2.2387 * x Now, substitute y back in the first equation: x + 2.2387x = total moles 3.2387x = total moles We also know that the total buffer concentration is 0.20 M: total moles = 0.20 mol/L * 2 L = 0.40 mol Substitute the total moles back into the equation: 3.2387x = 0.40 mol Therefore, x (moles of H3PO4) = 0.1235 mol y (moles of H2PO4-) = 0.2765 mol Now, we'll convert the moles to volume for each component: Volume of H3PO4 = \(\frac{0.1235}{0.2}\) L = 0.6175 L Volume of KH2PO4 = \(\frac{0.2765}{0.2}\) L = 1.3825 L

Step 5: Conclusion

To prepare a pH=2.50 buffer with the given solutions, use 0.6175 L of 0.20 M H3PO4 and 1.3825 L of 0.20 M KH2PO4. The sum of these volumes is approximately 2 L, which is the desired buffer volume.