Suppose that you have 0.500 L of each of the following solutions, and an unlimited supply of water. (Note: C9H7NHBr is a salt containing the ions C9H7NH+ and Br− and C9H7N is quinoline, an organic base with pKb = 6.24 at 298 K. If you like, you may represent C9H7NH+ as HB+ and C9H7N as B.)
0.113 mol L−1 C9H7NHBr (aq) 0.104 mol L−1 HBr(aq) 0.122 mol L−1 NaOH(aq)
(a)Provide simple instructions for preparing 1.00 L of a solution having pH = 7.00 at 298 K. Your instructions should include the volumes of the solutions required.
(b) What is the buffer capacity of the resulting solution? (The buffer capacity is the number of moles of NaOH that must be added to 1.0 L of solution to raise the pH by one unit.)
OK. Then what does that incomplete sentence mean " 0.113 mol ...0.104 HBr....0.122 NaOH? That tid bit is just thrown in there with no explanation of what it has to do with the problem. That MAY be the way to calculate the concn of the acid and base; it should be obvious that we can't calculate the mols acid and base with just liters and no concns listed. mols = M x L. We have L but no molarities as it stands unless that incomplete sentence is supposed to convey that information.
To prepare a 1.00 L solution with a pH of 7.00 at 298 K, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Since the given pKb for C9H7N is 6.24, we can calculate its pKa using the equation: pKa = 14 - pKb. Therefore, pKa = 7.76.
In order to make a buffer solution, we need to have a conjugate acid-base pair, which in this case is HB+ (C9H7NH+) and B (C9H7N). We can choose a ratio of HB+ and B that will give us a pH of 7.00.
(a) To prepare a 1.00 L solution with a pH of 7.00, we will need the following volumes of the given solutions:
1. Start by calculating the moles of HB+ and B required using the Henderson-Hasselbalch equation:
pH = pKa + log([B]/[HB+])
7.00 = 7.76 + log([B]/[HB+])
Taking the antilog of both sides:
[B]/[HB+] = 10^(7.00 - 7.76)
[B]/[HB+] = 0.125
2. Calculate the volume of 0.113 mol L-1 C9H7NHBr solution required to give the desired amount of HB+:
Volume of 0.113 mol L-1 C9H7NHBr = (0.125 * 0.500 L) / 0.113 mol L-1
Volume of 0.113 mol L-1 C9H7NHBr = 0.556 L
3. Calculate the volume of 0.122 mol L-1 NaOH solution required to give the desired amount of B:
Volume of 0.122 mol L-1 NaOH = (0.125 * 0.500 L) / 0.122 mol L-1
Volume of 0.122 mol L-1 NaOH = 0.514 L
4. Finally, add these volumes together with an unlimited supply of water to make a 1.00 L solution with a pH of 7.00:
Volume of HB+ solution + Volume of B solution + Unlimited water = 0.556 L + 0.514 L + Unlimited water
Total volume = 1.07 L (approximately)
(b) The buffer capacity of the resulting solution is defined as the number of moles of NaOH that must be added to 1.0 L of solution to raise the pH by one unit. In this case, the buffer capacity would be equal to the number of moles of B that will react with the added NaOH.
Given that the solution contains 0.514 mol of B (C9H7N), the buffer capacity will be 0.514 mol. Therefore, 0.514 moles of NaOH would be required to raise the pH by one unit in 1.0 L of solution.