How many moles of NaOH must be added to 1.0 L of 2.2 M HF to produce a solution buffered at each pH?
(a) pH = pKa
(b) pH = 4.24
(c) pH = 4.60
To answer this question, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the ratio of the concentrations of the conjugate acid and base. The equation is:
pH = pKa + log([A-]/[HA])
Where:
pH = desired pH of the buffer solution
pKa = dissociation constant of the weak acid
[A-] = concentration of the conjugate base
[HA] = concentration of the weak acid
In this case, the weak acid is HF and its conjugate base is F-. We want to create a buffer solution by adding NaOH, which will react with HF to form NaF.
(a) pH = pKa:
To create a buffer with a pH equal to the pKa of HF, we need the ratio of [A-] to [HA] to be 1. By rearranging the Henderson-Hasselbalch equation, we get:
[A-]/[HA] = 10^(pH - pKa)
Substituting the values, we can calculate the ratio:
[A-]/[HA] = 10^(pH - pKa) = 10^(-3.17) = 0.477
Since the volume is 1.0 L, the concentration of HF ([HA]) is given as 2.2 M. We can calculate the concentration of F- ([A-]) using the ratio:
[A-] = 0.477 * [HA] = 0.477 * 2.2 = 1.044 M
Therefore, we need to add 1.044 moles of NaOH to 1.0 L of 2.2 M HF to create a buffer solution with pH equal to the pKa.
(b) pH = 4.24:
Similarly, we can calculate the ratio [A-]/[HA]:
[A-]/[HA] = 10^(pH - pKa) = 10^(4.24 - 3.17) = 6.23
Again, the volume is 1.0 L and [HA] is 2.2 M. We can calculate [A-]:
[A-] = 6.23 * 2.2 = 13.706 M
Therefore, we need to add 13.706 moles of NaOH to 1.0 L of 2.2 M HF to create a buffer solution with pH equal to 4.24.
(c) pH = 4.60:
Using the same calculation:
[A-]/[HA] = 10^(pH - pKa) = 10^(4.60 - 3.17) = 10.314
With [HA] as 2.2 M, we can determine [A-]:
[A-] = 10.314 * 2.2 = 22.69 M
Hence, to create a buffer solution with pH of 4.60, it is necessary to add 22.69 moles of NaOH to 1.0 L of 2.2 M HF.