The ionization constant of HA is 1×10−6. What must be the ratio of [A−] to [HA] for the preparation of a buffer solution with a pH of 5.74?
Use the HH equation
To determine the ratio of [A-] to [HA] required for preparing a buffer solution with a pH of 5.74, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
First, let's rearrange the equation:
pH - pKa = log([A-]/[HA])
Taking the antilog of both sides, we get:
10^(pH - pKa) = [A-]/[HA]
The pKa is the negative logarithm of the ionization constant (Ka):
pKa = -log(Ka)
Given that the ionization constant (Ka) of HA is 1×10^-6, we can substitute this value into the equation:
pKa = -log(1×10^-6)
Using logarithm properties, we simplify this to:
pKa = 6 - log(1)
pKa = 6
Now, substitute the values into the revised Henderson-Hasselbalch equation:
10^(pH - pKa) = [A-]/[HA]
10^(5.74 - 6) = [A-]/[HA]
10^(-0.26) = [A-]/[HA]
[A-]/[HA] = 0.398
Therefore, the required ratio of [A-] to [HA] for the preparation of a buffer solution with a pH of 5.74 is approximately 0.398.