A buffer contains the weak acid HA abd its conjugated base A-. The weak acid has a Ka of 1.51 x10^-5 and the buffer has a pH of 4.25. Which statement is true about the relative concentrations of HA and A- in the buffer?
A) [HA]>[A-]
B) [HA]<[A-]
C) [HA] = [A-]
pH = pKa + log base/acid
pH = 4.26
pKa = 4.82
So log A/HA must be negative. Which (A or B) will make it negative?
To determine the relative concentrations of HA and A- in the buffer, we need to compare the pH of the buffer with the pKa of the weak acid (HA). Remember that the pKa is the negative logarithm (base 10) of the acid dissociation constant (Ka). The relationship between pH, pKa, [HA], and [A-] can be described by the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Given that the buffer has a pH of 4.25 and the Ka of HA is 1.51 x 10^-5, we can substitute these values into the equation:
4.25 = -log(1.51 x 10^-5) + log([A-]/[HA])
Rearranging the equation, we get:
log([A-]/[HA]) = 4.25 + log(1.51 x 10^-5)
Taking the antilog of both sides:
[A-]/[HA] = 10^(4.25 + log(1.51 x 10^-5))
Simplifying further:
[A-]/[HA] = (10^4.25) x (1.51 x 10^-5)
[A-]/[HA] = 1.77 x 10^(-1)
Now, since we are looking for the relative concentrations of HA and A-, we can compare their ratio. If [HA]>[A-], the ratio will be greater than 1. If [HA]<[A-], the ratio will be less than 1. And if [HA] = [A-], the ratio will be equal to 1.
Comparing the ratio 1.77 x 10^(-1) to 1, we can conclude that:
[HA] < [A-]
Therefore, the correct statement is:
B) [HA] < [A-]