a dibasic compound , B, has pkb2 = 6.00. Find the fraction in the form of BH+ at pH 9.00
To find the fraction of the dibasic compound B in the form of BH+ at pH 9.00, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
In this case, since B is a dibasic compound, we can consider it to be a weak acid, HA1 and HA2. The conjugate base for HA1 will be A-, and for HA2, it will be A.
Given that pKb2 = 6.00, we can use the relationship pKa + pKb = 14 to find pKa2:
pKb2 = 14 - pKa2
6.00 = 14 - pKa2
pKa2 = 14 - 6.00
pKa2 = 8.00
Now we consider the equilibrium reactions for HA1 and HA2:
HA1 ⇌ H+ + A-
HA2 ⇌ H+ + A
To find the fraction in the form of BH+ at pH 9.00, we need to use the Henderson-Hasselbalch equation for each equilibrium reaction.
For HA1:
pH = pKa1 + log ([A-] / [HA1])
9.00 = pKa1 + log ([A-] / [HA1])
We can rearrange this equation to solve for the ratio [A-] / [HA1]:
[A-] / [HA1] = 10^(pH - pKa1)
Now let's find the expression for HA2 using a similar approach.
For HA2:
pH = pKa2 + log ([A] / [HA2])
9.00 = pKa2 + log ([A] / [HA2])
Rearranging this equation:
[A] / [HA2] = 10^(pH - pKa2)
Now, we need to express the fractions based on the dissociation constants:
[A-] / [HA1] = [A] / [HA2] * [HA2] / [HA1]
Let's substitute the known values:
[A-] / [HA1] = 10^(pH - pKa2) * [HA2] / 10^(pH - pKa1)
[A-] / [HA1] = 10^(9.00 - 8.00) * [HA2] / 10^(9.00 - pKa1)
[A-] / [HA1] = 10 * [HA2] / 10^(9.00 - pKa1)
Finally, since [BH+] = [HA2] and [B] = [HA1], the fraction in the form of BH+ at pH 9.00 is:
[BH+] / [B] = [HA2] / [HA1]
[BH+] / [B] = [HA2] / [HA1] = 10 * [HA2] / 10^(9.00 - pKa1)
Please note that we need to know the value of pKa1 to obtain the exact fraction in the given form.
To find the fraction of a dibasic compound (B) in the form of BH+ at a certain pH (pH 9.00 in this case), we need to use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa of the compound:
pH = pKa + log ([A-]/[HA])
In this equation, [A-] represents the concentration of the deprotonated form of the compound (B) and [HA] represents the concentration of the protonated form (BH+).
Since B is a dibasic compound, we need to consider two deprotonation steps. The first dissociation removes a proton from BH+ to form B- (the first deprotonated form). The second dissociation then removes another proton from B- to form B2- (the second deprotonated form).
The pKa values corresponding to each deprotonation step can be determined using pKa = 14.00 - pKb.
For the first deprotonation step:
pKa1 = 14.00 - pKb1
For the second deprotonation step:
pKa2 = 14.00 - pKb2
To find the fraction of B in the form of BH+ at pH 9.00, we can apply the Henderson-Hasselbalch equation twice: once for the first deprotonation step and once for the second deprotonation step.
Step 1: Finding the fraction of B- (the first deprotonated form) at pH 9.00
Since B- is negatively charged and exists in negligible amounts at pH 9.00, we can ignore [A-], and the Henderson-Hasselbalch equation simplifies to:
pH = pKa1 + log ([A-]/[HA])
Since [A-] is negligible, the equation becomes:
pH = pKa1 + log (0/[HA])
pH = pKa1 + log(0) (any logarithm of 0 is undefined)
0 = pKa1 + log(0)
Since the logarithm of 0 is undefined, we cannot determine the fraction of B- in this case.
Step 2: Finding the fraction of B2- (the second deprotonated form) at pH 9.00
Again, we can ignore the negligible concentration of [A-], and the Henderson-Hasselbalch equation becomes:
pH = pKa2 + log ([A-]/[HA])
Since B2- is negatively charged, [A-] represents the concentration of B2-. [HA], on the other hand, represents the concentration of B-, the first deprotonated form.
To find [A-], we start with the fact that B- was formed from BH+ in the first step and then assumed to fully deprotonate in the second step to form B2-. This means that [A-] is equal to the concentration of B- formed in the first step. Therefore:
[A-] = [BH+] from first deprotonation step
Now, let's substitute these values into the equation:
pH = pKa2 + log ([A-]/[HA])
pH = pKa2 + log ([BH+]/[BH+]) (substituting [A-] with [BH+])
pH = pKa2 + log (1) ([BH+]/[BH+] = 1)
pH = pKa2
Therefore, the fraction of B2- (the second deprotonated form) at pH 9.00 is 10^(-pKa2).
In this case, pKa2 = 14.00 - pKb2 = 14.00 - 6.00 = 8.00
Hence, the fraction of B2- at pH 9.00 is 10^(-pKa2) = 10^(-8.00).