Glutamic acid is triprotic amino acid with pKa values of 2.23, 4.42, and 9.95. You
are given a 0.100 M solution of it in its fully protonated form (H3A). How many moles of KOH must be added to 100 mL of this solution to make a solution with a pH of 5.00.
I used 2.23 as the pka to obtain the concentrations H+ and H2A-. From there I keep getting the wrong answer when I add OH-
I would think you would use the 4.42 value for pH = 5.0 buffer.
To solve this problem, we need to consider the protonation states and equilibrium reactions of glutamic acid. Glutamic acid (H3A) has three ionizable protons, and its ionization reactions can be represented as follows:
H3A ⇌ H+ + H2A- (pKa1 = 2.23)
H2A- ⇌ H+ + HA2- (pKa2 = 4.42)
HA2- ⇌ H+ + A3- (pKa3 = 9.95)
First, let's determine the initial concentrations of H3A, H2A-, and A3- in the given 0.100 M solution of H3A. Since H3A is fully protonated, its initial concentration is 0.100 M.
Next, we need to know the equilibrium concentrations of H2A- and A3- at pH 5.00. At pH 5.00, we can assume that [H+] is 10^(-5.00) M.
To calculate the equilibrium concentrations, we can use the Henderson-Hasselbalch equation for each dissociation step:
pH = pKa + log([A-]/[HA])
For the first dissociation step:
5.00 = 2.23 + log([H2A-]/[H3A])
Rearranging the equation gives:
log([H2A-]/[H3A]) = 5.00 - 2.23
log([H2A-]/0.100) = 5.00 - 2.23
Using the properties of logarithms, we can solve for [H2A-]:
[H2A-] = 0.100 × 10^(5.00 - 2.23)
Similarly, for the second dissociation step:
pH = 4.42 + log([A3-]/[H2A-])
5.00 = 4.42 + log([A3-]/[H2A-])
Rearranging the equation gives:
log([A3-]/[H2A-]) = 5.00 - 4.42
log([A3-]/[0.100 × 10^(5.00 - 2.23)]) = 0.58
Using the properties of logarithms, we can solve for [A3-]:
[A3-] = [0.100 × 10^(5.00 - 2.23)] × 10^0.58
Now, we can calculate the amount of KOH required to neutralize the excess H+ in the solution. Since each OH- ion reacts with one H+ ion, the moles of KOH required will be equal to the moles of H+ present in the solution.
To calculate the moles of H+, we need to consider the equilibrium concentrations of H3A, H2A-, and A3-. The moles of H+ can be calculated using the equation:
[H+] = [H3A] + [H2A-] + 2[H2A-] + 3[A3-]
Substituting the known values, we get:
[H+] = 0.100 + [H2A-] + 2[H2A-] + 3[A3-]
Now, substitute the equilibrium concentrations of [H2A-] and [A3-] that we calculated earlier into the equation to find the moles of H+.
Finally, the moles of KOH required will be equal to the moles of H+, as each KOH molecule reacts with one H+ ion. You can convert the moles of H+ to the moles of KOH using stoichiometry.
I hope this explanation helps you understand how to approach the problem.