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Introduction
In the previous article, we talked about how a friend of mine sent me the following question (adapted). It’s more of an ‘interdisciplinary’ type of question, and it requires some skills from organic chemistry to solve it, as well. See if you are able to solve it:
“A 20mM histidine monohydrochloride solution (Mr = 209.63) is prepared. 20ml of it is used in a titration with 0.05M NaOH. It has an initial pH of 4.00. In a previous experiment, 2 pKa values of histidine were determined: 6.00 and 9.35. Calculate the pH of the solution after addition of 12ml of 0.05M NaOH.”
For reference, the structure of histidine is provided below (Fig. 1).
Fig. 1: Structure of histidine.
For those who are seeking a recap of acids and bases, check out our previous article, on exactly this topic. If you have already read that article, then let us get into the solving / understanding of this question!
Answer
Before we start, we will first analyse the question. We observe two pKas have been calculated; this is typical of an amino acid. When the pKa is at 6.00, the structure of histidine will look as in Fig. 1; however, when the pKa increases above 9.35, the NH3+ side chain is deprotonated.
We name these two structures at different pKas B and C. The structure before B, which is A, is not that significant and will not be discussed. The transition between A to B to C is reversible. At the equivalence point, only B is present (i.e. pKa = 6.00). When the pKa is equivalent to 9.35 (second equivalence point), only C is present.
We want to begin by calculating the volume of NaOH needed to bring the pKa to 6.00, so as to ensure that it is indeed possible. We can do this by looking at the number of moles of histidine in the mixture. No matter A, B or C, all of them react with NaOH in a 1:1 ratio at the correct pH.
The moles of histidine can be calculated by converting the millimolar to molar, and then multiplying this value by the volume in litres (so as to ensure that there are correct units). This is shown in Fig. 2.
20mM/1000 X 20ml/1000 = 4.00 x 10^-4 moles.
Fig. 2: Calculation of moles of histidine.
We continue by dividing the number of moles by the molarity of NaOH (mol/L) which is given in the question as 0.05, because mol/(mol/L) = L. Then we convert it to ml to find out the volume of NaOH needed to reach the equivalence point.
4.00 X 10^-4 / 0.05 = 8.00 X 10^-3 L = 8ml
Fig. 3: Calculation of volume of NaOH to reach equivalence.
Since only 8ml is needed to reach the equivalence point, it can thus be concluded that the pH of the solution is likely to be above pH 6.00. Now we know there is only B and C left in the mixture. They are also connected by a reversible reaction, indicating that we should use an I.C.E table.
But how do we know how much of B there is initially? We can do this easily because we have previously calculated that there are 4.00 x 10^4 moles of histidine. At pH 6.00, all have converted to B. Thus there are 4.00 X 10^4 moles of B at 6.00. We calculate concentration of B by dividing the moles by the volume, which is 28 (20 + 8, since 8ml of NaOH is required to achieve equivalence). We also convert ml to L.
(4.00 x 10^-4) / (28/1000) = 1.4286 x 10^-2 mol dm-3
Fig. 4: Calculation of [B] at equivalence.
Naturally, we also draw out the I.C.E table and equate y to the equilibrium concentration of H+ (which is lost from the amine side chain). Thus the table is drawn (Fig. 5) and the correct values for y are calculated (Fig. 6). Note that the Ka may be derived from pKa = 9.35.
Fig. 5: I.C.E table.
Ka = 10^-pKa = 10^-9.35 = 4.4668 x 10^-10
4.4688 x 10^-10 = y^2/1.4286 x 10^-2 - y = y^2/1.4286 x 10^-2
[B] = 1.4286 x 10^-2 - 2.5261 x 10^-6 = 1.4283 x 10^-2
[C] = y = 2.5261 x 10^-6
Fig. 6: Calculation of [B] and [C].
Now we can calculate the number of moles of B and C in order to know how many moles are deducted (due to acid-base reaction by NaOH). This is done in Fig. 7 (at this point there is still 28ml, with 4ml not added). We also calculate moles in 4ml NaOH.
Moles of B = (1.4283 x 10^-2) x (28/1000) = 4.00 x 10^-4
Moles of C = (2.5261 x 10^-6) x (28/1000) = 7.073 x 10^-8
Moles of 4ml NaOH = 0.05 x (4/1000) = 2.0 x 10^-4
Fig. 7: Calculation of moles.
Because each mole of NaOH will react in a 1:1 ratio with B to form C, we subtract moles of NaOH from B and add it to C. This will give us the final number of moles of B and C after 12ml of NaOH has been added. We then calculate the concentrations from there.
Moles of B = 4.00 x 10^-4 - 2.0 x 10^-4 = 2.00 x 10^-4
Moles of C = 7.073 x 10^-6 + 2.0 x 10^-4 = 2.0007 x 10^-4
Fig. 8: Final moles of B and C.
Concentration of B = (2.00 x 10^-4)/(32/1000) = 6.25 x 10^-3
Concentration of C = (2.0007 x 10^-4)/(32/1000) = 6.2522 x 10^-3
Fig. 9: Concentrations of B and C.
By the Henderson-Hasselbalch equation, pH = pKa + lg(A-/HA). Now, since we have the values for A-, which is C, and HA, which is B. The pKa is 9.35 (equivalence point, pH = pKa). We plug in the values as shown in Fig. 10 (eliminating the 10^-3, which is common).
pH = 9.35 + lg(6.522/6.25) = 9.37 (3sf)
Fig. 10: Calculation of pH.
So there you have it! The answer is 9.37.