quant chemistry

Question

When 5.00 mL of 0.1032 M NaOH was added to 0.1123 g of alanine in 100 mL of 0.100M potassium nitrate solution, the measured pH was 9.57. Using activity coefficients, determine pKa₂ for alanine?

Answer 1

Short answer: pKa2 = 9.84. Without the activity correction you would report 9.73, so the correction is worth 0.11 pH units - roughly a 30 percent error in Ka2. Here is the whole route.

1. Which ionization are you actually measuring?

Alanine dissolved in water is the zwitterion, +H3N-CH(CH3)-COO-, which I will write HA. It has two ionizations:

H2A+  <==>  H+  +  HA      pKa1 ~ 2.34   (loses the -COOH proton)
HA    <==>  H+  +  A-      pKa2 ~ 9.9    (loses the -NH3+ proton)

NaOH pulls the second proton, so what you have made is an HA / A- buffer and the constant being measured is Ka2. At pH 9.57 the fraction of alanine still sitting as H2A+ is about 10^-7 of the total, so it drops out completely - you never have to touch pKa1.

2. The buffer ratio

M(alanine) = 89.09 g/mol
n(alanine) = 0.1123 g / 89.09 g/mol = 1.2605 mmol
n(NaOH)    = 5.00 mL x 0.1032 M     = 0.5160 mmol

n(A-) = 0.5160 mmol          (every mole of OH- converts one HA to A-)
n(HA) = 1.2605 - 0.5160 = 0.7445 mmol

[A-] / [HA] = 0.5160 / 0.7445 = 0.6931

The volume cancels inside that ratio, so you do not need it here. You do need it for the ionic strength, which is the next step.

Worth noticing: you added 0.5160 / 1.2605 = 41 percent of an equivalent, not 50 percent. So you are just short of the half-equivalence point, [A-] is less than [HA], and pKa2 must come out ABOVE the measured pH of 9.57. That is your free one-line sanity check, and it kills any answer below 9.57 instantly.

3. Ionic strength

Total volume after the addition is 100 + 5.00 = 105 mL.

K+  and NO3- : 10.0 mmol each / 105 mL   = 0.0952 M each
Na+ and A-   : 0.5160 mmol each / 105 mL = 0.00491 M each

mu = 0.5 * sum(ci * zi^2)
   = 0.5 * (0.0952 + 0.0952 + 0.00491 + 0.00491)
   = 0.100 M

There is a small piece of luck here: diluting the KNO3 from 100 to 105 mL costs almost exactly what the sodium alaninate puts back, so mu lands on 0.100 M anyway. Do the arithmetic rather than assuming it - the coincidence is specific to these numbers.

This is why the KNO3 is in the problem at all. It is an inert swamping electrolyte. Its only job is to hold the ionic strength at a known, fixed value so the activity coefficients are defined and do not drift as you titrate. If it were absent, mu would change with every drop of NaOH and no single gamma would apply.

4. Activity coefficients

Extended Debye-Huckel at 25 C  (alpha = hydrated radius in pm):

              -0.51 * z^2 * sqrt(mu)
log gamma =  ------------------------
              1 + alpha * sqrt(mu) / 305

A-  :  z = -1,  alpha ~ 450 pm,  sqrt(mu) = 0.3165
       log gamma = -0.1614 / 1.467 = -0.110
       gamma(A-) = 0.78

HA  :  net charge = 0   ->   gamma(HA) = 1.00

The single most useful idea in this problem: the zwitterion is electrically neutral overall, so Debye-Huckel assigns it an activity coefficient of 1.00 and it drops straight out of the expression. Only the anion needs correcting. Students routinely try to look up a gamma for the zwitterion and stall - there is nothing to look up, because gamma depends on net charge, not on whether a molecule contains charges.

How much does the assumed size matter? Very little. alpha = 350 pm gives gamma 0.76 and pKa2 9.85; alpha = 600 pm gives gamma 0.80 and pKa2 9.83. The whole plausible range moves the answer by 0.02 units, so use the value in your own book and do not agonize over it. (One caveat for honesty: a real zwitterion carries a very large dipole moment and its gamma does drift a little from 1 in concentrated media. At mu = 0.1 that is a second-order effect and this problem is not asking for it.)

5. Henderson-Hasselbalch, written properly in activities

Ka2 = a(H+) * a(A-) / a(HA)

A glass pH electrode responds to the ACTIVITY of H+, not its concentration,
so pH = -log a(H+) = 9.57 goes in exactly as measured. No correction there.

pKa2 = pH  -  log[ gamma(A-)[A-] / ( gamma(HA)[HA] ) ]
     = 9.57 - log( 0.78 * 0.6931 / 1.00 )
     = 9.57 - log(0.5406)
     = 9.57 + 0.267
     = 9.84

Why bother - what the correction bought you

Ignoring activities entirely : pKa2 = 9.57 - log(0.6931) = 9.73
With activity coefficients   : pKa2 = 9.84
Literature, extrapolated to mu = 0 : pKa2 = 9.87

That is the payoff. The uncorrected number misses the accepted thermodynamic value by 0.14 units; the corrected one lands within 0.03. Activity coefficients are not decoration in this experiment - they are the difference between agreeing with the literature and not.

Four traps this problem is built around

1. Do not convert the pH into a concentration. The commonest wreck is writing [H+] = 10^-9.57 and then feeding it into a concentration-based Ka. That silently mixes two conventions: you have already got an activity from the meter, and using it as a concentration double-counts (or un-counts) the correction. Either work entirely in activities, as above, or work entirely in concentrations - and if you do the latter you must first convert the meter reading using gamma(H+), which is extra work for the same answer.

2. Do not assume you are at the half-equivalence point. Seeing NaOH added to an amino acid, it is tempting to set pH = pKa2 and report 9.57. You are at 41 percent neutralized, not 50 percent, and that shortcut is off by 0.27 units.

3. Use 105 mL, not 100 mL, for the ionic strength. The moles ratio is volume-independent, so it is easy to forget that mu is not.

4. Take the ratio from the balanced neutralization, not from the total alanine. The denominator is the UNREACTED alanine, 0.7445 mmol, not the 1.2605 mmol you weighed out. Using the total gives 0.409 and a pKa2 of 9.96.

The bigger picture

A pKa reported without an ionic strength is an incomplete number. The value you just computed, 9.84, is the thermodynamic constant, because the activity coefficients removed the medium effect. Had you stopped at 9.73 you would have a conditional constant that is only valid in 0.1 M KNO3 - still useful, but it belongs with the ionic strength stapled to it. This is exactly why analytical papers quote constants as "pKa = 9.84 (mu = 0.1 M, 25 C)" rather than as a bare number, and why buffer tables in biochemistry, where mu is often 0.15 M, disagree slightly with the physical chemistry tables that extrapolate to infinite dilution.