Advanced Thermodynamics   Quiz #5

(Open handouts & note sheet)

  1. OK, since nobody bothered to compute rainwater's natural pH at 25°C, we'll do it here. You'll need to know XCO2(g)=330 ppm, carbonic acid's pKA1=6.35 and pKA2=10.33, and that a saturated solution of CO2(aq) is 0.145 g CO2 / 100 cc.

KA1 = [H+][HCO3-]/[H2CO3]

And if we assume that the KA2 equilibrium is unimportant (due to the vast disparity in their K values), we know [H+] = [HCO3-] = x and

x2 = KA1×[H2CO3]

Now we need [H2CO3], and we don't even have to agonize over whether or not to include (initial value)-x since it's concentration is maintained throughout by its equilibrium with PCO2 in the air. Indeed the latter fixes [H2CO3] at its saturated value scaled by the ratio of the actual PCO2 in ratio to 1 atm (from which the saturated solution is taken). That's Raoult's Law (for ideal gases over ideal solutions).

The pressure ratio is the same as the mole fraction, since air is an ideal gas mixture, so

[H2CO3] = XCO2(g)×[H2CO3]sat'd
= 3.3×10-4 × 0.145 g CO2 / dL (1 mol CO2 / 44 g CO2)×(10 dL / L)
= 1.09×10-5 M

x2 = 10-6.35 × 1.09×10-5 = 4.86×10-12

x = [H+] = 2.2×10-6 or pH = 5.6

Now, of course, we should check that the [H+] contribution from KA2 is woefully smaller than this, and we did that in class, and it was; so we were right to ignore it.

  • Of course, the pH of neutral water at 25°C is 6.998 (you knew it couldn't be 7 exactly, right?). However, it lowers to 6.133 at 100°C!

    1. Estimate water's enthalpy of ionization in kJ/mol.

    • For neutral water, pH=pOH, and for all water, pH+pOH=pKw. Thus for neutral water, pKw=2pH. So much for freshman chemistry.

    Thermodynamically,

    ln(K2/K1) = - Delta H/R ( 1/T2 - 1/T1 )
    so
    Delta H = - R ln(K2/K1) / ( 1/T2 - 1/T1 ) = - R 2.303(pK1 - pK2) / ( 1/T2 - 1/T1 )

    at least as long as Delta H isn't a strong function of T over the range T1 to T2. Unfortunately, that is the case here since CP(water) is pretty large, and we're changing T by 25%! But we were asked for an estimate.

    Delta H = - 8.3145 J/mol K 2.303(2×6.998-2×6.133) / ( 1/373 - 1/298 )
    = +49.1 kJ/mol

  • Estimate water's pH at 0°C. (It's 7.469 when measured. Why don't we get quite that value here?)

    • Now we want pH = ½ pKw. So we rearrange the above to

    pK0 = pK1 - Delta H/2.303R ( 1/T1 - 1/T0 )
    pH0 = ½ pK0 = ½ pK1 - ½ Delta H/2.303R ( 1/T1 - 1/T0 )
    pH0 = pH1 -½ (+4.91×104 J/mol / 2.303×8.3145 J/mol K) ( 1/298 - 1/273 )
    pH0 = 7.392

    which isn't quite 7.469 since Delta H isn't quite constant over this 100° span. Close though!

    R = 8.3145 J/mol K

    Last modified 19 November 1997.