Tutorial Answers 3

(9)

Periodate is IO^4- and the central I has oxidation state 7. Reaction with the diol gives the cyclic intermediate shown below.

This undergoes an internal redox reaction with rupture of the C-C bond (standard periodate cleavage of 1,2- diols), delivering 2 methanals for each iodate IO₃ ^- (oxidation state 5 for I). Isotopic labelling (¹⁸O) would help to confirm that both O atoms of the diol are retained in the carbonyl compound. Periodate anions have Td resonance structures, so the mechanism shown above seems reasonable. Hydrated forms such as H₅IO₆are known at pH 1-6, and that species has been invoked as the principal reactant.

(10)

Enthalpy of vaporisation is simply [100.0 - 4.0] = 96.0 KJmol^-1.

(11)

Water and ice are in equilibrium at zero degrees Celsius. Hence (ii) is correct.

(12)

Enthalpy of reaction given by [2 x (-285.8 - 187.8)] = -196.0 kJ mol^-1 .

(13)

Tough one. It needs thought! The formula for the work done is (since the external pressure is not changed during the process):

w = - p_ex DV

where pis 1 atm. (the external pressure). The change in volume, DV, is just the volume of COemitted, since we start from a solid. Use ideal gas law (A- level syllabus) to get (at T = 700+273K).

DV = RT/p_ex

Hence,

w = - RT = - 8.31 x 973 = -8.09 kJmol^-1

The case where the piston's off has just the same work, since the evolved gas has to push air molecules out of the way regardless. Any student who gets all this right should be suitably congratulated.

(14)

The pH can be established using the equation given in the course and in this case is of the form:

(15)

The buffer is best when the ratio of acid to conjugate base is 1:1, and this pH corresponds to the pK. Hence, the buffer would be best at pH of 2.29 or 9.72. Rather obvious. In the second part, the equilibrium is

H₂A ====> HA^- + H⁺, K_a = 10^-2.29 = 5.13x10^-3

NB the second ionisation constant is over very much smaller, so further ionisation may be ignored.

So that pH = 2.29