THE RUDOLF REPORT

6. FORMATION AND STABILITY OF IRON BLUEan appropriate metal salt solution with Li 4 [Fe(CN) 6 ], probably acquiringa high rate of inclusions (lithium, water) as well. Thus, in spite ofthe four hour-long accumulation of the precipitation, the filtrate wouldcertainly still have contained colloidally dispersed Iron Blue. Sincethey finally determined the amount of free Fe 3+ in the filtrate by precipitatingit with ammonia as Fe(OH) 3 , they will undoubtedly alsohave precipitated the Fe 3+ of the colloidally dispersed Iron Blue, asammonia raises the pH value so much that Iron Blue is no longer stable(see chapter 6.6.1.).Therefore, they did not determine the solubility of Iron Blue, butthe measure of stability of the dispersion of fresh precipitations of thepigment.The solubility product of Pb 2 [Fe(CN) 6 ] given by Krleza et al. , 336which they used as a reference to determine the solubility products, ismuch lower than the one used by Tananaev et al.. If applied to Tananaev’scalculations, this produces a solubility of Iron Blue of only 0.05mg per liter. Krleza et al., however, find similar results for the solubilityof most of the metal cyanides analyzed, including Iron Blue. Sinceconventional methods of analysis, such as gravimetry and titration,tend to be unreliable when facing minute traces, one must but wonderabout these similar results.However, one can escape this dilemma by thoughtful reasoning.It is safe to say that Iron Blue is stable at a pH value of 7, i.e., in aneutral aqueous medium, so we take this as a minimum value. As mentionedearlier, a pH value of about 10 can be considered the upper limitof stability for Iron Blue, so we take this as maximum value for thefollowing calculations. At pH=7, and even more so at pH=10, the freeiron concentration is extremely low, since Fe(OH) 3 is nearly insoluble(see Table 5).At pH 7 and 10, respectively, a saturated Fe(OH) 3 solution has thefollowing free Fe 3+ concentration:c(Fe 3+ ) = K L(Fe(OH) 3 )c 3 (OH - )(2)pH=7: = 2.67×10-39 mol 4 l -410 -21 mol 3 l -3 = 2.67×10-18 mol l -1 (3)pH=10: = 2.67×10-39 mol 4 l -410 -12 mol 3 l -3 = 2.67×10-27 mol l -1 (4)173