[Alcoid]

Explain, with a smaller volume of which of the solutions, the particle can be coagulated:
0,05М Pb(SO4)2
0,003М SnCl2
0,25М PbCl2
0,07М CrCl3
0,03М Cr(NO3)2
0,0006М NaCl
0,02М MgCl2
0,07М AlCl3
0,11М BaCl2
0,05М Li3PO4
0,04М SnCl4
0,0004М Ba2XeO6
1М AsCl5
0,005М BeBr2
0,006М CaBr2

A particle of Prussian blue micelle { [mFe4[Fe(CN)6]3 ] nFe3+ (3n-x)Cl- } X+ xCl- * yH2O, it was diluted in 80 ml of distilled water to obtain a sol

[Borek]

You have to show your efforts to receive help, this is a forum policy.

[Alcoid]

My solution is this: that first you need to exclude non-electrolytes, that is, Pb (SO4) 2 Li3PO4 and Ba2XeO6 because they are insoluble, then we look at the charge of the granule - it is +, therefore, for coagulation, anions with the highest oxidation state are needed, but since we have everywhere CL and Br, then there is no particular choice, all have a charge of -1; And in theory it is necessary to start from the molar concentration, but I do not know correctly or not, I am arguing - if we assume that we take all solutions with an equal volume, say 100 ml, and calculate the coagulation threshold, NaCl will have the smallest one. That is, its smaller volume should be taken, since it has the lowest threshold? Sorry for the English, this is not my native language

[Alcoid]

You have to show your efforts to receive help, this is a forum policy.

thanks for the clarification

[Alcoid]

We have 0.05M Pb (SO4) 2
0.003M SnCl2
0.25M PbCl2
0.07M CrCl3
0.03M Cr (NO3) 2
0.0006M NaCl
0.02M MgCl2
0.07M AlCl3
0.11M BaCl2
0.05M Li3PO4
0.04M SnCl4
0.0004М Ba2XeO6
1M AsCl5
0.005M BeBr2
0.006M CaBr2
To begin with, we exclude all non-electrolytes, since they do not have a coagulating ability, - Pb (SO4) 2 Li3PO4 Ba2XeO4 AsCl5 PbCl2. The following solutions remain 0.003M SnCl2 0.07M CrCl3 0.03M Cr (NO3) 2 0.0006M NaCl 0.02M MgCl2 0.07M AlCl3 0.11M BaCl2 0.04M SnCl4 0.005M BeBr2 0.006M CaBr2. The coagulating ability is possessed by an electrolyte ion, which is opposite to the charge of the granule, that is, it coincides with the sign of the PI charge. Of all the electrolytes, there are Cl- Br- and NO3- anions, according to the Schulze-Hardy rule, the coagulating ability of the coagulant ion is the greater, the greater the charge of the ion, but since we all have the same charge, all the anions have the same coagulating ability. it follows that the coagulation threshold is the same for all solutions.

Then from the formula Cpc = Cel * Vel / Vsol + Vel, for the proposed solutions, whose coagulation thresholds are the same, it follows that the higher the concentration of the electrolyte, the less V of this solution is required. (proof - let the coagulation threshold for the 1st and 2nd electrolyte be the same and equal to 0.5 mmol / L, and Vsol 5 L. The concentration of the 1st solution is 0.6M, and the 2nd solution is 0.8M, and Vel is taken as X.

for the first solution 0.5 = 0.6 * x / 5 + x, hence x = 25l, that is, at a concentration of -0.6M, you need to take 25 liters of electrolyte
for the second solution 0.5 = 0.8 * x / 5 + x, hence x = 8.3 l, therefore, at a concentration of 0.8M, you need to take 8.3 l
CONCLUSION - the volume and concentration of the values ​​are inversely proportional, with an increase in concentration, less electrolyte volume is required for coagulation)

Therefore, it is necessary to choose a solution with a higher concentration - this is BaCl2 0.11M. Consequently, a particle can be coagulated with a smaller volume of BaCl2 with a concentration of 0.11M.

[Alcoid]

I think this is most correct