Dear all,

probably it is a silly question.

I need to create a solution with a fixed quantities of N, P, K starting from different sources: NH4NO3, (NH4)3PO4, KNO3.

Namely, I have to reach the concentration 4,28M for N, 0,846M for P and 1,275M for K.

I am starting my calculation, but I suppose that I need partial volume tables of this system to take account volume variation during the mixing.

Apparently it is a silly exercise but when we analyze our solution we observed strong deviation from the ideal solution.

As anyone performs this kind of solution? It is the creation of “liquid fertilizer” from solid ones. I want to generalize this calculation to different concentration to implement in a project that I follow.

Andrew PelEnlightened

# How to create multicomponent fixed quantity solution of inorganic salts?

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## John Wick

This was my approach, but the solution are quite concentrate, in the order of 10% w/w. So the variation of volume is necessary to take in account. Of course, I reached experimentally the correct amount by successive approximations but I would to generalize the procedure (of course, taking account instability region)

A)

The volume cannot vary in my method

A)

The solubilities for the NH4NO3, (NH4)3PO4, KNO3 salts in their joint/mixed aq. solutions can be predicted after those for each single salt saturated solution. The last are readily available in the literature. That may be achieved as follows:

i) Being SNH4NO3 the solubility for NH4NO3 given in terms of molarity for its single salt saturated solution, KspNH4NO3 = [NH4+]*·[NO3-]* = (SNH4NO3)2; where C*NH4NO3 = [NH4+]* = [NO3-]* = SNH4NO3. Here C stands for formality and saturation is denoted by *.

ii) Simillarly, for a (NH4)3PO4 single salt saturated solution: Ksp(NH4)3PO4 =([NH4+]*)3·[PO43-]*.

iii) While for a KNO3 single salt saturated solution: KspKNO3 = [K+]*·[NO3-]* = (SKNO3)2; where C*KNO3 = [K+]* = [NO3-]* = SKNO3.

iv) For NH4, cation hydrolysis translates as: NH4+ + 2H2O ⇌ NH4OH + H3O+. For NH4OH ⇌ NH4+ + OH-, KbNH4OH = 1.80·10-5 = [NH4+]·[OH-]/[NH4OH]. For the hydrolysis equilibrium: KhNH4+= [NH4OH]·[H3O+]/[NH4+] = Kw/KbNH4OH = 5.56·10-10 (Kw = 1.00 x 10-14); units were omitted. Ammonium balance writes: C(NH4)3PO4 = 3·{[NH4+] + [NH4OH]}.

v) For PO43-, anion hydrolysis writes: PO43- + H2O ⇌ HPO42- + OH-; KhPO43-= Kw/Ka3H3PO4 = 2.4·10-2, where Ka3H3PO4 stands for the 3rd dissociation constant of H3PO4. Phosphate balance writes: C(NH4)3PO4 = [PO43-] + [HPO42-].

vi) Species H2PO4-, H3PO4, NH3, and HNO3, can be neglected.

vii) Unlike the solubility’s for single salt solutions; KspNH4NO3, Ksp(NH4)3PO4, and KspKNO3 can be accepted to apply also to the joint salt solution. In this case, common ion effects should be accounted for.

viii) The solubility of each salt in the joint salts solution, also given in terms of molarity (here denoted by S†NH4NO3, S†(NH4)3PO4, and S†KNO3), can be predicted considering that for saturated joint (†) salt solutions we should have: S†NH4NO3 = C*NH4NO3, S†(NH4)3PO4 = C*(NH4)3PO4, and S†KNO3 = C*KNO3.

ix) Nitrate balance for the saturated joint salts solution writes: C*NH4NO3 + C*KNO3 = [NO3-].

x) Ammonium balance for the saturated joint salts solution writes: C*NH4NO3 + C*(NH4)3PO4 = 3·{[NH4+] + [NH4OH]}.

xi) Phosphate balance for the saturated joint salts solution still writes: C*(NH4)3PO4 = [PO43-] + [HPO42-].

xii) Each salt (interdependent) solubilizes can now be predicted for their joint solution. There are two degrees of freedom that can be fixed. We can arbitrarily fix the concentration of two of the salts ― below the saturation limit at each salt single solution ― to then predict the solubility of the other salt in the joint solution.

xiii) The considered salts are strong electrolytes, so that they can be taken as completely dissociated.

xiv) For this kind of prediction, the available solubility data must often be first converted for M ≡ mol/dm3 (solute : solution) units.

xv) Activity coefficients (and related ionic strength effect) were not ― but can be ― accounted for.

## aron

Assuming that the desired end concentrations are possible given the starting materials and that they are compatible in solution at those concentrations, could you not weigh the calculated masses into a container and dilute to the target volume?

## andrew

The volume cannot vary in my method

## avery

This was my approach, but the solution are quite concentrate, in the order of 10% w/w. So the variation of volume is necessary to take in account. Of course, I reached experimentally the correct amount by successive approximations but I would to generalize the procedure (of course, taking account instability region)