PhreeqcUsers Discussion Forum

Registrations currently disabled due to excessive spam. Please email phreeqcusers at gmail.com to request an account.
Welcome Guest
 

  • Forum Home
  • Login
  • Register

  • PhreeqcUsers Discussion Forum »
  • Beginners »
  • PHREEQC basics »
  • Calcium hydroxyphosphate(Ca5(PO4)3OH)
« previous next »
  • Print
Pages: [1]   Go Down

Author Topic: Calcium hydroxyphosphate(Ca5(PO4)3OH)  (Read 1454 times)

melaniewalker

  • Contributor
  • Posts: 1
Calcium hydroxyphosphate(Ca5(PO4)3OH)
« on: 11/08/23 11:27 »
Thank you for reading my post. I'm now  using phreeqc to simulate precipitation in urine. It's basically adding calcium to an ammonia buffer and to forming Calcium hydroxyphosphate(Ca5(PO4)3OH). The initial phosphorus concentration is 0.3g/L and the initial ammonium is 7g/L. But when I add 3g/L Ca(OH)2 in the solution and the pH up to about 8.5, the concentration of PO4(3-) is still high(0.14g/L), which in my actual experiment the phosphorus concentration will reach to a very low level.

I checked the difference, found that the logk of Ca5(PO4)3OH is -3.42 and the literature reported that the log k of  Ca5(PO4)3OH is -57.536. So  this may be the cause of the error between the model and the reality.
Now I'm confused. How can I solve this problem.

Thank you for answer me!
Logged

dlparkhurst

  • Global Moderator
  • *****
  • Posts: 4034
Re: Calcium hydroxyphosphate(Ca5(PO4)3OH)
« Reply #1 on: 11/08/23 15:38 »
First, you need to consider the chemcial equations for each of the log K values.

Here is the definition in the phreeqc.dat database. It is written in terms of the HPO4-2 ion:

Hydroxyapatite
   Ca5(PO4)3OH + 4 H+ = H2O + 3 HPO4-2 + 5 Ca+2
   -log_k    -3.421
   -delta_h -36.155 kcal
   -Vm 128.9

Here is the definition in the sit.dat database. It is written in terms of the H2PO4- ion:

Hydroxyapatite
Ca5(OH)(PO4)3     = 5.000Ca+2     - 7.000H+     + 3.000H2(PO4)-     + 1.000H2O
     log_k    14.350     #84NAN
     delta_h -178.487    #kJ/mol       
     # Enthalpy of formation:           -6730.143        #kJ/mol       
     -analytic -1.69195E+1 0E+0 9.32301E+3 0E+0 0E+0

And here is the reaction in minteq.v4.dat. It is written in terms of PO4-3:

Hydroxylapatite
   Ca5(PO4)3OH + H+ = 5Ca+2 + 3PO4-3 + H2O
   log_k   -44.333
   delta_h   -0   kJ

To compare these definitions you will have to convert to an ion of your choice by adding the PO4 reactions and log Ks. The following from sit.dat could be used to convert sit.dat to the phreeqc.dat form:

- 1.000H+     + 1.000H2(PO4)-     = HPO4-2
     log_k    -7.210     
     delta_h   3.600     #kJ/mol       
     # Enthalpy of formation:           -1299        #kJ/mol        #89COX/WAG
     -analytic -6.57931E+0 0E+0 -1.88041E+2 0E+0 0E+0

Note there may be a different log K for this reaction in each of the databases, so that would add some uncertainty when you are trying to compare the values.

More important, there is just a large range in log K estimates for apatite even after you convert them to a common reaction. The range is larger than the uncertainties caused by the various log Ks for PO4-3-> HPO4-2 -> H2PO4- reactions used in the conversions.

So, I don't have a good answer. You will have to evaluate which literature value is most accurate and most appropriate for your experiment. You could estimate a log K from your experiments and compare it to the literature values.
Logged

  • Print
Pages: [1]   Go Up
« previous next »
  • PhreeqcUsers Discussion Forum »
  • Beginners »
  • PHREEQC basics »
  • Calcium hydroxyphosphate(Ca5(PO4)3OH)
 

  • SMF 2.0.19 | SMF © 2021, Simple Machines | Terms and Policies
  • XHTML
  • RSS
  • WAP2