Calculate Solubility of Ca5(PO4)3OH: Ksp & g/100mL

  • Thread starter Thread starter higherme
  • Start date Start date
  • Tags Tags
    Solubility
Click For Summary

Discussion Overview

The discussion centers around calculating the solubility of Ca5(PO4)3OH in grams per 100 mL of water, and whether this is equivalent to determining its solubility product constant (Ksp). Participants explore the relationship between solubility and Ksp, as well as the steps necessary to arrive at the solubility value.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant questions whether calculating solubility is the same as finding Ksp, expressing confusion about the units involved.
  • Another participant clarifies that the question is asking for the mass of Ca5(PO4)3OH that can dissolve in 100 mL of water at standard room temperature.
  • A suggestion is made to write down the expression for Ksp as a preliminary step in the calculation.
  • A participant provides a detailed calculation of Ksp and solubility, including the dissociation equation and the derived solubility value in grams per 100 mL.
  • One participant expresses confidence in the correctness of the calculation provided by another, but does not confirm it as definitive.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the relationship between solubility and Ksp, and there is uncertainty regarding the interpretation of the question. Some participants agree on the calculation steps, while others express confusion about the units and concepts involved.

Contextual Notes

There are unresolved assumptions regarding the conditions under which the solubility is calculated, such as temperature and pressure, as well as the interpretation of Ksp in relation to solubility units.

higherme
Messages
126
Reaction score
0
The question is:

Calculate the solubility, in g/100mL, of Ca5(PO4)3OH.

when it asks for solubility, is that the same as asking for the Ksp of Ca5(PO4)3OH??

so basically, i have to find the concentration of Ca2+, PO4^3- and OH- and then find the Ksp ?

BUT, i thought Ksp have no units... how can it be in g/100mL then?

any one help please?
 
Physics news on Phys.org
What it is asking for is "how many grams of [itex]Ca_5(PO_4)_3OH[/itex] can you dissolve in 100 mL of water at standard room temperature?"
 
Try writing down the expression for Ksp as a first step.
 
Ca5(PO4)3OH <----> 5Ca2+ + 3PO4^3- + OH-

ksp = [Ca2+]^5 [PO4)3-]^3 [OH-]
ksp = (5X)^5 (3x)^3 (x)
6.80E-37 = 84375x^9
x = 2.72E-5 M = [Ca5(PO4)3OH ]

(2.72E-5 mol/L)* (502.307 g/mol) = 0.01366275g/L = 0.0136627g/1000mL

divide the top and bottom by 10 to get per 100mL:
0.0136627g/1000mL

1.37 E-3 g/100mL <---- my answer

am i doing it right? thanks
 
That should do it.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 14 ·
Replies
14
Views
13K
Replies
4
Views
2K
Replies
2
Views
2K
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
9K
Replies
1
Views
2K