##K_{sp}## value and solubility

In summary, Ksp is not always an indicator of solubility. You must use the solubility equation for the ionization ratio in order to compare solubility.
  • #1
mooncrater
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Homework Statement


I read this line in my textbook:
"##PbS,CdS## are precipated in dilute solution only due to higher ##K_{sp}##"
And I googled about the relation between solubility and ##K_{sp}## and I found that:
"More is the ##K_{sp}## of a salt more it is soluble in a solution(I think water)"
So are the two statements in accordance to each other?If no then which one is wrong?

Homework Equations

The Attempt at a Solution


I am confused. I do think that one of these lines should be wrong and I would prefer the first one. But I am not sure about it because of which I have asked it here.
 
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  • #2
Like any equilibrium constant, for the reaction aA + bB <===> cC + dD, K = [C]^c[D]^d/ ([A]^a^b).

Therefore, if your reaction is solid <===> solute then a higher Ksp will mean that more solute will dissolve in the solution. (Note that in the equilibrium constant you don't include the solids, so in this case Ksp = [solute])

I'm also confused about the first statement, but I know the above is true
 
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  • #3
x86 said:
Therefore, if your reaction is solid <===> solute then a higher Ksp will mean that more solute will dissolve in the solution.
So, using your statement in this line
mooncrater said:
"##PbS,CdS## are precipated in dilute solution only due to ## higher ## ##K_{sp}##"
I may say that instead of ##higher## it should be ##lower##.
What do you think?
 
  • #4
Yes, that makes sense. If Ksp is really low then there will be very low solubility.
 
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  • #5
You may already know this, but be careful using Ksp as an indication of solubility. You can use Ksp to compare solubility only if the salt ionization ratios are the same. If not, you'll need to use the solubility equation for the ionization ratio. Here's an example where you can get into trouble... Consider single salt saturated aqueous solutions of MgCO3 (Ksp = 1 x 10-5) and BaF2 (Ksp = 1 x 10-6). Is MgCO3 more soluble than BaF2 because it has a larger Ksp-value? NO!

Solubility of MgCO3 = (Ksp)1/2 = (1x10-5)1/2 = 0.0032M b/c MgCO3 has a 1:1 ionization ratio

Solubility of BaF2 = (Ksp/4)1/3 = (1 x 10-6/4)1/3 = 0.0063M b/c BaF2 has a 1:2 ionization ratio

Solubility of BaF2 > Solubility of MgCO3 ... Just an FYI...
Have a good day.:wink:
 
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1. What is the ##K_{sp}## value and how is it related to solubility?

The ##K_{sp}## value, also known as the solubility product constant, is a measure of the maximum concentration of a solute that can be dissolved in a solvent at a given temperature. It is related to solubility because a higher ##K_{sp}## value indicates a higher solubility of the solute in the solvent.

2. How is the ##K_{sp}## value determined?

The ##K_{sp}## value can be determined experimentally by measuring the concentration of the dissolved solute in a saturated solution and using it to calculate the equilibrium constant. It can also be calculated using thermodynamic data for the solute and solvent.

3. Can the ##K_{sp}## value change?

The ##K_{sp}## value is a constant at a given temperature and pressure, but it can change if the temperature or pressure is changed. In general, the ##K_{sp}## value increases with increasing temperature and decreases with increasing pressure.

4. How does the ##K_{sp}## value affect the solubility of a substance?

The ##K_{sp}## value directly affects the solubility of a substance. A higher ##K_{sp}## value indicates a higher solubility, while a lower ##K_{sp}## value indicates a lower solubility. This is because the ##K_{sp}## value is a measure of the equilibrium between the dissolved solute and undissolved solid, and a higher ##K_{sp}## value means that more solute can be dissolved in the solvent.

5. How can the ##K_{sp}## value and solubility be used in practical applications?

The ##K_{sp}## value and solubility are important in various practical applications, such as in industrial processes, pharmaceuticals, and environmental studies. They can be used to determine the appropriate conditions for maximum solubility of a substance, to predict the formation of precipitates in a solution, and to understand the behavior of different compounds in various environments.

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