# Ksp

Solubility Equilibria | Precipitation.

Tip: This isn't the place to ask a question because the teacher can't reply.

1 of 3 videos by Dr. Hayek

## Key Questions

• ${K}_{s p}$ is called solubility product constant, or simply solubility product. In general, the solubility product of a compound represents the product of molar concentrations of ions raised to the power of their respective stoichiometric coefficients in the equilibrium reaction.

Here's an example to better demonstrate the concept. Let's consider the saturated solution of silver chloride ($A g C l$), where an equilibrium exists between the dissolved ions and undissolved silver chloride according to the following reaction:

$A g C {l}_{\left(s\right)} r i g h t \le f t h a r p \infty n s A {g}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-}$

Since this is an equilibrium reaction, we can write the equilibrium constant for it:

$K = \frac{\left[A {g}^{+}\right] \cdot \left[C {l}^{-}\right]}{\left[A g C l\right]}$. Now, the concentrations of solids are either unknown or assumed to be constant, so this reaction becomes

$K \cdot \left[A g C l\right] = {K}_{s p} = \left[A {g}^{+}\right] \cdot \left[C {l}^{-}\right]$

The magnitude of ${K}_{s p}$ directly indicates the solubility of the salt in water, since ${K}_{s p}$ is derived from the concentrations of ions in equilibrium reactions. Thus, higher concentrations of ions mean greater solubility of the salt.

When trying to write the equation for ${K}_{s p}$, you need to know how to break the compound into ions (identify the monoatomic and polyatomic ions), how many moles of each ion are formed, and the charge on each ion.

• In a saturated solution the solid is in equilibrium with its ions e.g :

$C a C {O}_{3 \left(s\right)} r i g h t \le f t h a r p \infty n s C {a}_{\left(a q\right)}^{2 +} + C {O}_{3 \left(a q\right)}^{2 -}$

The expression for ${K}_{s p}$ is:

${K}_{s p} = \left[C {a}_{\left(a q\right)}^{2 +}\right] \left[C {O}_{3 \left(a q\right)}^{2 -}\right]$

We don't include the concentration of the solid as this is assumed constant.

So if we know the concentration of the ions you can get ${K}_{s p}$ at that particular temperature.

Using ${K}_{s p}$ enables you to find the limit of concentration before a solid forms.

An interesting example of this occurs in nature. The organisms in coral can extract calcium and carbonate ions from seawater by active transport. They then excrete them such that their local concentrations cause the value of ${K}_{s p}$ to be exceeded. This shifts the equilibrium to the left causing the calcium carbonate to deposit on the coral.

• ${K}_{s p}$ is related to molarity as follows:

For a saturated solution where an ionic solid such as silver chloride is in equilibria with its aqueous ions we can write:

$A g C {l}_{\left(s\right)} r i g h t \le f t h a r p \infty n s A {g}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-}$

For which ${K}_{s p} = \left[A {g}_{\left(a q\right)}^{+}\right] \left[C {l}_{\left(a q\right)}^{-}\right]$

This relationship is temperature dependent.

## Questions

• · 1 week ago
• · 3 weeks ago
• · 1 month ago
• · 2 months ago
• · 2 months ago
• · 3 months ago
• · 6 months ago
• · 7 months ago
• · 7 months ago
• · 7 months ago
• · 8 months ago
• · 8 months ago
• · 8 months ago
• · 8 months ago
• · 9 months ago
• · 10 months ago
• · 10 months ago
• · 10 months ago
• · 10 months ago
• · 10 months ago
• · 11 months ago
• · 11 months ago
• · 11 months ago
• · 11 months ago
• · 11 months ago
• · 12 months ago
• · 12 months ago
• · 1 year ago
• · 1 year ago
• · 1 year ago
• · 1 year ago
• 1 year ago
• · 1 year ago
• · 1 year ago
• · 1 year ago
• · 1 year ago
• · 1 year ago
• · 1 year ago
• · 1 year ago
• 1 year ago