# What does Ksp in chemistry stand for?

Jan 15, 2018

${K}_{\text{sp}}$..where $\text{sp "-=" solubility product}$

#### Explanation:

This is just another equilibrium expression, and represents the solubility of a sparingly soluble or insoluble salt, $M {X}_{n}$, in some solution (usually water)....

For a binary salt....say $M {X}_{2}$, we would represent the solubility equilibrium as:

$M X \left(s\right) r i g h t \le f t h a r p \infty n s {M}^{+} + 2 {X}^{-}$

As for any equilibrium, we can write (and quantify) this equilibrium:

$\frac{\left[{M}^{+}\right] {\left[{X}^{-}\right]}^{2}}{\left[M X \left(s\right)\right]}$ $=$ ${K}_{s p}$

But $\left[M X \left(s\right)\right]$ is meaningless, as you cannot have the concentration of a solid, so we are left the solubility expression:

${K}_{s p} = \left[{M}^{+}\right] {\left[{X}^{-}\right]}^{2}$, i.e. $\text{the so-called solubility product}$

${K}_{s p}$ values have been measured for a great variety of insoluble and sparingly soluble salts . Why? Because suppose you were working with solutions of precious metal salts, i.e. those of gold, or rhodium, or iridium. You don't want to throw precious metals away. Likewise, if you had lead, or cadmium, or mercury salts, you don't want to throw these metals away, for the reason that you might poison the waterways.

${K}_{s p} , \text{ lead chloride } = 1.62 \times {10}^{-} 5$ at $25$ ""^@C. A temperature is specified because a hot solution can normally hold more solute than a cold one.

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

And, ${K}_{s p} = \left[P {b}^{2 +}\right] {\left[C {l}^{-}\right]}^{2} = 1.62 \times {10}^{-} 5$.

If we say $\left[P {b}^{2 +}\right] = S$, then ${K}_{s p} = \left(S\right) {\left(2 S\right)}^{2}$.

i.e. ${K}_{s p} = 4 {S}^{3}$.

And thus $S$ $=$ ""^3sqrt{{(1.62xx10^-5)/(4)} $=$ ??*mol*L^-1.

I leave it to you to solve for the solubility of lead chloride in water in $g \cdot {L}^{-} 1$ under standard conditions.