Chemistry 3.6
Demonstrate understanding of equilibrium principles in aqueous systems, External, 5 credits
Solubility
There is no such thing as an insoluble substance. A small amount of substance will always dissolve. The correct term for a substance that dissolves in this way is sparingly soluble.
Similar to equilibrium constants that are written to describe the concentrations of aqueous/gaseous species in an equilibrium reaction, Solubility Constants can be written to show the amount of substance that dissolves:
Example:
CaCO_3(s)\rightleftharpoons{Ca}^{2+}(aq)+CO_3^{2-}(aq)
K_s=[Ca^{2+}(aq)][CO_3^{2-}(aq)], where [ion] denotes the concentration of the ion in aqueous solution.
A large K_s value represents a highly soluble substance, while a small K_s value represents a sparingly soluble substance.
s represents solubility in mol L^{-1}, and the K_s value can be found in two ways, depending on the ratio of ions in the final solution:
For a 1:1 ratio of ions, K_s=s²
For a 2:1 or 1:2 ratio of ions, K_s=4s³
These can be rearranged to find the solubility of each ion.
In many cases, solubility will be given in gL^{-1}, so this must be converted to molL^{-1}.
molL^{-1}=\frac{gL^{-1}}{gmol^{-1}}
A Precipitate is a solid suspended in aqueous solution, which forms from two liquid substances. Precipitation occurs when the ionic product is larger than the solubility constant of the reaction. This means that the equilibrium will be as such that the solid formed will not dissolve.
Example:
200mL of 0.001molL^{-1} Na_2SO_4 is mixed with 200mL of 0.001molL^{-1} BaCl_2. K_s(BaSO_4)=1.1×10^-10.
BaSO_4 (s)\rightleftharpoons{Ba^{2+}(aq)}+SO_4^{2-}(aq)
K_s=[Ba^{2+}(aq)][SO_4^{2-}(aq)]
[Ba^{2+}]=\frac{0.001}{2}=5×10^{-4}molL^{-1}
[SO_4^{2-}=\frac{0.001}{2}=5×10^{-4}molL^{-1}
IP=(5×10^{-4})²=2.50×10^{-7}
IP>K_s so a precipitate will form.
The Common Ion Effect describes how a compound is less soluble when in the presence of another compound with a common ion (two identical ions). This is because the ionic product will increase (due to higher concentration of the ion), so it will become larger than the solubility constant, causing precipitation.
Example:
100mL of 0.002moL^{-1} \space CaSO_4 and 100mL of 0.2molL^{-1} \space Na_2SO_4 are mixed. K_s(CaSO_4)=2.4×10^{-5}.
K_s=[Ca^{2+}(aq)][SO_4^{2-}(aq)]
[Ca^{2+}]=\frac{0.002}{2}=0.001molL^{-1}
[SO_4^{2-}]=\frac{0.002}{2}+\frac{0.2}{2}=0.101molL^{-1}
IP=0.001×0.101=1.01×10^{-4}
IP>K_s so a precipitate will form.
It is important to note that in basic solutions, there will be OH^- ions and in acidic solutions there will be H_3O^+ ions present, which will cause the common ion effect to occur.