What is the solubility in mol/L of silver iodide, #"AgI"# ?
Its #K_(sp)# value is #8.3*10^-17# .
Its
1 Answer
Explanation:
Silver iodide,
#"AgI"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "I"_ ((aq))^(-)#
Notice that every mole of silver iodide that dissolves produces
The expression for the solubility product constant,
#K_(sp) = ["Ag"^(+)] * ["I"^(-)]#
If you take
#K_(sp) = s * s = s^2#
This means that the molar solubility of this salt will be equal to
#s = sqrt(K_(sp)) = sqrt(8.3 * 10^(-17)) = color(darkgreen)(ul(color(black)(9.11 * 10^(-9)"M")))#
In other words, you can only hope to dissolve