Solubility, in chemistry, is the ability of a solute (the chemical being dissolved) to achieve dissolution in a solvent (the liquid that the solute dissolves into, often water). When we refer to a compound’s solubility, we usually mean the most that can be dissolved in a given volume (usu. 1L).<\/p>\n
How can you tell if the solute truly has dissolved?\u00a0 Knowing for sure can be tricky, but at the high school level the likely tell of failure to dissolve is fallout at the bottom of the beaker.\u00a0 If you see an accumulation of particles there, then those particles have failed to dissolve.\u00a0 The accumulation is called a precipitate.\u00a0 So if a precipitate forms, the solute has not fully dissolved, although some of it may have.<\/p>\n
A significant part of Chemistry 12 focuses on the solubility of substances that aren’t very soluble. Small concentrations can be dissolved, but how much? What’s the tipping point between a sample’s being fully soluble in a given volume of water, versus a precipitate forming?<\/p>\n
Example 1:<\/strong> The solubility product, ksp<\/em>, of magnesium hydroxide, Mg(OH)2<\/sub>, is 8.9 x 10^(-12). Give the mass that can dissolve in 1.5L of water.<\/p>\n Solution:<\/strong><\/p>\n We approach the problem by writing, first, the dissolution equation:<\/p>\n Mg(OH)2<\/sub> \u21cc Mg2+<\/sup>aq<\/sub> + 2OH–<\/sup>aq<\/sub>, ksp=8.9 x 10^(-12)<\/p>\n ksp<\/em> means solubility product<\/em>. Its expression is as follows:<\/p>\n [Mg2+<\/sup>][OH–<\/sup>]2<\/sup> = 8.9 x10^(-12)<\/p>\n The square brackets mean concentration: moles per litre. The exponent 2 on the OH–<\/sup> term is due to its coefficient 2 in the dissolution equation. Dissolution equations need to be balanced because their coefficients become exponents in the ksp expression.<\/strong><\/p>\n At first, we need to find how many moles of Mg(OH)2<\/sub> will dissolve in 1.0L of water. We use a let statement (so often a good idea when solving word problems):<\/p>\n let x=[Mg2+<\/sup>] Then, from the dissolution equation, 2x=[OH–<\/sup>] We substitute into the ksp expression:<\/p>\n x(2x)2<\/sup> = 8.9 x 10^(-12)<\/p>\n Remember to square the 2 as well as the x:<\/p>\n 4x3<\/sup> = 8.9 x 10^(-12)<\/p>\n Divide both sides by 4:<\/p>\n x3<\/sup>=2.225 x 10^(-12)<\/p>\n Take the cube root:<\/p>\n x=1.31 x 10^(-4) = [Mg2+<\/sup>]<\/p>\n By the dissociation equation, the concentration of Mg2+<\/sup> is equal to the concentration of Mg(OH)2<\/sub>. Therefore, the maximum [Mg(OH)2<\/sub>] that we can expect is 1.31 x 10^(-4) moles\/L.<\/p>\n Now, we need to find how many grams of Mg(OH)2<\/sub> will be present in 1.5L of a 1.31mol\/L solution. (My article from Oct 28<\/a> may be helpful to read). We proceed by finding the molar mass (aka M.M.) of Mg(OH)2<\/sub>:<\/p>\n M.M. Mg(OH)2<\/sub>= 24.3 + 2(16.0) + 2(1.01)= 58.32g<\/p>\n 1.31×10^(-4)mol\/L x 1.5L x 58.32g\/mol = 0.011g<\/p>\n So, apparently, as much as 0.011g Mg(OH)2<\/sub> can dissolve in 1.5L water.<\/p>\n This post has opened several discussions which I’ll continue moving forward:)<\/p>\n Sources<\/em>:<\/p>\n Mortimer, Charles E. Chemistry<\/em>. Belmont: Wadsworth, Inc., 1986.<\/p>\n Hebden, James. Chemistry: Theory and Problems<\/em>, book two. Toronto: McGraw-Hill \u00a0\u00a0\u00a0Ryerson, 1980.<\/p>\n