These is a 1:4 molar ratio between the molar solubility and the iodate ion concentration.Įxample #5: Calculate the K sp for Mg 3(PO 4) 2, given that its molar solubility is 3.57 x 10 -6 mol/L. There is a 1:1 molar ratio between the molar solubility and the cerium(IV) ion concentration. Putting the values into the K sp expression, we obtain: This means that, when 2.52 x 10¯ 8 mole per liter of Hg 2Br 2 dissolves, it produces 2.52 x 10¯ 8 mole per liter of Hg 2 2+, as well as 5.04 x 10¯ 8 mole per liter of Br¯ in solution.
There is a 1:1 ratio between Hg 2Br 2 and Hg 2 2+, BUT there is a 1:2 ratio between Hg 2Br 2 and Br¯. However, it will give the wrong K sp expression and the wrong answer to the problem. IT IS NOT!!! If you decide that you prefer 2Hg +, then I cannot stop you. When Hg 2Br 2 dissolves, it dissociates like this: The F¯ concentration is TWICE the value of the amount of CaF 2 dissolving.Įxample #3: Determine the K sp of mercury(I) bromide (Hg 2Br 2), given that its molar solubility is 2.52 x 10¯ 8 mole per liter. Please note, I DID NOT double the F¯ concentration. This means that, when 2.14 x 10¯ 4 mole per liter of CaF 2 dissolves, it produces 2.14 x 10¯ 4 mole per liter of Ca 2+ and it produces 4.28 x 10¯ 4 mole per liter of F¯ in solution. K sp = (5.71 x 10¯ 7) (5.71 x 10¯ 7) = 3.26 x 10¯ 13Įxample #2: Determine the K sp of calcium fluoride (CaF 2), given that its molar solubility is 2.14 x 10¯ 4 moles per liter.ġ) When CaF 2 dissolves, it dissociates like this:ģ) There is a 1:1 molar ratio between CaF 2 and Ca 2+, BUT there is a 1:2 molar ratio between CaF 2 and F¯. This means that, when 5.71 x 10¯ 7 mole per liter of AgBr dissolves, it produces 5.71 x 10¯ 7 mole per liter of Ag + and 5.71 x 10¯ 7 mole per liter of Br¯ in solution.Ĥ) Putting the values into the K sp expression, we obtain: In like manner, there is a 1:1 molar ratio between dissolved AgBr and Br¯ in solution.
Given this value, how does one go about calculating the K sp of the substance? Here is a skeleton outline of the process:ġ) Write the chemical equation for the substance dissolving and dissociating.ģ) Insert the concentration of each ion and multiply out.Įxample #1: Determine the K sp of silver bromide, given that its molar solubility is 5.71 x 10¯ 7 moles per liter.ġ) When AgBr dissolves, it dissociates like this:ģ) There is a 1:1 molar ratio between the AgBr that dissolves and Ag + that is in solution. In the case of AgBr, the value is 5.71 x 10¯ 7 moles per liter. For very soluble substances (like sodium nitrate, NaNO 3), this value can be quite high, exceeding 10.0 moles per liter of solution in some cases.įor insoluble substances like silver bromide (AgBr), the molar solubility can be quite small. The molar solubility of a substance is the number of moles that dissolve per liter of solution. Although most mercury compounds are now known to be poisonous, eighteenth-century physicians used calomel as a medication.ChemTeam: Calculating the Ksp from Molar SolubilityĬalculating the K sp from the Molar Solubility
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