So the equilibriumĬoncentration of fluoride anion must be 4.2 times 10 Therefore the equilibriumĬoncentration of the fluoride anion would just be twice There's twice as many fluoride ions in solutionĪs there are Ca2+ ions. We're looking at the disillusion equation, the mole ratio of Ca2+ Plug in the equilibrium concentration of fluoride anion. So here's our expression with 2.1 times 10 to the negative 4th plugged in, and next we need to So that can be plugged in for the equilibrium concentration of Ca2+. For a Ksp expression, theseĪre equilibrium concentrations, and we already know theĬoncentration of Ca2+ at equilibrium is 2.1 timesġ0 to the negative 4th. So Ksp is equal to, there'sĪ one as a coefficient in front of Ca2+ so it'dīe the concentration of Ca2+ raise to the first power, times the concentration of fluoride anion and since there's a two as a coefficient, this is the concentration The balanced equation to write the Ksp expression. Write Ca2+ in aqueous solution and to balance everything So we would write CaF2 solid,Īnd we know that calcium forms a 2+ cation, so we would The first step is to write out the dissolution equation for calcium fluoride. The Ksp for calcium fluoride at 25 degrees Celsius. Reached and the equilibrium concentration of Ca2+ ions is measured to be 2.1 times 10 to the negative 4th M. Let's say we have some solidĬalcium fluoride that we add to pure water at 25 degrees Celsius. Has the highest solubility out of these three salts. Of these ions at equilibrium, which means that more of the The higher the value for Ksp, the higher the concentration For some insight into why this is true, let's look at the KspĮxpression for silver chloride, which we can get from The highest Ksp value of these three, silver chloride is Produce the same number of ions, the higher the value of Ksp, the higher the solubility of the salt. And for silver iodide, it's 8.3 times 10 to the negative 17th. Times 10 to the negative 10th, for silver bromide, it's 5.0 Here are the Ksp valuesįor the three salts at 25 degrees Celsius. Relative to the other three by comparing Ksp values. Three ions in solution, we can determine its solubility Would give one Pb2+ ion and two chloride anions in solution. However, a salt like Lead(II)chloride produces three ions in solution. And we could write out similarĮquations for silver bromide and silver iodide, so they all One Cl- ion for a total of two ions in solution. Let's look at the dissolutionĮquation for silver chloride to see why this is true. For example, silverĬhloride, silver bromide, and silver iodide all Predict the relative solubilities of salts that produce the same Would be moles per one liter or you could just write M. Number of moles of the solid that dissolve to form one litter Usually the units for solubilityĪre in grams per liter. Refers to the amount of solid that dissolves to formĪ saturated solution. Like sodium chloride, that indicates a soluble salt that dissolves easily in water. If the Ksp value is greater than one, like it is for something When the Ksp value is much less than one, that indicates the salt For example, at 25 degrees Celsius, the Ksp value for barium sulfate is 1.1 times 10 to the negative 10th. Ksp has only one value for a given salt at a specific temperature. For solubility equilibria, we would write Ksp where sp Out of equilibrium constant expressions, we would not Raised to the first power times the concentration of sulfate also raised to the first power. So we would write theĮquilibrium constant K is equal to the concentration of Ba2+ and since there's a coefficient of one in the balanced equation, it'd be the concentration And from the balanced equation, we can write an equilibriumĬonstant expression. Shows the dissolution of a salt barium sulfate. Ions and sulfate anions solution are constant. And when the system is at equilibrium, the concentrations of Ba2+ These types of equilibria are referred to as solubility equilibria. Is equal to the rate of precipitation, the And it's possible for the Ba2+ ion to combine with the sulfateĪnion to form a precipitate, of barium sulfate. So barium sulfate canĭissolve to form Ba2+ ions and sulfate anions in solution. Here sitting on the bottom of the beaker. Sulfate remains undissolved and so we'll draw that So we're gonna form some Ba2+ ions and some sulfate anions. A small amount of the barium sulfate dissolves in the water andįorms Ba2+ ions in solution and sulfate ions in solution. And to the beaker, weĪdd some barium sulfate. Let's say we have a beaker of distilled water at 25 degrees Celsius.
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