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This post explains how the molar solubility of a slightly soluble compound is calculated. All that is needed is the solubility product constant value and the balanced dissociation equilibrium equation for the compound in question. The key to understanding this type of chemistry problem is a careful analysis of molar solubility (which I call "S") as it relates to the amount of solid compound dissociated.
The actual amount of slightly soluble compound dissociating is exceedingly small compared to the amount of compound used. Therefore, its "molarity" is considered to be constant and so is excluded from the molar solubility calculation. Because the "molarity" of a slightly soluble solid compound is not part of the calculation, it is natural to also ignore it when it comes to truly understanding the chemistry behind the problem.
An understanding of the chemistry for a molar solubility determination requires us to consider the number of formula units solid compound dissociated compared to the numbers of formula units cations and anions produced. For every "S" formula units of solid compound dissolving there is a fraction, "S", of total formula units ions produced. The mole ratios of the balanced solid compound formula (necessarily equivalent to the mole ratios of the balanced equilibrium equation) are utilized to obtain this fraction, "S", of the total number of ions present.
If we assume one liter of solution, then we are free to think of the molar solubility, "S", as the mole fraction of total ions produced, "S", equivalent to the mole quantity of solid compound dissociated, "S".
But, keeping in mind that the mole is just a convenient counting unit then, with the aid of Avogadro's Number, we are back to considering, "S", in terms of formula unit numbers. So, it seems we are back to where we started: "S" is a particular fraction of total hydrated ion formula units which equals the quantity of formula units solid compound dissociated. But, once again, "S", is also the molar solubility of a general slightly soluble compound: Let's call it MxAy. The following diagram illustrates the calculation of molar solubility for our general compound.
In the process of showing the calculation for molar solubility, we have also produced a formula to obtain the same for any "sufficiently" slightly soluble ionic compound.
So there you have it!: The determination of molar solubility for a slightly soluble ionic compound, given the composition of said compound and its Ksp value.
As always, thank you for reading!
A Publication of http://ExcellenceInLearning.biz
This post explains how the molar solubility of a slightly soluble compound is calculated. All that is needed is the solubility product constant value and the balanced dissociation equilibrium equation for the compound in question. The key to understanding this type of chemistry problem is a careful analysis of molar solubility (which I call "S") as it relates to the amount of solid compound dissociated.
The actual amount of slightly soluble compound dissociating is exceedingly small compared to the amount of compound used. Therefore, its "molarity" is considered to be constant and so is excluded from the molar solubility calculation. Because the "molarity" of a slightly soluble solid compound is not part of the calculation, it is natural to also ignore it when it comes to truly understanding the chemistry behind the problem.
An understanding of the chemistry for a molar solubility determination requires us to consider the number of formula units solid compound dissociated compared to the numbers of formula units cations and anions produced. For every "S" formula units of solid compound dissolving there is a fraction, "S", of total formula units ions produced. The mole ratios of the balanced solid compound formula (necessarily equivalent to the mole ratios of the balanced equilibrium equation) are utilized to obtain this fraction, "S", of the total number of ions present.
If we assume one liter of solution, then we are free to think of the molar solubility, "S", as the mole fraction of total ions produced, "S", equivalent to the mole quantity of solid compound dissociated, "S".
But, keeping in mind that the mole is just a convenient counting unit then, with the aid of Avogadro's Number, we are back to considering, "S", in terms of formula unit numbers. So, it seems we are back to where we started: "S" is a particular fraction of total hydrated ion formula units which equals the quantity of formula units solid compound dissociated. But, once again, "S", is also the molar solubility of a general slightly soluble compound: Let's call it MxAy. The following diagram illustrates the calculation of molar solubility for our general compound.
In the process of showing the calculation for molar solubility, we have also produced a formula to obtain the same for any "sufficiently" slightly soluble ionic compound.
So there you have it!: The determination of molar solubility for a slightly soluble ionic compound, given the composition of said compound and its Ksp value.
As always, thank you for reading!
A Publication of http://ExcellenceInLearning.biz