Our episode starts with addressing the question: What IS solubility?
Our episode starts with addressing the question: What IS solubility? (1:19). The episode introduces solubility equilibria with respect to the Solubility-Product Constant, K, expression (1:57) and discussion how its magnitude relates to solubility (3:05). If you have the Ksp you can calculate the molar solubility - in mol/L - as well as the mass solubility in g/L (4:35). Comparing K to Q let’s you determine if a precipitate forms (5:58).
Question: How does a common ion affect the solubility of a salt?
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Hi and welcome to the APsolute Recap: Chemistry Edition. Today’s episode will recap Solubility Equilibria.
Lets Zoom Out:
Unit 7 - Equilibrium
Topic 7.11 - Introduction to Solubility Equilibria
Big idea - Scale, Proportion, and Quantity
Most of our listeners will be too young to have experienced it, but let this episode be a warning: Kidney stones! Kidney stones are a crystalline solid that can form in your urinary tract and they HURT. Big time. One of the main components of kidney stones is calcium phosphate, which is, of course, insoluble in water. But is it really? How absolute is that statement of insolubility? Well… you know, Chemistry is rarely black and white. There is a lot of grey.
Let’s zoom in:
What does being soluble ACTUALLY mean? Solubility is how much of an ionic compound will dissolve and form a saturated solution. Substances that we label as “insoluble” are usually not completely insoluble, but they dissolve to such a small amount that we say it is pretty much insoluble. There are different cut off values. One, for example, is - if less than 0.01 mol of the compound dissolves in 1 L, then it is insoluble. And then there are, of course, slightly soluble salts and and and…
When we put our salt into water, the dissolution reaches equilibrium: The rate of the formation of aqueous ions and solid precipitate is equal. To quantify how much of a salt dissolves, we can use the Solubility-Product Constant, (Capital K subscript sp)Ksp. As discussed in episode 32, the expression of the constant for this equilibrium is the concentration of products to their coefficients divided by the concentration of reactants to their coefficients. But let’s use an example: Our calcium phosphate. Writing the equation of the dissolution, we have: solid calcium phosphate in equilibrium with three Ca2+ (two plus) ions and two phosphate ions. Writing the expression for the solubility product constant we have: the concentration of Ca2+ to the third times the concentration of phosphate to the second. That’s it. Why not dividing it by the solid calcium phosphate? … Exactly: solids and pure liquids are not part of the equilibrium expression.
Kidney stones wouldn’t be an issue if they were soluble. So the Ksp value for calcium phosphate must be really small, indicating that the concentration of aqueous Calcium cations and Phosphate anions is really, really low. Looking it up, we actually find a tabulated value of 2.07 x 10-33 at 25℃. That is TIIINY. As a guideline, we say that salts with a Ksp greater than 1 are soluble.
If you have the Ksp you can calculate the molar solubility - in mol/L - as well as the mass solubility in g/L. It is important here that you can distinguish between those two and that you keep track of your units. UNITS ARE FRIENDS! They help us! Don’t neglect them!
First step: Write the balanced equation for the dissolution of the compound. Second step: Write the equilibrium expression for the solubility-product constant. Don’t forget: No solids! Third step: Set up our RICE or ICE table, with an initial concentration of 0 for the aqueous ions, use your stoichiometric coefficients for the change, set up your equilibrium concentrations in terms of x. Then plug it into your expression and calculate your “x”. Follow these steps carefully. Students often get confused because the coefficient shows up twice: As superscript in the equilibrium expression as well as in the “change” of our RICE table! On your study guide, do this for calcium phosphate! Since we are using RICE tables, our x is in terms of Molarity - mol/L. To get the solubility in g/L, we have to convert our moles to grams by multiplying with the Molar Mass of the salt.
A common problem in AP Chemistry asks if a precipitate forms when you are combining two aqueous solutions. For this, we can again use our reaction quotient Qsp and compare it to Ksp. Reminder: If Q is smaller than Ksp, then no precipitate forms, my solution is unsaturated. If Q is greater than Ksp a precipitate will form. In your calculations, you start by combining two aqueous solutions. That CHANGES the concentration because you are adding two aqueous solutions. So use M one V one equals M two V two to determine the new ion concentrations! Then you can plug those into your Qsp expression - which is set up the same way as your Ksp expression and calculate the value of Qsp. By comparing it now with Ksp you can determine if the precipitate forms.
To recap:
Solubility is how much of an ionic compound will dissolve and form a saturated solution. In a solution you have an equilibrium between the solid salt and the aqueous ions. To quantify how much of a salt dissolves, we can use the Solubility-Product Constant, Ksp. Using Ksp we can set up the equilibrium expression, a RICE table and calculate our molar solubility. To determine if a precipitate forms when combining two aqueous solutions, we can calculate the reaction quotient Qsp and compare it to Ksp.
Coming up next on the APsolute RecAP Chemistry Edition: Acids and Bases: The Basics
Today’s Question of the day is about the common-ion effect.
Question: How does a common ion affect the solubility of a salt?