yego.me
💡 Stop wasting time. Read Youtube instead of watch. Download Chrome Extension

Worked example: Calculating solubility from Kₛₚ | Equilibrium | AP Chemistry | Khan Academy


3m read
·Nov 10, 2024

  • [Instructor] Let's calculate the molar solubility of calcium fluoride if the Ksp value for calcium fluoride is 3.9 times 10 to the negative 11th at 25 degrees Celsius.

The first step is to write the dissolution equation for calcium fluoride. So, solid calcium fluoride will dissolve in solution to form aqueous calcium two plus ions and fluoride anions. And to balance that out, we need to make sure and include a two in front of the fluoride anions.

The next step is to set up an ICE table, where I stands for initial concentration, C stands for the change in concentration, and E stands for equilibrium concentration. Before any of the solid calcium fluoride dissolves, the initial concentrations of calcium two plus ions and fluoride anions in solution is zero. So we can go ahead and put a zero in here for the initial concentration of the ions in solution.

Some of the calcium fluoride will dissolve, and we don't know how much. So I like to represent that by writing -X on the ICE table, where X is the concentration of calcium fluoride that dissolves. Looking at the mole ratios, it's a one-to-one mole ratio between calcium fluoride and calcium two plus ions. So if we're losing X for the concentration of calcium fluoride, we must be gaining X for the concentration of calcium two plus ions.

And since it's a one-to-two mole ratio for calcium two plus ions to fluoride anions, if we're gaining +X for calcium two plus, we must gain plus +2X for fluoride anions. So the equilibrium concentration of calcium two plus ions is zero plus X, or just X, and the equilibrium concentration of fluoride anions will be zero plus 2X, or just 2X.

The next step is to write the Ksp expression from the balanced equation. So Ksp is equal to the concentration of calcium two plus ions, and since there's a coefficient of one in the balanced equation, that's the concentration of calcium two plus ions raised to the first power, times the concentration of fluoride anions, and since there is a coefficient of two in the balanced equation, it's the concentration of fluoride anions raised to the second power.

Pure solids are not included in equilibrium constant expression. So we're going to leave calcium fluoride out of the Ksp expression. The concentration of ions in our Ksp expression are equilibrium concentrations. Therefore we can plug in X for the equilibrium concentration of calcium two plus and 2X for the equilibrium concentration of fluoride anions.

We can also plug in the Ksp value for calcium fluoride. So that would give us 3.9 times 10 to the negative 11th is equal to X times 2X squared. Next we need to solve for X. So, 3.9 times 10 to the negative 11th is equal to X times 2X squared.

Well, 2X squared is equal to 4X squared times X is equal to 4X cubed. So to solve for X, we need to divide both sides by four and then take the cube root of both sides. So we'd take the cube root of the left side and the cube root of X cubed. That gives us X is equal to 2.1 times 10 to the negative fourth.

And looking at our ICE table, X represents the equilibrium concentration of calcium two plus ions. So 2.1 times 10 to the negative fourth molar is the equilibrium concentration of calcium two plus ions. For the fluoride anions, the equilibrium concentration is 2X. So two times 2.1 times 10 to the negative fourth is 4.2, let me go ahead and write that down here, 4.2 times 10 to the negative fourth molar for the equilibrium concentration of fluoride anions.

Our goal was to calculate the molar solubility of calcium fluoride. And molar solubility refers to the concentration of our salt that dissolved to form a saturated solution at equilibrium. So if X refers to the concentration of calcium two plus ions at equilibrium, looking at our mole ratios, that's also the concentration of calcium fluoride that dissolved.

Therefore, 2.1 times 10 to the negative fourth molar is also the molar solubility of calcium fluoride. Technically at a constant temperature of 25 degrees, the concentration of a solid doesn't change. And so you'll see most textbooks not to put in -X on the ICE table. I like to just put it in though to remind me that X in this case does refer to the molar solubility.

More Articles

View All
Ideas, Products, Teams, and Execution with Dustin Moskovitz (How to Start a Startup 2014: Lecture 1)
Welcome! Can I turn this on? Baby, all right. Hit people here. Can you guys hear me? Is the mic on? No? Maybe you can ask them to turn it on. Maybe we can get a big—there we go. All right! Maybe we can get a bigger auditorium; we’ll see. So welcome to CS…
5 habits that make you feel incredible
Maybe it’s been a while since you felt like you’ve had your mojo. Maybe you remember times in the past, maybe several years ago, where you actually felt incredible, a lot more incredible than you do now. I feel like all of us, at some point in our lives, …
Business Lessons From The Ancient Greeks
Business today is much more than just making profits. In fact, companies are now expected to have a positive impact on society, the environment, and the lives of their employees. And what better way to do that than through philosophy? One of the best ways…
Help support Khan Academy
Hi everyone, Sal Khan here from Khan Academy, and I just wanted to remind you that we are a not-for-profit, and we can only exist through donations from folks like yourself. Our goal is for everyone to reach their potential. Potential is everywhere; unfo…
Multivariable chain rule and directional derivatives
So in the last video, I introduced the vector form of the multivariable chain rule. Just to remind ourselves, I’m saying you have some kind of function f, and in this case, I said it comes from a 100-dimensional space. You might imagine, well, I can’t im…
Save the Ocean, Save Ourselves | Sea of Hope: America's Underwater Treasures
There’s been this arc to my career in the sense that in the beginning I just wanted to make beautiful pictures. But I began more and more to see all these problems happening in the ocean. Fewer fish in the places I used to see many fish, or not as many sh…