At 3:42 why do you raise the concentration of the reactants to the power of the coefficient? In what video is that explained?

It is explained in this video set: https://www.khanacademy.org/science/chemistry/chemical-equilibrium

What would you do if you were asked to find the ppm of the cu2+ ion or the OH- ion?

Ppm means: "how many in a million?" as in, "How many grams of Cu²⁺ in a million grams of solution"? We know that [Cu²⁺] = 1.8 × 10⁻⁵ g/L. We also know that 1 L of solution = 1000 mL, and the density of water is 1 g/mL, so we have 1000 g of solution. ∴ [Cu²⁺] = 1.8 × 10⁻⁵ g/1 L = 1.8 × 10⁻⁵ g/10³ g We want to know how many grams of Cu²⁺ in 10⁶ g of solution, so we multiply the denominator by 1000 to get 10⁶ and the numerator by 1000 to keep the fraction constant. [Cu²⁺] = (1.8 × 10⁻⁵ g/10³ g) × (10³/10³) = 1.8 × 10⁻² g/10⁶ g = 1.8 × 10⁻² ppm or 0.18 ppm or 180 ppb.

4:57 how did we get x times 4x squared, and then 4x cubed?

Jay misspoke, he should have said x times 2x squared which results in 4x cubed. Small math error on his part. Hope that clears it up.

At around 4:53, why do you times the 2x by 2 to get 4x? Thanks

You aren't multiplying, you're squaring. 2 times 2 is 4 and x times x is x^2, so 2x times 2x equals 4x^2. Then, multiplying that by x equals 4x^3.

At 5:46, is there some reason not to type (5.5*10^-21)^(1/3) instead of scrolling through the catalog?

He is using a calculator simulator, so it might be a bit different from a normal graphing calculator

At 4:55 while he was solving the problem, he mentioned that we should multiply the solubility "X" by 2 "mole ratio" to get the solubility right. I wonder why we don't do so while calculating the equilibrium expressions generally, and the Ksp specifically, using the molarity "concentration".

You actually would use the coefficients when solving for equilibrium expressions. Looking back over my notes that I took over the Khanacademy MCAT prep videos I don't see any examples with this, but doing just a little research you can confirm that the coefficients are incorporated when determining any equilibrium expression (even if it is just 1).

How would you take the cube root of something WITHOUT a calculator? Is there a trick to it?

Not really. I hate plugging a non-Khan video but i`ve looked for it and I can`t find it. Start here, she does a very good job of math without a calculator. Good luck. https://www.youtube.com/watch?v=4ApPRHD1ahQ&feature=youtu.be PS. I would explain it myself if I thought I could do a better job than her.

How do you know what values to put into an ICE table?

Ice table stands for: Initial concentrations (M) Change Equilibrium concentrations (M) so whatever values you have that you can put in that table you can.

Main content

AP®︎/College Chemistry

Course: AP®︎/College Chemistry > Unit 13

Lesson 3: Solubility equilibria

Worked example: Calculating solubility from Kₛₚ

Name: Worked example: Calculating solubility from Kₛₚ
Uploaded: 2014-12-10T22:50:52Z
Description: A compound's molar solubility in water can be calculated from its Kₛₚ value at 25°C. To do so, first prepare an ICE (Initial, Change, and Equilibrium) table showing the equilibrium concentrations of the ions in terms of x, the molar solubility of the compound. Then, plug these expressions into the solubility-product expression for the compound and solve for the value of x.

Google Classroom

A compound's molar solubility in water can be calculated from its Kₛₚ value at 25°C. To do so, first prepare an ICE (Initial, Change, and Equilibrium) table showing the equilibrium concentrations of the ions in terms of x, the molar solubility of the compound. Then, plug these expressions into the solubility-product expression for the compound and solve for the value of x. Created by Jay.

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Michael
Posted 9 years ago. Direct link to Michael's post “At 3:42 why do you raise ...”
At
3:42
why do you raise the concentration of the reactants to the power of the coefficient? In what video is that explained?
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(3 votes)
Answer
- Matthew Belliveau
  Posted 9 years ago. Direct link to Matthew Belliveau's post “It is explained in this v...”
  It is explained in this video set:
  
  https://www.khanacademy.org/science/chemistry/chemical-equilibrium
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  (7 votes)
Shariq Khan
Posted 7 years ago. Direct link to Shariq Khan's post “What would you do if you ...”
What would you do if you were asked to find the ppm of the cu2+ ion or the OH- ion?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Ernest Zinck
  Posted 7 years ago. Direct link to Ernest Zinck's post “Ppm means: "how many in a...”
  Ppm means: "how many in a million?" as in, "How many grams of Cu²⁺ in a million grams of solution"?
  We know that [Cu²⁺] = 1.8 × 10⁻⁵ g/L.
  We also know that 1 L of solution = 1000 mL, and the density of water
  is 1 g/mL, so we have 1000 g of solution.
  ∴ [Cu²⁺] = 1.8 × 10⁻⁵ g/1 L = 1.8 × 10⁻⁵ g/10³ g
  We want to know how many grams of Cu²⁺ in 10⁶ g of solution, so we multiply the denominator by 1000 to get 10⁶ and the numerator by 1000 to keep the fraction constant.
  [Cu²⁺] = (1.8 × 10⁻⁵ g/10³ g) × (10³/10³) = 1.8 × 10⁻² g/10⁶ g = 1.8 × 10⁻² ppm or 0.18 ppm or 180 ppb.
  Comment on Ernest Zinck's post “Ppm means: "how many in a...”
  (4 votes)
Division Joy
Posted 3 years ago. Direct link to Division Joy's post “4:57 how did we get x tim...”
4:57
how did we get x times 4x squared, and then 4x cubed?
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(2 votes)
Answer
- Richard
  Posted 3 years ago. Direct link to Richard's post “Jay misspoke, he should h...”
  Jay misspoke, he should have said x times 2x squared which results in 4x cubed. Small math error on his part.
  
  Hope that clears it up.
  Comment on Richard's post “Jay misspoke, he should h...”
  (3 votes)
Sophie Butt
Posted 8 years ago. Direct link to Sophie Butt's post “At around 4:53, why do yo...”
At around
4:53
, why do you times the 2x by 2 to get 4x? Thanks
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(2 votes)
Answer
- regan85922
  Posted 7 years ago. Direct link to regan85922's post “You aren't multiplying, y...”
  You aren't multiplying, you're squaring. 2 times 2 is 4 and x times x is x^2, so 2x times 2x equals 4x^2. Then, multiplying that by x equals 4x^3.
  Button navigates to signup page
  (2 votes)
An_Awesome_Person
Posted 7 years ago. Direct link to An_Awesome_Person's post “At 5:46, is there some re...”
At
5:46
, is there some reason not to type (5.5*10^-21)^(1/3) instead of scrolling through the catalog?
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(2 votes)
Answer
- Tony Tu
  Posted 6 years ago. Direct link to Tony Tu's post “He is using a calculator ...”
  He is using a calculator simulator, so it might be a bit different from a normal graphing calculator
  Button navigates to signup page
  (2 votes)
Nada Youssef
Posted 9 years ago. Direct link to Nada Youssef's post “At 4:55 while he was solv...”
At
4:55
while he was solving the problem, he mentioned that we should multiply the solubility "X" by 2 "mole ratio" to get the solubility right. I wonder why we don't do so while calculating the equilibrium expressions generally, and the Ksp specifically, using the molarity "concentration".
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Seth Sturgill
  Posted 7 years ago. Direct link to Seth Sturgill's post “You actually would use th...”
  You actually would use the coefficients when solving for equilibrium expressions. Looking back over my notes that I took over the Khanacademy MCAT prep videos I don't see any examples with this, but doing just a little research you can confirm that the coefficients are incorporated when determining any equilibrium expression (even if it is just 1).
  Button navigates to signup page
  (2 votes)
dalila.barba
Posted 5 years ago. Direct link to dalila.barba's post “What is the difference be...”
What is the difference between the solubility and the solubility product constant?
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(2 votes)
Answer
bpoole9925
Posted 9 years ago. Direct link to bpoole9925's post “How would you take the cu...”
How would you take the cube root of something WITHOUT a calculator? Is there a trick to it?
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(1 vote)
Answer
- Jennifer Ness
  Posted 9 years ago. Direct link to Jennifer Ness's post “Not really. I hate plugg...”
  Not really. I hate plugging a non-Khan video but ive looked for it and I cant find it. Start here, she does a very good job of math without a calculator. Good luck. https://www.youtube.com/watch?v=4ApPRHD1ahQ&feature=youtu.be PS. I would explain it myself if I thought I could do a better job than her.
  Comment on Jennifer Ness's post “Not really. I hate plugg...”
  (3 votes)
Jerry J. Français II
Posted 8 years ago. Direct link to Jerry J. Français II's post “How do you know what valu...”
How do you know what values to put into an ICE table?
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(1 vote)
Answer
- Zenu Destroyer of Worlds (AK)
  Posted 8 years ago. Direct link to Zenu Destroyer of Worlds (AK)'s post “Ice table stands for: Ini...”
  Ice table stands for:
  Initial concentrations (M)
  Change
  Equilibrium concentrations (M)
  so whatever values you have that you can put in that table you can.
  Comment on Zenu Destroyer of Worlds (AK)'s post “Ice table stands for: Ini...”
  (3 votes)
Cameron Katz
Posted 8 years ago. Direct link to Cameron Katz's post “How do you know when to m...”
How do you know when to make the initial concentration for OH- 0 versus making it 1.0x10^-7? Do you only make it 1.0x10^-7 if the problems states that the compound is already in solution?
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(2 votes)
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- Ernest Zinck
  Posted 8 years ago. Direct link to Ernest Zinck's post “If you have a slightly so...”
  If you have a slightly soluble hydroxide, the initial concentration of OH⁻ before the hydroxide dissolves is 0.
  At this level of study, you can ignore any OH⁻ produced by the water.
  Comment on Ernest Zinck's post “If you have a slightly so...”
  (1 vote)

Video transcript

- [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.