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# Solubility Equilibrium

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Author: carolyn fruin
##### Objective:

Learning Targets:

• I can calculate the ion-product of a salt.
• Based on the value of the solubility product, I can determine if a substance is soluble or slightly soluble in water.
• Given the molar solubility of a substance, I can determine the concentration of it's ions.
• Given the Ksp, I can determine the molar solubility of a substance.
• I can determine whether or not a precipitate will form when two solutions are mixed.
(more)
Tutorial

## PhET Lab Simulation - Solubility Equilibrium

This simulation, Salts and Solubilities, explores the solubility of salts and allows the student to gather data and calculate the Ksp for particular salts. Use the document below as a guide for your lab report.

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## Student Reading Guide: Solubility Equilibrium

Use this guide as you read pages 577-584 in the Modern Chemistry textbook.

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## Video Homework

The following videos are examples of Ksp problems. Using the handout from class (also loaded below), follow along as the solutions are discussed. Feel free to start, stop or rewind the videos as necessary so you understand each step as it takes place.

Be prepared to ask questions in class on the predetermined date.

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## Determining the Molar Solubility given Ksp

The final math step in this video is incorrect. Her answer from the calculator should be 1.5E-3. Thanks AN for pointing this out so quickly!

## Example - Molar Solubility from Ksp

The solubility of a salt can be calculated from Ksp.

Let's try an example calculation problem to demonstrate the relationship between the solubility and the solubility product of a salt.

Calculate the solubility of CaF2 in mol/L (Ksp = 4.0 x 10-8)

First, write the BALANCED REACTION:

Next, set up the SOLUBILITY PRODUCT EQUILIBRIUM EXPRESSION:

In the above equation, however, we have two unknowns, [Ca2+] and [F-]2. So, we have to write one in terms of the other using mole ratios. According to the balanced equation, for every one mole of Ca2+ formed, 2 moles of F- are formed. To simplify things a little, let's assign the the variable x for the solubility of the Ca2+:

If we SUBSTITUTE these values into the equilibrium expression, we now only have one variable to worry about, x:

We can now SOLVE for x:

We assigned x as the solubility of the Ca2+, [Ca2+] which is equal to the solubility of the salt, CaF2. However, we must double this if we are finding the the solubility of F-, [F-].

Pb3(PO4)2

## Example - Ksp from Molar Solubility

We can calculate the Ksp from the molar solubility of a salt.

The solubility of AgCl in pure water is 1.3 x 10-5 M. Calculate the value of Ksp.

First, write the BALANCED REACTION:

AgCl (s) Ag+(aq) + Cl-(aq)

Next, set up the SOLUBILITY PRODUCT EQUILIBRIUM EXPRESSION:

Ksp = [Ag+] [Cl-]

It is given in the problem that the solubility of AgCl is 1.3 x 10-5. Since the mole ratio of AgCl to both Ag+ and Cl- is 1:1, the solubility of each of the ions is equal to the solubility of AgCl. SUBSTITUTE the solubility given into the equilibrium expression to get Ksp:

Ksp = (1.3 x 10-5)(1.3 x 10-5)

Ksp = 1.69 x 10-10

Ag2CrO4

## Solving for Q: Will a Precipitate Form?

Sodium hydroxide added to magnesium nitrate.

## Example - Predicting a Precipitate Formation

The Ksp for CaCO3 is 5.0 x 10-9 .

Ksp = [Ca2+] [CO32-]

so in a saturated solution

5.0 x 10-9 = [Ca2+] [CO32-]

This means that 5.0 x 10-9 is the highest the product of the concentrations can be. Any more Ca2+ or CO32- added will precipitate , the solution can’t dissolve any more!

In an experiment, a chemist mixed a 2.3x10-4 mole Ca(OH)2 and 8.8x10-2 mole Na2CO3 in order to make a liter of solution. Will a precipitate form?

First, the precipitate that could form is CaCO3(s).

And, we know If the product of concentrations of Ca2+ and CO3-2 is a bit larger than the Ksp then a precipitate will form. If the value is smaller, then the concentrations of the ions are too dilute to produce a precipitate.

[Ca2+] = 2.3 x 10-4 mol/L and [CO32-] = 8.8 x 10-2 mol/L.

the table shows CaCO3 Ksp = 5.0 x 10-9.

Calculate Q

Q = [Ca2+] [CO32-]

= (2.3 x 10-4) (8.8 x 10-2)

= 2.024 x 10-5

Compare the Q with the real value for Ksp

Q = 2.024 x 10-5

Ksp = 5.0 x 10-9

Q > Ksp

Notice: product of the ion concentrations(Q) is MORE than Ksp

Solution can't hold this many ions, so the excess ions will precipitate as a result.

## Another Worked Example

Solubility Equilibrium

Lanthanum iodate, La(IO3)3, (F.W. 663.62) has Ksp=1.0 x 10‑11. How many grams lanthanum iodate will dissolve in 1.00 L pure water?

The equilibrium equation is

La(IO3)3 (s) ⇔ La3+ (aq) + 3IO3- (aq)

Using the table method,

 La(IO3)3 (s) La3+ (aq) IO3- (aq) initial Solid 0 0 change -x +x +3x final Solid x 3x

Solve the Ksp equation for the molar solubility, x

Now find the weight in dissolved in 1.00 L pure water

## Solubility Equilibrium Worksheet 7.6

Practice problems for section 18-4, solubility equilibrium. This is a required practice worksheet.

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Paul Chick over 1 year ago

On the bottom of the first page of the Student Reading Guide should it say "Therefore the Ksp for CaF2 is" rather than Ag2CrO4?

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carolyn fruin answered over 1 year ago

Yes! Didn't fix the typo. If you'd like an e-copy, email me and I can send the word version.

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