Lesson 1: Solubility of Salts

Part b: Solubility and K_sp

Part a: The Solubility Product Constant, K_sp
Part b: Solubility and K_sp
Part c: Solubility and Common Ion Effects

 

The Big Idea

This lesson takes the guesswork out of solubility problems. ICE tables plus a K_sp equation give you a clear path to find ion concentrations and molar solubility - even in the most difficult algebraic cases.

 

Solving K_sp Problems

When a salt (i.e., ionic compound) dissolves, it dissociates into ions. As discussed in Lesson 1a, some salts dissolve more than others. The Solubility Product Constant or K_sp (introduced in Lesson 1a) provides a measure of the extent to which an insoluble salt dissolves and dissociates. The K_sp value and the formula of the salt determine the ion concentrations when the salt is in equilibrium with the undissolved solid.

 
Knowing the K_sp and the formula of the salt, the maximum ion concentrations can be calculated. The calculation of ion concentrations was demonstrated in Lesson 1a for saturated solutions of two simple salts – AgCl and PbSO₄. We refer to these as “simple salts” because a single formula unit contains only two ions – one cation and one anion. When a salt contains more than two ions, the process becomes algebraically more complicated. In this part of Lesson 1, we will learn how to solve problems with such “complicated salts”. It requires only a couple of additional details. Otherwise, the problem-solving strategy remains the same.
 
 

 

How to Determine Ion Concentrations for Insoluble Salts

The following method can be used to determine the ion concentrations from the K_sp value.

1. Write the balanced chemical equation for the dissociation of the solid.
2. Write the K_sp equation for the reaction.
3. Set up an ICE table. Use the variable x to represent the amount of solid that dissolves.
4. Substitute the K_sp value and the expressions for the equilibrium concentrations of the ions (x, 2x, 3x, etc.) into the K_sp equation.
5. Use algebra to solve for the x and for the ion concentrations.

   

 

Example 1 – Ion Concentrations of PbBr₂

The K_sp for lead(II) bromide is 6.6 × 10^-6 at 25°C. Determine the ion concentrations at equilibrium.
 

Solution:

Step 1:  The equation for the dissociation of PbBr₂ is:  
Step 2:  The K_sp equation is:  
Step 3:  The completed ICE table is shown below. Before dissociation, the ion concentrations are 0 M; that’s the Initial row. An unknown amount of ions are produced (“x” and “2x”); the coefficients in the balanced equation are used for this Change row. The Equilibrium row of an ICE table is always determined by adding the Change row to the Initial row. We wish to determine x. Since solids are not included in the equilibrium constant expression, we do not worry about their concentrations; that explains why the first column is greyed out.
   
Step 4:  The K_sp value and the last row of the ICE table are substituted into the equation written in Step 2.  
Step 5:  Algebra is used to solve for x. The value of x is needed to determine the ion concentrations. Perform the algebra as slowly as your comfort level requires; do not rush! Note that the square of 2x requires that both the 2 and the x are squared, resulting in 4x² (interpret as 2•x•2•x). And when 4x² is multiplied by x, the result is 4•x³ (interpret as x•2•x•2•x). Solving for x requires that the x³ first be isolated by itself on one side of the equation; this is done by dividing both sides of the equation by 4. Then take the cube root of both sides of the equation.
   
Once the value of x is determined, the ICE table can be used to determine the concentrations at equilibrium for the two ions. The last row indicates that the lead ion concentration is x and the bromide ion concentration is 2•x. Thus, …
   

 
 

Example 2 – Ion Concentrations of Li₂CO₃

The K_sp for lithium carbonate is 8.2 × 10^-4 at 25°C. Determine the ion concentrations at equilibrium.
 

Solution:

Step 1:  The equation for the dissociation of Li₂CO₃ is:  
Step 2:  The K_sp equation is:  
Step 3:  The completed ICE table is shown below. Before dissociation, the ion concentrations are 0 M; that’s the Initial row. An unknown amount of ions are produced (“2x” and “x”); the coefficients in the balanced equation are used for this Change row. The Equilibrium row of an ICE table is always determined by adding the Change row to the Initial row. We wish to determine x. Since solids are not included in the equilibrium constant expression, we do not worry about their concentrations; that explains why the first column is greyed out.
   
 
Step 4:  The K_sp value and the last row of the ICE table are substituted into the equation written in Step 2.  
Step 5:  Algebra is used to solve for x. The value of x is needed to determine the ion concentrations. The algebra of Example 2 is similar to that of Example 1. We perform the steps in sequential order in an effort to solve for x.
   
Once the value of x is determined, the ICE table can be used to determine the concentrations at equilibrium for the two ions. The last row indicates that the lithium ion concentration is 2•x and the carbonate ion concentration is x. Thus, …
 

  

Example 3 – Ion Concentrations of Ca₃(PO₄)₂

The K_sp for calcium phosphate is 2.1 × 10^-33 at 25°C. Determine the ion concentrations at equilibrium.
 
 

Solution:

Step 1:  The equation for the dissociation of Ca₃(PO₄)₂ is:  
Step 2:  The K_sp equation is:  
Step 3:  The completed ICE table is shown below. Before dissociation, the ion concentrations are 0 M; that’s the Initial row. An unknown amount of ions are produced (“3x” and “2x”); the coefficients in the balanced equation are used for this Change row. The Equilibrium row of an ICE table is always determined by adding the Change row to the Initial row. We wish to determine x. Since solids are not included in the equilibrium constant expression, we do not worry about their concentrations; that explains why the first column is greyed out.
   
Step 4:  The K_sp value and the last row of the ICE table are substituted into the equation written in Step 2.  
Step 5:  Algebra is used to solve for x. The value of x is needed to determine the ion concentrations. Do the algebra as slowly as you need to and be sequential, taking one step at a time. The cube of 3x requires that both the 3 and the x are raised to the third power, resulting in 27x³ (think of this as 3•x•3•x•3•x). The squaring of the 2x is done just like Examples 1 and 2. And when 27x³ is multiplied by 4x², the result is 108•x⁵ (think of this as 3•x•3•x•3•x•2•x•2•x). Solving for x requires that the x⁵ first be isolated by itself on one side of the equation. Then take the fifth root of both sides of the equation.
   
Once the value of x is determined, the ICE table can be used to determine the concentrations at equilibrium for the two ions. The last row indicates that the calcium ion concentration is 3•x and the phosphate ion concentration is 2•x. The unrounded value of x is used in these calculations. The final results are rounded to the second significant digit. Thus, …
   

 
It is worth noting that many student difficulties with solving K_sp problems stem from difficulties with writing formulae for ionic compounds and with writing dissociation equations. To remedy these issues, we recommend the following pages:
   

 
 

What is Molar Solubility?

Every salt, whether soluble or insoluble, has a molar solubility. Molar solubility describes the maximum number of moles of the salt that will dissolve per 1.0 L of water. The value depends upon the temperature of water. Our discussion and calculations will be restricted to a temperature of 25.0°C.
 
Molar solubility values can be determined from knowledge of the K_sp and the formula of the salt. The method of calculating the solubility from the K_sp is identical to the process of calculating ion concentrations at equilibrium. In every ICE table that we’ve used, there was a -x in the Change in [ ] row in the salt column. This -x indicated that x moles per liter of the salt dissolved (and dissociated). This is the definition of molar solubility – the number of moles of salt that dissolve per liter. Determining the value of x is equivalent to determining the molar solubility of that salt.
   
 

 

Example 4 – Molar Solubility of AgI

The K_sp for silver(I) iodide is 8.5 × 10^-17 at 25°C. Determine the molar solubility of AgI.
 

Solution:

Step 1:  The equation for the dissociation of AgI is:  
Step 2:  The K_sp equation is:  
Step 3:  The ICE table is completed similar to other problems on this page. The completed table is shown below.
   
 
Step 4:  The K_sp value and the last row of the ICE table are substituted into the equation written in Step 2.  
Step 5:  Algebra is used to solve for x. The value of x is equal to the molar solubility.
   
 

 

Example 5 – Molar Solubility of PbI₂

The K_sp for lead(II) iodide is 9.8 × 10^-9 at 25°C. Determine the molar solubility of PbI₂.
 

Solution:

Step 1:  The equation for the dissociation of PbI₂ is:  
Step 2:  The K_sp equation is:  
Step 3:  The ICE table is completed similar to other problems on this page. The completed table is shown below.
   
Step 4:  The K_sp value and the last row of the ICE table are substituted into the equation written in Step 2.  
Step 5:  Algebra is used to solve for x. Details associated with the algebraic manipulations are identical to Example 1 above. The value of x is equal to the molar solubility.
   
 

 

How to Determine the K_sp from the Molar Solubility

Just as the molar solubility can be determined from the K_sp, the K_sp value for a salt can also be determined from its molar solubility. The following step-by-step method provides guidance on how to do it:
 

1. Write the balanced chemical equation for the dissociation of the solid.
2. Write the K_sp equation for the reaction.
3. Set up an ICE table. Use the variable x to represent the amount of solid that dissolves. Complete the entire ICE table.
4. Substitute the expressions for the equilibrium concentrations of the ions (x, 2x, 3x, etc.) into the K_sp equation.
5. The value of x in the K_sp equation is the molar solubility. Substitute this value into your K_sp equation (from step 4) and use your calculator to solve for K_sp. Be cautious with an exponents.

 
 
 

Example 6 – Determining the K_sp from the Molar Solubility

The molar solubility of silver(I) oxalate (Ag₂C₂O₄) is 1.1x10^-4 mol/L. Use the above procedure to determine the K_sp for Ag₂C₂O₄.
 

Solution:

Step 1:  The equation for the dissociation of Ag₂C₂O₄ is:  
Step 2:  The K_sp equation is:  
Step 3:  The ICE table is completed similar to other problems on this page. The completed table is shown below.
   
Step 4:  The expressions in the last row of the ICE table are substituted into the K_sp equation written in Step 2.  
Step 5:  Algebra is used to simplify the K_sp equation. Details associated with the algebraic manipulations are identical to Example 1 above.
   
The value of x is equal to the molar solubility. The 1.1x10^-4 value is substituted into the above equation for x.
   
 
 

Mass Solubility in g/L

The solubility of a salt is often given in grams per liter instead of moles per liter. This is referred to as the mass solubility of the salt. These two solubility units are related to one another by the molar mass of the salt. Conversion between the two units can be performed by using the molar mass of the salt.
   
 
 

Next Up

Our focus in Lesson 1b has been on the solubility of a salt in pure water. In all ICE tables, the initial concentrations of the ions were 0 M. In Lesson 1c, we will investigate situations in which the salt is dissolved in a solution that contains one or both of the ions. Before advancing forward, take some time to solidify your understanding of the concepts on this page by using one or more of the suggestions in the Before You Leave section below.
 
 

Before You Leave - Practice and Reinforcement

Now that you've done the reading, take some time to strengthen your understanding and to put the ideas into practice. Here's some suggestions.
 

• Our Calculator Pad section is the go-to location to practice solving problems. You’ll find plenty of practice problems on our Solution Equilibria page. Check out the following four problem sets:  K_sp and Ion Concentrations 1 || K_sp and Ion Concentrations 2 || K_sp and Solubility 1 || K_sp and Solubility 2
• The Check Your Understanding section below includes questions with answers and explanations. It provides a great chance to self-assess your understanding.
• Download our Study Card on K_sp and Molar Solubility. Save it to a safe location and use it as a review tool. (Coming Soon)

 
 

Check Your Understanding of Solubility and K_sp

Use the following questions to practice the skill of solving K_sp problems. Tap the Check Answer buttons when ready.
 
1. Explain why it is written “Doesn’t Matter” in the first column of the ICE tables?
 

 
2. ScF₃ is a partially soluble salt that has a K_sp of 5.81 x10^-34. Use a dissociation equation, an ICE table, a K_sp expression, and good algebra skills to determine the concentrations of ions when ScF₃ is dissolved in pure water.
 

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3. On planet Exwizee, the binary ionic compound X₂Y₃ is a partially soluble salt composed of X³⁺ and Y^2- ions. It has a K_sp of 3.81 x 10^-15. Use a dissociation equation, an ICE table, a K_sp expression, and good algebra skills to determine the concentrations of ions when X₂Y₃ is dissolved in pure water.
 

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4. Copper(II) phosphate has a K_sp of 1.40 x 10^-37. Use a balanced chemical equation for its dissociation, an ICE table, a K_sp expression, and good algebra skills to determine …
a. the concentrations of its ions at equilibrium.
b. the molar solubility of the salt.
c. the solubility in grams/liter
 

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5. The K_sp of Ag₂CO₃ is 8.46 x 10^-12. Determine the molar solubility of silver(I) carbonate.
 

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6. On Planet Exwizee, X₂Y is a binary, ionic compound that is only slightly soluble in water. It has a molar solubility of 4.06 x 10^-4. Determine the ion concentrations at equilibrium and the K_sp of X₂Y.
 

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7. Suppose that an insoluble salt is composed of the cation M with a +a charge (i.e., M^a+) and the anion A with the -b charge (i.e., A^b-).
a. Write the chemical formula for the salt.
b. Write the balanced equation for the salt dissociating in water.
c. Complete an ICE table for its dissociation using the symbols a, b, and x).
d. Derive an equation that relates the solubility of the salt (x) to the K_sp and the values of a and b.
e. Show how your equation can be used to determine the solubility (i.e., the value of x) for any one of the five examples on this page.
 
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