Learning Targets:
This tutorial contains the complete set of materials in order to help you learn the information necessary for mastery of SOLUBILITY EQUILIBRIUM. As always, you need to work daily to complete the activities and notes provided here. READING THE TEXTBOOK is always a crucial part of your learning.
During class time, there will be sufficient time to ask the instructor for clarification on the content as well as assistance on assigned homework problems.
Completed homework packet includes all flipped notes, completed reading guides, worksheets, checkpoints and PhET simulation. Staple all materials together and be prepared to turn them in before the exam.
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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.
Use this guide as you read pages 577584 in the Modern Chemistry textbook.
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.
The final math step in this video is incorrect. Her answer from the calculator should be 1.5E3. Thanks AN for pointing this out so quickly!
The solubility of a salt can be calculated from K_{sp}.
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 CaF_{2} in mol/L (K_{sp} = 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, [Ca^{2}^{+}] 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 Ca^{2+} formed, 2 moles of F^{} are formed. To simplify things a little, let's assign the the variable x for the solubility of the Ca^{2}^{+}:
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 Ca^{2}^{+}, [Ca^{2}^{+}] which is equal to the solubility of the salt, CaF_{2}. However, we must double this if we are finding the the solubility of F^{}, [F^{}].
We can calculate the K_{sp} 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 K_{sp}.
First, write the BALANCED REACTION:
AgCl (s) Ag^{+}(aq) + Cl^{}(aq)
Next, set up the SOLUBILITY PRODUCT EQUILIBRIUM EXPRESSION:
K_{sp}_{ } = [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 K_{sp}:
K_{sp} = (1.3 x 10^{5})(1.3 x 10^{5})
K_{sp} = 1.69 x 10^{10}
Sodium hydroxide added to magnesium nitrate.
The Ksp for CaCO_{3} is 5.0 x 10^{9} .
Ksp = [Ca^{2}^{+}] [CO_{3}^{2}]
so in a saturated solution
5.0 x 10^{9} = [Ca^{2}^{+}] [CO_{3}^{2}^{}]
This means that 5.0 x 10^{9} is the highest the product of the concentrations can be. Any more Ca^{2}^{+} or CO_{3}^{2}^{} 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 Na_{2}CO_{3} in order to make a liter of solution. Will a precipitate form?
First, the precipitate that could form is CaCO_{3}(s).
And, we know If the product of concentrations of Ca^{2}^{+} and CO_{3}^{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.
[Ca^{2}^{+}] = 2.3 x 10^{4} mol/L and [CO_{3}^{2}^{}] = 8.8 x 10^{2} mol/L.
the table shows CaCO_{3} K_{sp} = 5.0 x 10^{9}.
Calculate Q
Q = [Ca^{2}^{+}] [CO_{3}^{2}^{}]
= (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}
K_{sp} = 5.0 x 10^{9}
Q > K_{sp}
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.
Solubility Equilibrium
Lanthanum iodate, La(IO_{3})_{3}, (F.W. 663.62) has K_{sp}=1.0 x 10^{‑11}. How many grams lanthanum iodate will dissolve in 1.00 L pure water?
The equilibrium equation is
La(IO_{3})_{3} (s) ⇔ La^{3}^{+} (aq) + 3IO_{3}^{} (aq)
Using the table method,

La(IO_{3})_{3} (s) 
La^{3}^{+} (aq) 
IO_{3}^{} (aq) 
initial 
Solid 
0 
0 
change 
x 
+x 
+3x 
final 
Solid 
x 
3x 
Solve the K_{sp} equation for the molar solubility, x
_{}
Now find the weight in dissolved in 1.00 L pure water
_{}
Practice problems for section 184, solubility equilibrium. This is a required practice worksheet.