Back To CourseChemistry 101: General Chemistry
14 chapters | 132 lessons | 11 flashcard sets
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Nikki has a master's degree in teaching chemistry and has taught high school chemistry, biology and astronomy.
Every once in a while you may notice that your faucet isn't flowing so well or that your shower head is spraying unevenly. Upon closer inspection you might notice that there's some white 'buildup' obstructing the flow of water. What's with that? Where did this gunk come from? Unless you put it there, the only place this buildup could've come from is the water itself.
Even though it looks clear, the water coming into our houses is loaded with different dissolved ions and compounds. When the temperature or concentration of the solution changes, these ions and compounds emerge out of solution in their solid form. The temperatures or concentrations at which these chemicals, or solutes, emerge stem from each solute's solubility, or ability to be dissolved in a given volume of a solvent at a given temperature.
Solubility for a compound is expressed in terms of molarity (mol/L) and is temperature dependent. Generally, increasing temperature increases a substance's solubility. When you run cold water through a faucet or decrease the flow to a trickle, the dissolved chemicals become less soluble and form solids - a.k.a. pesky buildup.
A substance is considered to be soluble when its solubility is greater than 1 g per 100 g of solvent. This means that 1 g of substance can completely dissolve and dissociate into ions in 100 g of solvent. Sodium chloride, NaCl, is very soluble in water. A substance is considered to be insoluble when its solubility is less than 0.1 g per 100 g of solvent. Calcium sulfate, CaSO4, one of the main ingredients in water build up, is insoluble in water. Anything in between is considered slightly soluble.
When a solution of a given volume has the maximum amount of a solid dissolved in it, it is said to be saturated. For soluble compounds, saturation occurs at high concentrations. For insoluble compounds, saturation happens at very low concentrations.
Let's go to our virtual lab and do some little experiments with solubility. For any compound that goes into solution, we know that it will break down into its constituent ions. For example, if we put compound AB2 into water, it will break into 1 A^2+ ion and 2 B^- ions.
While these ions are swimming around in solution, there is a possibility that they will collide with each other and reform into a solid. We can show the tendency of this reaction to go both forwards and backwards by using a double headed arrow.
Eventually, a dynamic equilibrium will be reached. We can write an equilibrium expression for this by using the law of mass action. If you're a little fuzzy on how to do that, watch the video on dynamic equilibrium and determining the value of K, the equilibrium constant.
Remember that when writing laws of mass action you don't have to consider solids or liquids because their quantities cannot be expressed in terms of concentration. So, we get to leave out the solid form of AB2.
Ksp = [A^2+][B^-]^2
Similar to other equilibrium expressions you may have encountered, the symbol K is used for the equilibrium constant. We use Ksp to show that this equilibrium is specific to solubility. The proper term for Ksp is solubility product constant, or solubility constant. It has no units!
Just for fun, let's write the Ksp expression for MgF2. We know that MgF2 will break down into 1 Mg^2+ ion and 2 F^- ions and that the reverse process is also possible. The law of mass action for this expression will be Ksp = [Mg^2+][F^-]^2.
Every chemical has a specific Ksp value for a given temperature. These values are usually given for 25 degrees Celsius, or room temperature. The smaller a Ksp value is, the lower the solubility of a compound. Just like with Keq, small values of Ksp suggest that the reaction is dominant in the reverse direction or reactant heavy.
At times, it may be necessary to determine Ksp for a particular solubility problem, other times, concentrations of dissolved ions may need to be determined based on the Ksp value. Either way, we already have the tools to solve either kind of problem.
Ksp values are based on saturated solutions, or solutions containing the maximum concentration of ions in solution. Solutions that are not saturated don't have Ksp values because they are not at equilibrium.
One can determine Ksp values given concentrations of ions in a saturated solution. One can also determine the maximum concentration of ions for a compound with a given Ksp value. We'll do some example problems for each of these situations.
So, here we are in lab. We need to determine the value of Ksp for a saturated solution of AgCl at 25 degrees Celsius. The concentration of both Ag^+ and Cl^- ions is 1.26 * 10^-5 mol/L. This is a pretty small number, and this solution is saturated! That means that AgCl (silver chloride) is fairly insoluble.
We know that AgCl is in a dynamic equilibrium with its constituent ions, Ag^+ and Cl^-, which can be expressed as AgCl(s) <==> Ag^+(aq) + Cl^-(aq).
The equilibrium expression is Ksp = [Ag^+][Cl^-].
To find Ksp, we insert our concentrations of Ag^+ and Cl^- into our expression and solve.
Ksp = (1.26 * 10^-5)(1.26 * 10^-5) = 1.59 * 10^-10
1.59 * 10^-10 is a really small number! It reinforces our earlier notion that this is an insoluble compound.
Now, imagine that we have the Ksp value for a specific compound. But this time, we want to determine the concentrations of ions present. Let's start with an easy one: calcium carbonate (CaCO3). Calcium carbonate has a Ksp of 8.7 * 10^-9 at 25 degrees Celsius.
We know that calcium carbonate breaks into calcium and carbonate ions, and we know that the equilibrium expression is Ksp = [Ca^2+][CO3^2-].
Calcium and carbonate ions dissociate from calcium carbonate in a 1:1 ratio, so we should have equal concentrations of each ion in solution. We can substitute x for the concentrations of each ion and plug in our known value of Ksp, then solve for x!
Ksp = (x)(x) = 8.7 * 10^-9
We find the value of x by taking the square root of 8.7 * 10^-9. x = 9.3 * 10^-5.
Since we set x equal to the concentrations of the ions, we know that the concentration of Ca^2+ and CO3^2- are each 9.3 * 10^-5 mol/L.
One more, and we're done! Let's find the concentrations of lead (II) chloride, PbCl2, at 25 degrees Celsius. The Ksp for this compound is 1.6 * 10^-5. That's a small number, so we'll expect the concentrations of our dissolved ions will be small amounts.
Like we've done before, let's write an equation showing how PbCl2 will break into ions (and reform). The law of mass action for this dissociation is Ksp = [Pb^2+][Cl^-]^2.
We know that each PbCl2 compound breaks into 1 Pb^2+ ion and 2 Cl^- ions. This means that the concentration of Cl^- ions will be twice that of Pb^2+. If I set the concentration of Pb^2+ equal to x, then I must set the concentration of Cl^- to 2x. I can now plug these values into my equilibrium expression along with my value of Ksp.
Ksp = (x)(2x)^2 = 1.6 * 10^-5
Simplifying, I find that 4x^3 = 1.6 * 10^-5, and therefore x = 0.016.
This means that my concentration of Pb^2+ is 0.016 mol/L and the concentration of Cl^- is 0.032 mol/L. Those are tiny numbers, just like we thought.
Solubility is the ability to be dissolved in a given volume of a solvent at a given temperature. Solubility for a compound is expressed in terms of molarity (mol/L) and is temperature dependent. A substance is considered to be soluble when its solubility is greater than 1 g per 100 g of solvent. A substance is considered to be insoluble when its solubility is less than 0.1 g per 100 g solvent.
When a solution of a given volume has a maximum amount of a solid dissolved in it, it is said to be saturated. For solutions at equilibrium, the law of mass action is written only in terms of the ions produced. The equilibrium constant is written as Ksp and identified as the solubility product constant (or the solubility constant) for a solution at equilibrium.
Ksp can be calculated by writing the law of mass action for a solution and inputting the concentration of ions into the equation. This can be done using ion concentrations or based on solubility data. The concentrations of ions can be determined given only the Ksp value for a solution at equilibrium. In this case, write the law of mass action. Set the unknown concentrations equal to x. Make sure to account for stoichiometric ratios. Solve for x.
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Back To CourseChemistry 101: General Chemistry
14 chapters | 132 lessons | 11 flashcard sets