Solubility Product Practice Problems – Stan's Page
The solubility product (Ksp) is used to calculate equilibrium concentrations of the ions in The net ionic equation only shows the precipitation reaction. B To solve this problem, we must first calculate the ion product—Q. Consider the reaction the reaction MX(s) M+(aq) + X-(aq) Qsp is the solubility priduct of the concentration of ions in solution of a dissolving ionic solid. Students have difficulties in understanding theconcept on the macroscopic submicroscopic level based the relationship Qsp and Ksp with precipitate, and the.
So we have ions in solution and we have a large amount of undissolved lead two chlorides. All right, in part A, our job is to calculate the solubility of lead two chloride in water at 25 degrees Celsius. And first, we're going to find the solubility in grams per liter.
All right, so in grams per liter. If we move our decimal place one, two, three that's. So that is the solubility. And the solubility is the number of grams of solute in one liter of a saturated solution.
Solubility and the common-ion effect (video) | Khan Academy
So if you had one liter of water, you could only dissolve about 4. So this is why this is only a slightly soluble ionic compound. A small amount dissolves. All right, we could also find the solubility in moles per liter, which would be the molar solubility. So we know grams, we know the grams is. To find moles, we need to know the molar mass. So for lead two chloride we have lead with a molar mass of So we have To that we'd have to add two times So two times If we add that to That's the molar mass of PbCl2.
So if we divide the grams by the grams per mole, we divide grams by We'd get one over one over moles. So this would tell us how many moles. So let's get out the calculator and let's do this. So if I round that, we would get. We're trying to find the molar solubility, so we need to divide moles by liters.
And we already saw the liters was. We'll go ahead and use the rounded number here. So this is equal to. Moles over liters is molarity. So this is the molar solubility of lead two chloride in water at 25 degrees Celsius. You have to make sure to specify the temperature because, obviously, if you change the temperature, you change how much can dissolve in the water. All right, so that's the idea of solubility and molar solubility. In part B our goal is to calculate the solubility product constant, Ksp, at 25 degrees Celsius for lead two chloride.
Ksp is really just an equilibrium constant.
So let's think about a solubility equilibrium. Let's think about this picture right up here. So we have a saturated solution of lead two chloride and our solution is in contact with our solid, lead two chloride, here. And at equilibrium the rate of dissolution is equal to the rate of precipitation.
So the rate at which the solid turns into ions is the same as the rate in which the ions turn back into the solid. So let's go ahead and represent that here. PbCl2, lead two chloride is our solid. And our ions are Pb two plus in solution and Cl minus.
We need to balance this, so we need a two here in front of our chloride anion, and everything else would get a one. So if we're trying to find our equilibrium constant, Ksp, we need to start with an ice table. So we're going to start with an initial concentration.
So an initial concentration, then we need to think about the change, and finally, we can find equilibrium concentrations. So let's pretend like nothing has dissolved yet.
Common Ions and Complex Ions
So let's pretend like we haven't made our solution, our saturated solution yet. So our initial concentrations would be zero for our products.
All right, next, we need to think about how much of our lead two chloride dissolves. The ion product Q is analogous to the reaction quotient Q for gaseous equilibria.
The solution is unsaturated, and more of the ionic solid, if available, will dissolve. The solution is saturated and at equilibrium.
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The solution is supersaturated, and ionic solid will precipitate. The Relationship between Q and Ksp.
16.3: Precipitation and the Solubility Product
The process of calculating the value of the ion product and comparing it with the magnitude of the solubility product is a straightforward way to determine whether a solution is unsaturated, saturated, or supersaturated. More important, the ion product tells chemists whether a precipitate will form when solutions of two soluble salts are mixed.
Its solubility product is 1. The pathway of the sparingly soluble salt can be easily monitored by x-rays. Will barium sulfate precipitate if Recall that NaCl is highly soluble in water.
Solubility Product Practice Problems
Ksp and volumes and concentrations of reactants Asked for: Write the balanced equilibrium equation for the precipitation reaction and the expression for Ksp. Determine the concentrations of all ions in solution when the solutions are mixed and use them to calculate the ion product Q. Compare the values of Q and Ksp to decide whether a precipitate will form. A The only slightly soluble salt that can be formed when these two solutions are mixed is BaSO4 because NaCl is highly soluble.
The equation for the precipitation of BaSO4 is as follows: True chemical equilibrium can only occur when all components are simultaneously present. A solubility system can be in equilibrium only when some of the solid is in contact with a saturated solution of its ions. Failure to appreciate this is a very common cause of errors in solving solubility problems. Undersaturated and supersaturated solutions If the ion product is smaller than the solubility product, the system is not in equilibrium and no solid can be present.
Such a solution is said to be undersaturated.