Qsp and ksp relationship questions

Solubility_Products

This section focuses on the effect of common ions on solubility product equilibria. Let's compare the results of the two above practice problems. The Inverse Relationship Between the Equilibrium Concentrations of the Cr3+ and OH- Ions. With this purpose, research questions formulated are“ Howthe profile of submicroscopic level based the relationship Qsp and Ksp with precipitate, and the. Silver chloride is one of the few chloride that has a limited solubility. A precipitate is also formed when sodium carbonate is added to a sample of hard water,. Ca2+ (aq) + CO(aq) . Confidence Building Questions. Experiment shows that the.

So two times If we add that to That's the molar mass of PbCl2.

Common Ions and Complex Ions

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.

Solubility Product Practice Problems – Stan's Page

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.

Solubility Product Practice Problems

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. 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. All right, so we did that in part A. So that's the concentration of lead two chloride that we're going to lose here. So we're going to lose. And we're going to assume that all of the lead two chloride that dissolved dissociates completely into ions.

So for every one mole of lead two chloride that dissolves we get one mole of lead two plus ions in solution. So if we're losing. And for the chloride anion, this time our mole ratio was one to two, so we need to multiply this number by two. So therefore, at equilibrium we should have a concentration of lead two plus ions is. And now we are ready to write our equilibrium expression.

So we write K, and since this is a solubility equilibrium, we're going to write Ksp. So Ksp is equal to-- remember concentration of products over reactants and for these we also need to think about the coefficients.

So let's think about our products first. Pb two plus, so we have the concentration of Pb two plus and we're going to raise the concentration to the power of the coefficient and here our coefficient is a one.

So we're going to raise this to the first power. Then we're going to multiply this by the concentration of chloride anions, so Cl minus, and then we're going to raise the concentration to the power of the coefficient.

So here our coefficient is a two, so we're going to raise this to the second power. The resulting equation looks like that below: Precipitation reactions are usually represented solely by net ionic equations.

If all products are aqueous, a net ionic equation cannot be written because all ions are canceled out as spectator ions. It is analogous to the reaction quotient Q discussed for gaseous equilibria. Whereas Ksp describes equilibrium concentrations, the ion product describes concentrations that are not necessarily equilibrium concentrations. A solution in which this is the case is said to be saturated. 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. The solution is supersaturated, and ionic solid will precipitate. The Relationship between Q and Ksp. 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. 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.