The solubility of KCl is 3.7 at 20^∘ C . Two beakers contain 100 . mL of saturated KCl solution:

The solubility of KCl is 3.7 at 20^∘ C . Two beakers contain...

The solubility of $\mathrm{KCl}$ is 3.7 $\mathrm{at} 20^{\circ} \mathrm{C}$ . Two beakers contain $100 .$ mL of saturated $\mathrm{KCl}$ solution: $100 . \mathrm{mL}$ of 6.0 $\mathrm{MHCl}$ is added to the first beaker and $100 . \mathrm{mL}$ of 12$M \mathrm{HCl}$ to the second.(a) Find the ion-product constant of $\mathrm{KCl}$ at $20^{\circ} \mathrm{C} .$ (b) What mass, if any, of $\mathrm{KCl}$ will precipitate from each beaker?

Key Concepts

Saturated Solution

A saturated solution is one in which the maximum amount of solute has been dissolved at a particular temperature, resulting in a dynamic equilibrium between the dissolved ions and the undissolved salt. In such a solution, any addition of more solute will result in precipitation, keeping the ion concentrations constant as determined by the Ksp.

Ksp (Solubility Product Constant)

The solubility product constant (Ksp) is the equilibrium constant for the dissolution of a sparingly soluble salt. It is defined as the product of the concentrations of the ions each raised to the power of its stoichiometric coefficient. Ksp provides a measure of the salt’s solubility under equilibrium conditions and is used to predict whether precipitation will occur when the ionic concentrations in a solution change.

Common Ion Effect

The common ion effect involves the addition of an ion that is already present in the equilibrium system, shifting the balance according to Le Chatelier's principle. When a common ion is added, the equilibrium shifts to form more of the undissolved salt, thereby reducing its solubility. This is important when considering the addition of a source of a common ion, such as HCl providing Cl–, to a solution of a salt like KCl.

Ion Product

The ion product is the calculated product of the current concentrations of the dissociated ions in solution at any moment, evaluated in the same form as the Ksp expression. Comparing the ion product to the Ksp helps determine whether a solution will remain unsaturated, become exactly saturated, or shift to a supersaturated state leading to precipitation.

Transcript

00:01 Question number 113 is in reference to a saturated solution of calcium chloride, to which some hydrochloric acid is added.

00:11 If it's saturated, then we can assume that we have reached an equilibrium where no more potassium chloride can dissociate into potassium ions and chloride anions.

00:24 So we have two solutions, 100 milliliters each.

00:29 They both have 3 .7 molar potassium chloride.

00:35 To one, we add 100 mil liters of 6 molar hydrochloric acid.

00:40 To the other, we add 100 mil liters of 12 molar hydrochloric acid.

00:46 The equilibrium that we're considering is potassium chloride going to potassium ion and chloride anion.

00:52 The ksp, of course, will be equal to the concentration of these two products, where this is a solid, so it's excluded.

01:02 That then will be equal to the concentration of potassium and the concentration of chloride when the solution has become saturated or reached equilibrium.

01:11 That will be 3 .7 for both.

01:15 So we end up with a ksp value of 13 .69.

01:20 The chloride concentration that we have, once we have added 6 molar hcl with an additional 100 milliliters of hcl, will simply be the molarity of the chloride in the 100 milliliters, so 3 .7, multiplied by 100 mil liters or 0 .1 molar, plus the moles of chloride that were added from the 100 milliliters of 6 molar hcl.

01:51 So we multiply the molarity times the volume to get the additional moles.

01:57 So this is the moles we had, this is the moles we added, this is the total.

02:01 Of the total moles of chloride we now have.

02:03 We divide that by the new volume to get the new concentration, the new volume being 200 milliliters or 0 .2 liters.

02:12 We get 4 .85 molar.

02:14 We can do the same thing with the potassium ion.

02:17 We have 3 .7 molar potassium ion at a volume of 0 .1 liters.

02:23 We divide that by the new volume to 0 .2 liters or 200 mil liters.

02:29 And we get a concentration of 1 .85 molar.

02:33 To determine whether or not precipitation is going to occur, we need to calculate the reaction quotient.

02:39 The reaction quotient is calculated the exact same way ksp is...