Does any solid PbCl2 form when 3.5 mg of NaCl is dissolved in 0.250 L of 0.12 M Pb(NO3)2 ?

Name: Problem 87, Chapter 19: Chemistry: The Molecular Nature of Matter and Change
Uploaded: 2020-06-18T03:50:42-08:00
Duration: 6 min 36 s
Channel: Nicole Powell
Description: Does any solid PbCl2 form when 3.5 mg of NaCl is dissolved in 0.250 L of 0.12 M Pb(NO3)2 ?

Does any solid PbCl2 form when 3.5 mg of NaCl is dissolved...

Key Concepts

Precipitation Reactions

Precipitation reactions occur when two soluble salts react in solution to form an insoluble solid. Understanding these reactions involves knowing which combinations of ions produce compounds with low solubility, leading to the formation of a precipitate when the product of the ion concentrations exceeds their solubility limits.

Solubility Product (Ksp)

The solubility product constant, Ksp, is an equilibrium constant that applies to the dissolution of a sparingly soluble ionic compound. It quantifies the maximum product of the ion concentrations that can exist in solution at equilibrium, and it is essential for predicting whether a precipitate will form when different ionic solutions are mixed.

Reaction Quotient (Ion Product) vs. Ksp

The reaction quotient, often called the ion product when discussing solubility, is calculated using the current concentrations of the ions in solution. By comparing the ion product to the Ksp value of a compound, one can determine if the system is at equilibrium or if a precipitate will form: precipitation occurs if the ion product exceeds the Ksp.

Stoichiometry and Concentration Calculations

Accurate stoichiometric calculations and unit conversions are critical for determining the actual concentrations of ions in a solution. This forms the basis for comparing the ion product with the Ksp and deciding whether the solution is saturated, unsaturated, or supersaturated with respect to a given sparingly soluble salt.

Transcript

00:01 We want to know whether a precipitate will form if we dissolve 3 .5 milligrams of nacl in 0 .250 liters of 0 .12 molar pb n032.

00:12 The precipitate that would form is pbcl2, and if we write out the equation for a saturated solution of this, it would partially dissociate to pb2 plus and 2cl minus.

00:36 The ksp for pbcl2, which we can look up in the textbook, is 1 .7 times 10 to the negative 5th.

00:46 And our qsp expression is the concentration of pb2 plus times the concentration of cl minus raised to the second power because of this coefficient of 2 before the cl minus.

01:01 We want to remember that if the qsp equals the ksp, then we have a saturated solution without a precipitate.

01:23 If the qsp is greater than the ksp, we have a saturated solution with a precipitate.

01:42 And if the qsp is less than the ksp, we have an unsaturated solution, which means we have no precipitate.

02:01 This is because a greater qsp means that we have more of these ions from this qsp expression.

02:14 So the equilibrium will shift to the left by lashali's principle.

02:19 If we have a qsp that's less than the ksp, then we have less of these ions, meaning the reaction will shift to the right.

02:29 And so we won't have any of this solid precipitate.

02:32 If the qsp is equal to the ksp, then our system is in equilibrium.

02:37 So it'll be a saturated solution, but no solid precipitable form.

02:43 So now let's do the calculations so that we know which of these is true.

02:48 Starting with our concentration of pb2 plus, we can get that from our given concentration of pbno32.

02:59 So we have 0 .12 molar pbn032, and there is a one -toin ratio between pbno32 and pb, n032, and there is a one -toin ratio between pbno32 and pb...

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