Calculate the maximum mass (in grams) of each metal sulfide that will dissolve in 1.0 L of a so

Calculate the maximum mass (in grams) of each metal sulfide...

Key Concepts

Solubility Product (Ksp)

The solubility product is an equilibrium constant that applies to the dissolution of sparingly soluble salts. It represents the product of the molar concentrations of the ions each raised to the power of its stoichiometric coefficient in the dissolution reaction. In calculations involving sparingly soluble compounds, Ksp helps determine how much of the compound can dissolve before the solution becomes saturated.

Common Ion Effect

The common ion effect occurs when a solution already contains one of the ions produced by the dissolution of a sparingly soluble salt. This pre-existing ion shifts the dissolution equilibrium, typically reducing the solubility of the salt. In problems of solubility equilibrium, the presence of a common ion must be considered as it affects the maximum amount of compound that can dissolve.

Chemical Equilibrium

Chemical equilibrium refers to the state in a reversible reaction where the rates of the forward and reverse reactions are equal, and the concentrations of the reactants and products remain constant over time. Understanding equilibrium is crucial in solubility questions because the balance between the solid phase and its dissolved ions determines the extent of dissolution.

Stoichiometry in Dissolution Reactions

Stoichiometry in dissolution reactions involves the quantitative relationship between the dissolved compound and the ions produced in solution. This concept is essential when calculating molar solubilities and later converting them into mass, as it ensures that the proportions of ions produced from the sparingly soluble salt are correctly accounted for according to the balanced dissolution equation.

Transcript

00:01 This question is asking us to take a look at solution chemistry and the concept of solubility using ksp values or equilibrium constants for solubility product.

00:13 We're given this information that we have a solution that has a sulfide concentration of 5 times 10 in the negative 5 molar.

00:21 And we're asked to determine how much lead ion or lead sulfide compound could dissolve at that concentration.

00:31 And so what we want to basically take a look at is express this as an equilibrium expression.

00:38 So if we started with the reaction, if you will, or an equilibrium expression of what's happening with a lead sulfide compound, as it dissolves, it's going to furnish for us some lead 2 plus cations and sulfide anions.

00:52 And we looked up here, i looked up the value of, and we're going to assume that this process takes place at room temperature 25 degrees c.

01:00 The equilibrium constant for that solubility product is 3 times 10 in the negative 28.

01:06 So a pretty low value at that temperature.

01:11 So here's basically what we're furnished with and what we need to do is we know the sulfide ion concentration.

01:22 And again, if we wrote the equilibrium expression for this reaction, we'd have these two ions being present, and that equals that solubility product.

01:31 So let's plug in that known value for sulfide ion and determine then based on over here the solubility product itself will substitute that in.

01:46 We'll also substitute in the sulfide concentration that we were provided with at 5 .00 times 10 in the negative 5 and solve for then what's the molar concentration or the amount of lead 2 plus ion that would exist or could exist or could dissolve under these conditions with that concentration of sulfide ion being present.

02:17 So as we take a look at this calculation, we end up getting a molar concentration of lead 2 plus cations that's 6 .00 times 10 of the negative 24 molar.

02:30 By the way, i just added a couple of digits here so that we'd end up with three significant figures as we could appropriately do for the equilibrium constant for the lead sulfide.

02:43 So we can report this in three significant figures and be consistent with the other numbers.

02:48 So now that we know that, that we're talking about a one -liter solution, that there's a one -to -one relationship between the number of moles of lead sulfide i have and the lead 2 plus cations furnished.

03:02 So if i have that many moles of lead 2 -plus, that must mean they were furnished by the same number of moles of lead sulfide.

03:12 We're asked to determine how much mass of lead sulfide that is.

03:17 So the final step in the process then is to take the molar mass of lead sulfide, which is 239 .27 and determine the mass of lead sulfide that dissolved.

03:31 And so when we do that, we end up getting 1 .44 times 10 of the negative 21 grams of lead sulfide.

03:39 So a very small number...

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