Calculate the solubility of LaF 3 in grams per liter in (b) 0.010 M KF solution.

Hello everyone today we are being asked to calculate the molar ability of lead chloride and 0.25 molar of lead nitrate. And the first thing I wanna do is you want to draw this dissolution of this lead nitrate when it's placed in water. So we say that lead nitrate in the acquis form is going to disassociate into lead two plus Aquarius as well as to nitrate ions also in the acquis form. And then we can also say that the concentrations of this lead nitrate is equal to the concentration of regular lead, since it is just a dissociation from that. And that is what gives us our 0.25 molar. And so we say that lead chloride is P. B. C. L. Two. Let me say this is a reversible reaction it's going to dissociate into lead two plus Aquarius as well as to chloride ions also in the quickest form. And we're going to set up our ice table here which is I C. E. And so I stands for the initial concentration. Of course we don't have an initial concentration for lead chloride but we do have it for lead which is going to be 0.25 molar. And for chloride we have zero we can represent each ion as X. So we have X for lead and then we have two X. For chloride two because there's two chlorine. And so when we add this we can say that we have 20.25 Plus X. So we have .25 for our equilibrium. And so for chloride you say that this is two X. And so the reason why it's 20.25 for lead and not 0.25 plus X. Is because here X. Is negligible since our molar concentration is much greater than X. And so we set our K sp equation up. We say that K. S P is equal to our concentration of lead two plus times our concentration of chloride. And since we have that coefficient of two, that becomes our exponent and R K S P for this specific case for lead is going to be So we're gonna write a K. S. P for our lead Chloride is going to be 1.17 times 10 to the -5. And so we plugged that into the equation We get 0.25 times x squared. Of course when we factor in the X squared we get 1.17 times 10 to the negative five is equal to 0.25 times for X squared. Of course when we combine these numbers we get 1.17 times 10 to the negative five is equal to one X squared. Or just X squared. We can then take the square root of both sides and we get that X is equal to 3.42 times at 10 to the negative third molar. Therefore this is going to be our final answer. That's the most liability. I hope this helped. And until next time

Ch.17 - Additional Aspects of Aqueous Equilibria

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Brown 14th Edition

Ch.17 - Additional Aspects of Aqueous Equilibria

Problem 58b

Chapter 17, Problem 58b

Calculate the solubility of LaF₃ in grams per liter in (b) 0.010 M KF solution.

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Identify the dissolution reaction of LaF_3: LaF_3(s) \rightleftharpoons La^{3+}(aq) + 3F^{-}(aq).

Write the expression for the solubility product constant (K_{sp}) of LaF_3: K_{sp} = [La^{3+}][F^{-}]^3.

Recognize that the presence of KF provides a common ion (F^-), which affects the solubility of LaF_3. The concentration of F^- from KF is 0.010 M.

Assume the solubility of LaF_3 is 's' mol/L. Then, [La^{3+}] = s and [F^-] = 0.010 + 3s.

Substitute these concentrations into the K_{sp} expression and solve for 's', considering that 3s is negligible compared to 0.010 M.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Solubility Product Constant (Ksp)

The solubility product constant (Ksp) is an equilibrium constant that applies to the solubility of sparingly soluble ionic compounds. It is defined as the product of the molar concentrations of the ions, each raised to the power of their coefficients in the balanced equation. For LaF3, Ksp can be used to determine how much of the compound can dissolve in a given solution, which is essential for calculating its solubility in the presence of other ions.

Common Ion Effect

The common ion effect refers to the decrease in solubility of a salt when a common ion is added to the solution. In this case, adding KF introduces F- ions, which shifts the equilibrium of LaF3 dissolution to the left, reducing its solubility. Understanding this effect is crucial for calculating the solubility of LaF3 in a KF solution, as it directly influences the concentration of La3+ and F- ions in equilibrium.

Stoichiometry of Dissolution

Stoichiometry of dissolution involves understanding the molar ratios of the ions produced when a salt dissolves in water. For LaF3, the dissolution can be represented as LaF3(s) ⇌ La3+(aq) + 3F-(aq). This stoichiometric relationship is vital for calculating the concentrations of La3+ and F- ions in solution, which are necessary for applying Ksp and determining the solubility in the presence of a common ion.