Two compounds with general formulas AX and AX 2 have Ksp = 1.5⨉10 -5 . Which of the two compounds has the higher molar solubility?

Hello everyone today we are being asked to decide between which compound M. Two X or mx three would have the lower molar ability. And we're given the K. S. P. Which is 8.1 times 10 to the negative six for each value. So the first thing we wanna do is we want to draw each compound associated as well as an ice table for each. So we're gonna start with M two X. We're going to stay in the solid form that dissociates into two moles of M plus in the acquis form as well as one X two minus in the acquis forum here. When we set up our ice tables here, our initial concentrations for each one of these are going to be zero. We don't have one for solids because we don't complete the ice table for them. So we're going to place an X. There with the change, we're gonna have plus two X. Since we have two of one of em and then we have just one of X. So we're just gonna say X. E. Is going to be the initial plus the change or I. Plus C. So we're gonna have to X. For em and then we're gonna have X. For X. And so we're giving the K. S. P. For both values. And when we set that up it looks like this K. S. P. Is equal to the concentration of em which we have two of those. So we have two X. And since we have a coefficient of two in front of em that becomes the exponent and for regular X. we just have X is the exponent plugging in the K s p form value, we have 8.1 times 10 to the negative six gives us four X cubed. When we incorporated these square, will we take the cube root? So we're going to take the cube root of both of these values and we divide by four. So we're going to divide both of these by four. After taking the cue route we should get X is equal to 0.1265. When we do the same for M X three we see that mx three is going to associate as a solid Into M three plus acquis As well as three x minus in the acquis form. We set up our ice table here, I c E for our solids, we don't worry about that. And for initial concentrations we both have zero, our change in X is going to be plus X for em and then plus three X for X. Since we have three of them, that's the coefficient E. Is going to be I plus C. So we have X. And then we have three X. Here we do the same as before we set up the K S. P values for the M we have X and for three X for the product when we dissociated, we have three X, we're gonna cube that. Since the exponent is three plugging in our case p values, we get 8.1 times 10 to the negative six is equal to 27 X to the fourth. And so what we're gonna do here is we're gonna take the fourth root here. We're also gonna divide by 27. After doing so that's gonna give us a value of X. is equal to 0. 340. So we look at these numbers and we say which one is lower and of course .01265 is going to be lower, have a lower molars eligibility and therefore M2X is going to have the lowest moller saw the ability, I hope this helped and until next time.

Ch.18 - Aqueous Ionic Equilibrium

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Tro 6th Edition

Ch.18 - Aqueous Ionic Equilibrium

Problem 97

Chapter 18, Problem 97

Two compounds with general formulas AX and AX₂ have Ksp = 1.5⨉10^-5. Which of the two compounds has the higher molar solubility?

Verified step by step guidance

Identify the dissociation reactions for both compounds. For AX, the reaction can be written as AX(s) ⇌ A+(aq) + X-(aq). For AX2, the reaction can be written as AX2(s) ⇌ A+(aq) + 2X2-(aq).

Write the expression for the solubility product constant (Ksp) for each compound. For AX, Ksp = [A+][X-]. For AX2, Ksp = [A+][2X-]^2.

Assume the molar solubility of AX is 's' mol/L, then [A+] = s and [X-] = s. Substitute these into the Ksp expression for AX to get Ksp = s^2.

Assume the molar solubility of AX2 is 's' mol/L, then [A+] = s and [2X-] = 2s. Substitute these into the Ksp expression for AX2 to get Ksp = s(2s)^2 = 4s^3.

Compare the expressions derived for Ksp in terms of 's' for both compounds. The compound with the simpler expression (AX) will generally have a higher molar solubility than the compound with the more complex expression (AX2) when Ksp values are the same.

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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 quantifies the solubility of a sparingly soluble ionic compound. 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. A higher Ksp value indicates greater solubility of the compound in solution.

Molar Solubility

Molar solubility refers to the maximum amount of a solute that can dissolve in a given volume of solvent at equilibrium, expressed in moles per liter (mol/L). It is directly related to Ksp, as the molar solubility can be calculated from the Ksp value for a given compound, allowing for comparison between different compounds.

Comparative Solubility of Compounds

When comparing the solubility of compounds with different stoichiometries, such as AX and AX2, it is essential to consider how the dissociation of each compound affects the concentration of ions in solution. For example, AX dissociates into one ion, while AX2 dissociates into two, which influences the Ksp expression and ultimately the molar solubility of each compound.