In a certain mystery liquid, the compounds #Al_2S_3#, #K_3N#, #Na_2S#, and #RbBr# each have a value of #Ksp = 1.0 xx 10^-10#. Which of these solids should be the least soluble in the mystery liquid? (Choices in answer).

Answer 1

The answer is indeed A. rubidium bromide

The choices are as follows:

A. #RbBr# B. #Na_2S# C. #K_3N# D. #Al_2S_3# E. All four solids should be equally soluble in the liquid since their #K_(sp)#’s are all equal
The idea behind this problem is very simple - you have to use the definition of the solubility product constant, #K_(sp)#, to determine which compound has the greatest molar solubility.

The degree of dissociation of an ionic compound is a measure of its solubility; a more soluble ionic compound will result in higher concentrations of cations and anions in solution.

#K_(sp)# simply tells you where the dissociation equilibrium lies, or, more precisely, how far to the left it lies.

Examine the dissociation equilibria of the compounds you have.

#Al_2S_(3(s)) rightleftharpoons 2Al_((aq))^(3+) + 3 S_((aq))^(2-)#, #K_(sp1) = [Al^(3+)]^2 * [S^(2-)]^3#
#K_3N_((s)) rightleftharpoons 3K_((aq))^(+) + N_((aq))^(3-)#, #K_(sp2) = [K^(+)]^3 * [N^(3-)]#
#Na_2S_((s)) rightleftharpoons 2Na_((aq))^(+) + S_((aq))^(2-)#, #K_(sp3) = [Na^(+)]^2 * [S^(2-)]#
#RbBr_((s)) rightleftharpoons Rb_((aq))^(+) + Br_((aq))^(-)#, #K_(sp4) = [Rb] * [Br]#
If the #K_(sp)# values are equal for all these reactions, then you could predict that the compound that produces that smallest number of cations and anions per liter will be the least soluble.
Even without doing any calculations, you can predict that #Al_2S_3# will be the most soluble and #RbBr# will be the least soluble.
You can determine the molar solubility of each of those compounds by replacing the concentrations of the cations and anions with #x#.

You'll get this

#K_(sp1) = (3x)^2 * (2x)^3 = 72x^5#
#K_(sp2) = (3x)^3 * x = 27x^4#
#K_(sp3) = (2x)^2 * x = 4x^3#
#K_(sp4) = x * x = x^2#
Now it becomes obvious that rubidium bromide will have the lowest molar solubility, i.e. the smallest #x# value.
This means that one mole of #RbBr# will produce less #Rb^(+)# and #Br^(-)# ions per liter than 1 mole of #Na_2S#, which in turn will produce less ions in solution than one mole of #K_3N#, which in turn will produce less ions in solution than #Al_2S_3#.
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Answer 2

Aluminum sulfide (Al2S3) would be the least soluble in the mystery liquid.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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