# 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).

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#.*

*Sign up to view the whole answer*

*By signing up, you agree to our Terms of Service and Privacy Policy*

*Sign up with email*

*Answer 2Sign up to view the whole answerSign up with email*

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

By signing up, you agree to our Terms of Service and Privacy Policy

*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.*

*Trending questions*

*For an equilibrium reaction, how can #K_c#, and #K_P# be related?**Which one of the following equilibrium reactions corresponds to the β2 constant for the complexation reaction described above? Part a) Ag+(aq) + 2Cn^-1(aq)**How much #KBr# should be added to #1# #L# of #0.05# #M# #AgNO_3# solution just to start precipitation of #AgBr#? #K_(sp)# of #AgBr# = #5# x #10^-13#.**At 25°C in a closed system, ammonium hydrogen sulfide exists as the following equilibrium. NH4HS(s)<--->NH3(g) + H2S(g) When a sample of pure NH4HS(s) is placed in an evacuated reaction vessel and allowed to come to equilibrium at 25°C, total pressure is 0.660 atm. What is the value of Kp?**When I ask my calculator to divide #0.35# by #( 0.15 ) * ( 0.30 )^2# it gives me the result #0.21#. The correct result should be about #26#. What's going wrong?*

*Not the question you need?*

*HIX TutorSolve ANY homework problem with a smart AI*

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7