Tutorial 3 Unit 3 - Solubility The Answer is A Answer A is the primary condition for solubility equilibrium. Remember the solution is also called saturated. B, of course is wrong because these opposing processes are continuing to happen at equilibrium (remember dynamic ), but their rates are equal so the concentration of the dissolved ions remains constant. Looking at answer C, a solute is a solid (remember by definition there has to be some solid present to have a true saturated solution). The concentration of a solid is huge (ions are packed tightly together). There is also no relation between the concentration of a solute and of a solvent. Hopefully you can see that D is impossible because the dissolved solute is only part of the solution and the mass of a part cannot be greater than the mass of the whole. The Answer is D 1
These are irrelevant! Since they re asking for the concentration of Fe 3+ and the concentrations of the solutions are given (eg. 0.050 M), the volumes given for each solution (eg. 0.40 L) are completely irrelevant! If we do a balanced dissociation equation for D : 0.010M 0.020M (1) Fe 2 (C 2 O 4 ) 3(s) 2 Fe 3+ (aq) + 3 C 2 O 4 2- (aq) With the same method, if you dissociate the equation: in A [Fe 3+ ] = 0.050 M in B [Fe 3+ ] = 0.040 M in C [Fe 3+ ] = 0.040 M 2/1 2
The Answer is B Whenever you re asked to make any comparisons of solubility you should immediately turn to your solubility table and your Ksp table in the Data booklet. You may need to use one or both of these tables. If you check your solubility table, you will find that BaS is soluble and the others all have low solubilities. So A cannot be our answer. Since CuS, FeS and ZnS all have low solubilities, we use the Ksp table to find out which one is lowest. Remember that if all compounds are the same type (AB or AB 2 ) (These compounds are all type AB ie. they have one of each ion) : The higher the Ksp, the higher the solubility. So looking up the Ksp s on the table: Compound Ksp CuS 6.0 x 10-37 FeS 6.0 x 10-19 ZnS 2.0 x 10-25 We can see that CuS (Answer B ) has the lowest Ksp and therefore the lowest solubility. Or Ksp = [ion][ion] If the [ion] is high then lots of the compound dissolved = then it is highly soluble. Which is obviously not low solubility So small [ions] = small numbers = small Ksp = low solubility 3
The Answer is D You will see that the two possible products are SrS and Mg(OH) 2. Looking these up on the solubility table we see that SrS is soluble and Mg(OH) 2 has low solubility. Therefore Mg(OH) 2 will be the solid (precipitate) and SrS will be aqueous and dissociated. You have to be careful how you read this question. They are asking for the complete ionic equation. Remember that A is not correct because they have the SrS as a solid. B is the correct formula equation but that s not what they re asking for. C has the wrong precipitate and D is the correct complete ionic equation. Recall that if they asked for the net ionic equation, it would be: Mg 2+ (aq) + 2 OH - (aq) Mg(OH) 2(s) 4
The Answer is B A and C both dissociate to form OH - ions. Increasing the [OH - ] will shift the equilibrium to the left due to LeChatelier s Principle (LCP). Remember, this is called the Common Ion Effect. D dissociates into Fe 2+ and NO 3 -. Increasing the [Fe 2+ ] will again cause a shift to the left due to LCP. B dissociates into Na + and S 2-. Looking at the solubility table beside Sulphide, we see that S 2- forms a precipitate with Fe 2+. (Fe 2+ (aq) + S 2- (aq) FeS (s) ). The addition of Na 2 S, therefore will decrease the [Fe 2+ ] and cause the equilibrium above to shift to the right. This, of course will increase the solubility of the Fe(OH) 2(s). 5
The Answer is A The diagram in this question simply gives us the information that our unknown forms a precipitate with Ag + but not with Sr 2+. So again we consult our trustworthy old solubility table and go through the suggested answers. Looking at Hydroxide, we see that OH - forms a precipitate (compound of low solubility) with Ag + but not with Sr 2+. Therefore, this is our answer. In B NO 3 - does not precipitate with any cation. In C, PO 4 3- precipitates both Ag + and Sr 2+. In answer D, SO 4 2- also precipitates both Ag + and Sr 2+. 6
The Answer is C Please, NEVER confuse the solubility (which is the moles of a compound which will dissolve in a Litre of solution) with the Ksp (the solubility product). We first have to calculate the Ksp of the compound and then see which one matches on the Ksp table. The answers are all type AB compounds (one of each ion), so we can temporarily call our mystery compound AB (s) Equation for the Solubility equilibrium: The 7.1 x 10-5 M written on top is the molar solubility: ( -7.1 x 10-5 M is written above the solid AB because the solubility is how many moles of AB will dissolve per Litre. Remember the actual concentration of a solid doesn t change.) -7.1 x 10-5 M +7.1 x 10-5 M +7.1 x 10-5 M AB (s) A 2+ (aq) + B 2- (aq) (Don t worry if the charges aren t correct at this point) The Ksp expression is: Ksp = [A 2+ ] [B 2- ] So Ksp = (7.1 x 10-5 ) 2 Ksp = 5.0 x 10-9 Looking on the Ksp table, 5.0 x 10-9 is the Ksp for CaCO 3. So C is the correct answer. 7
Jan 2000 The answer is B Looking up Al 2 (SO 4 ) 3 on the solubility table tells us that it is soluble. The question does not mention anything about the solution being saturated, so this is simply an individual ion concentration problem. The [Al 3+ ] and [SO 4 2- ] can be obtained by using the balanced dissociation equation and the coefficient ratios as follows: x 2/1 0.25M 0.50 M Al 2 (SO 4 ) 3(s) 2Al 3+ (aq) + 2-3SO 4 (aq) x 3/1 0.25M 0.50 M 0.75 M Al 2 (SO 4 ) 3(s) 2Al 3+ (aq) + 2-3SO 4 (aq) 8
The answer is C For this one, we need to look at the Solubility Table. It s important here to realize that the compounds that could form precipitates are not the ones listed here, but the products that would result by mixing each pair. You must determine the possible products in each case and use the solubility table to find which products form precipitates and which ones don t. 2KOH + CaCl 2 2 KCl + Ca(OH) 2 soluble low solubility 3Zn(NO 3 ) 2 + 2K 3 PO 4 6KNO 3 + Zn 3 (PO 4 ) 2 soluble low solubility Sr(OH) 2 + (NH 4 ) 2 S 2NH 4 OH + SrS Na 2 SO 4 + Pb(NO 3 ) 2 2NaNO 3 + PbSO 4 soluble low solubility For this reaction (C) both products are soluble so no soluble soluble precipitate will be produced. The answer is D 9
The equilibrium equation describing a saturated solution of CaSO 4 is: CaSO 4(s) Ca 2+ (aq) + SO 4 2- (aq) Adding the H 2 SO 4 will increase the [SO 4 2- ]. This will cause the equilibrium to shift to the left, which will produce more solid CaSO 4. The answer is A Remember solubility and Ksp are NOT the same thing! The solubility equilibrium equation for CdS is: CdS (s) Cd 2+ (aq) + S 2- (aq) If we call the solubility -2.8 x 10-14 M, we can show the changes in concentration as the reaction reaches equilibrium: -2.8 x 10-14 M +2.8 x 10-14 M +2.8 x 10-14 M CdS (s) Cd 2+ (aq) + S 2- (aq) The Ksp expression is: Ksp = [Cd 2+ ] [S 2- ] substituting the concentrations: Ksp = (2.8 x 10-14 ) 2 = 7.8 x 10-28 10
The answer is C First of all we must find the molar solubility of FeS. We do this by writing the equation for the solubility equilibrium of FeS and using the Ksp expression. Since we don t know the molar solubility, we call is s : -s +s +s FeS (s) Fe 2+ (aq) + S 2- (aq) Ksp = [Fe 2+ ] [S 2- ] and substituting s for the concentrations: Ksp = s 2 Looking up the Ksp for FeS on the Ksp table, it is 6.0 x 10-19. So the molar solubility, s = Ksp = 6.0 19 x 10 = 7.75 x 10-10 M Remember: moles M L We can now use: moles = M x L to calculate the moles: Moles = 7.75 x 10-10 M x 0.2000 L = 1.5 x 10-10 moles 11
The answer is D If you look on the solubility table you will notice that IO - 3 and BrO - 3 are not on the table. The next step would be to look at the Ksp table. Comparing the Ksp s of the possible precipitates that could form: Possible precipitate Ksp AgCl 1.8 x 10-10 AgBr 5.4 x 10-13 AgIO 3 3.2 x 10-8 AgBrO 3 5.3 x 10-5 We see that AgBrO 3 has the highest Ksp. Since these compounds are all the same type (AB compounds), the higher the Ksp, the higher the solubility. So AgBrO 3 has the highest solubility of the four. Therefore if only one of these does not form a precipitate, the AgBrO 3 must be the one or BrO 3 - is the ion that does not precipitate with Ag +. 12
The answer is A When you are asked about compounds dissolving precipitates in a saturated solution, the first thing to do it write the solubility equilibrium equation for the compound of LOW solubility. In this case, that is CaC 2 O 4 : CaC 2 O 4(s) Ca 2+ (aq) + C 2 O 4 2- (aq) First, why B, C and D don t work! CaC 2 O 4 and H 2 C 2 O 4 both release the C 2 O 4 2- ion to the solution. This increases the [C 2 O 4 2- ] in the equilibrium above, causing a shift to the left, forming more solid precipitate, rather than dissolving it. The Ca 2+ in Ca(NO 3 ) 2 would increase the [Ca 2+ ] in the equilibrium above also causing a shift to the left, again forming more solid precipitate, rather than dissolving it. Looking at the solubility table, OH - forms a compound of low solubility with Ca 2+. So when NaOH is added to saturated CaC 2 O 4, the OH - precipitates the Ca 2+, forming Ca(OH) 2(s). This decreases the [Ca 2+ ] in the equilibrium above, causing it to shift to the right, thus dissolving the solid CaC 2 O 4. 13
The Answer is A The Answer is B The Answer is D 14
The Answer is C The Answer is B The Answer is C 15
The Answer is B The Answer is C The Answer is B 16
The Answer is A The Answer is A 17
The Answer is B The Answer is C The Answer is D 18
Tutorial 3 Unit 3 - Solubility (student handout) 19
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