Tutorial 3 Unit 3 - Solubility

Chemistry 12Unit 3 - Tutorial

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

Since they’re asking for the concentration of Fe3+ 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) Fe2(C2O4)3(s) 2 Fe3+(aq) + 3 C2O42-(aq)

With the same method, if you dissociate the equation:

in “A”  [Fe3+] = 0.050 M
in “B”  [Fe3+] = 0.040 M
in “C”  [Fe3+] = 0.040 M

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 AB2) (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

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: Mg2+(aq) + 2 OH-(aq)  Mg(OH)2(s)

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 Fe2+ and NO3- . Increasing the [Fe2+] will again cause a shift to the left due to LCP.

“B” dissociates into Na+ and S2-. Looking at the solubility table beside “Sulphide”, we see that S2- forms a precipitate with Fe2+.
(Fe2+(aq) + S2-(aq)  FeS(s)).

The addition of Na2S, therefore will decrease the [Fe2+] and cause the equilibrium above to shift to the right. This, of course will increase the solubility of the Fe(OH)2(s).

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 Sr2+. 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 Sr2+. Therefore, this is our answer. In “B” NO3- does not precipitate with any cation. In “C”, PO43- precipitates both Ag+ and Sr2+. In answer “D”, SO42- also precipitates both Ag+ and Sr2+.

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 10-5M” written on top is the molar solubility: (“-7.1 x 10-5M” 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 10-5M +7.1 x 10-5M +7.1 x 10-5M

AB (s)  A2+(aq) + B2-(aq) (Don’t worry if the charges aren’t correct at this point)

The Ksp expression is: Ksp = [A2+] [B2-]

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 CaCO3. So “C” is the correct answer.

Jan 2000

The answer is B

Looking upAl2(SO4)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 [Al3+] and [SO42-] can be obtained by using the balanced

dissociation equation and the coefficient ratios as follows:

x 2/1

0.25M 0.50 M

Al2(SO4) 3(s)  2Al3+(aq) + 3SO42-(aq)

x 3/1

0.25M 0.50 M 0.75 M

Al2(SO4) 3(s)  2Al3+(aq) + 3SO42-(aq)

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 + CaCl2 2 KCl + Ca(OH)2

soluble low solubility

3Zn(NO3)2 + 2K3PO4  6KNO3 + Zn3(PO4)2

soluble low solubility

Sr(OH)2 + (NH4)2S  2NH4OH + SrS For this reaction (C) both products are soluble so no

soluble soluble precipitate will be produced.

Na2SO4 + Pb(NO3)2  2NaNO3 + PbSO4

soluble low solubility

The answer is D

The equilibrium equation describing a saturated solution of CaSO4 is:

CaSO4(s) ↔ Ca2+(aq) + SO42-(aq)

Adding the H2SO4 will increase the [SO42- ].

This will cause the equilibrium to shift to the left, which will produce more solid CaSO4.

The answer is A

Remember solubility and Ksp are NOT the same thing! The solubility equilibrium equation for CdS is: CdS(s) ↔ Cd2+(aq) + S2-(aq)

If we call the solubility “-2.8 x 10-14M”, we can show the changes in concentration as the reaction reaches equilibrium:

-2.8 x 10-14M +2.8 x 10-14M +2.8 x 10-14M

CdS(s) ↔ Cd2+(aq) + S2-(aq)

The Ksp expression is:

Ksp = [Cd2+] [S2-] substituting the concentrations:

Ksp = (2.8 x 10-14)2 = 7.8 x 10-28

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) ↔ Fe2+(aq) + S2-(aq)

Ksp = [Fe2+] [S2-] and substituting “s” for the concentrations:

Ksp = s2

Looking up the Ksp for FeS on the Ksp table, it is 6.0 x 10-19.

So the molar solubility, s = = = 7.75 x 10-10 M

Remember:

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

The answer is D

If you look on the solubility table you will notice that IO3- and BrO3- 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
AgIO3 / 3.2 x 10-8
AgBrO3 / 5.3 x 10-5

We see that AgBrO3 has the highest Ksp. Since these compounds are all the same type (AB compounds), the higher the Ksp, the higher the solubility. So AgBrO3 has the highest solubility of the four. Therefore if only one of these does not form a precipitate, the AgBrO3 must be the one or BrO3- is the ion that does not precipitate with Ag+.

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 CaC2O4 :

CaC2O4(s) ↔ Ca2+ (aq) + C2O42-(aq)

First, why B, C and D don’t work! CaC2O4 and H2C2O4 both release the C2O42- ion to the solution. This increases the [C2O42-] in the equilibrium above, causing a shift to the left, forming more solid precipitate, rather than dissolving it.

The Ca2+ in Ca(NO3)2 would increase the [Ca2+ ] 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 Ca2+ . So when NaOH is added to saturated CaC2O4, the OH- precipitates the Ca2+ , forming Ca(OH)2(s) . This decreases the [Ca2+ ] in the equilibrium above, causing it to shift to the right, thus dissolving the solid CaC2O4.

The Answer is A

The Answer is B

The Answer is D

The Answer is C

The Answer is B

The Answer is C

The Answer is B

The Answer is C

The Answer is B

The Answer is A

The Answer is B

The Answer is C

The Answer is D

Tutorial 3 – Unit 3 - Solubility

(student handout)

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