Unit F FR (part B) Solubility Equilibrium,...

1. Answer the following questions about the solubility of some fluoride salts of alkaline earth metals.

(a) A student prepares 100. mL of a saturated solution of MgF2 by adding 0.50 g of solid MgF2 to 100. mL of distilled water at 25ºC and stirring until no more solid dissolves. (Assume that the volume of the undissolved MgF2 is negligibly small.) The saturated solution is analyzed, and it is determined that [F−] in the solution is 2.4 × 10−3 M. (i) Write the chemical equation for the dissolving of solid MgF2 in water.

(ii) Calculate the number of moles of MgF2 that dissolved.

(iii) Determine the value of the solubility-product constant, Ksp , for MgF2 at 25ºC.

(b) A beaker contains 500. mL of a solution in which both Ca2+(aq) and Ba2+(aq) are present at a concentration of 0.10 M at 25ºC. A student intends to separate the ions by adding 0.20 M NaF solution one drop at a time from a buret. At 25ºC the value of Ksp for CaF2 is 3.5 × 10−11; the value of Ksp for BaF2 is 1.8 × 10−6. (i) Which salt will precipitate first, CaF2 or BaF2 ? Justify your answer.

For parts (b ii) and (b iii) below, assume that the addition of the NaF solution does not significantly affect the total volume of the liquid in the beaker.

(ii) Calculate the minimum concentration of F−(aq) necessary to initiate precipitation of the salt selected in part (b i).

(iii) Calculate the minimum volume of 0.20 M NaF that must be added to the beaker to initiate precipitation of the salt selected in part (b i).

(c) There are several ways to dissolve salts that have limited solubility. Describe one procedure to redissolve the precipitate formed in part (b).

Unit F FR (part B) Solubility Equilibrium, Ksp (pg 1 of 16)

ANSWER #1

(a i) Remember that this is the “same,” only backwards to the way that we used to write precipitation reactions.

MgF2 ⇋ Mg2+ + 2 F−

(a ii) 2.4 ×10−3molF−

1.0L× 0.100L × 1molMgF2

2molF− = 1.2 ×10−4molMgF2

(a iii) [Mg2+] = ½ [F−] = ½[2.4 × 10−3] = 1.2 × 10−3 M

Ksp = [Mg2+][F−]2 substitute: Ksp = [1.2 × 10−3] [2.4 × 10−3]2 = 6.9 × 10−9

(b i) CaF2 will precipitate first. Its Ksp value is smaller (and the salt is the same number of ions; MX2), thus the ion-concentration product [Ca2+] [F−]2 will be the first to exceed the Ksp value. When a Q exceeds Ksp the solution will begin to precipitate.

(b ii) A precipitate will form when the product of the ion concentrations exceeds the Ksp, The concentration of the cations are very high and thus we can treat this problem as a “common ion” problem. Substituting in the high concentration of the cation, and solving for the concentration that the F− ion has to get to (adding it drop by drop) until it is high enough to cause a precipitate.

Ksp = [Ca2+][F−]2 Substitute: 3.5 × 10−11 = [0.10][F−]2

and solve: 3.5 ×10−10 = F−⎡⎣ ⎤⎦2

3.5 ×10−10 = F−⎡⎣ ⎤⎦ F−⎡⎣ ⎤⎦ = 1.9 ×10−5M

(b iii) Assuming that the volume of the NaF that is added (drop by drop) does not effectively change the total volume, we can use the dilution equation, (M1V1 = M2V2) to calculate the volume of NaF that must have been added.

(0.20 M)(V1) = 1.9 × 10−5 M)(0.500 L) V1 = 4.7 × 10−5 L which is 0.047 ml, (which is essentially the volume of one drop.)

(c) Any procedure that would move the equilibrium reaction towards products would work. This might include any of the following:

• adding water,• changing the temperature (without knowing if the ∆H of the dissolving process is exothermic or endothermic, we can

not know if raising or lowering that would help − if dissolving this salt is endothermic, raising would help dissolve the precipitate, and if dissolving this salt is exothermic, cooling will help dissolve the precipitate.)

• adding acid, H+ (In the next unit, we will learn more about why the acid will help dissolve the precipitate.)


2. Answer the following questions about the solubility and reactions of the ionic compounds M(OH)2 and MCO3, where M represents an unidentified metal.(a) Identify the charge of the M ion in the ionic compounds above.

Save part (i) for unit G

(b) At 25ºC, a saturated solution of M(OH)2 has a pH of 9.15.

(i) Calculate the molar concentration of OH−(aq) in the saturated solution

(ii) Write the solubility-product constant expression for M(OH)2

(iii) Calculate the value of the solubility−product constant, Ksp , for M(OH)2 at 25°C.

(c) For the metal carbonate, MCO3, the value of the solubility-product constant, Ksp, is 7.4 × 10−14 at 25°C. On the basis of this information and your results in part (b), which compound, M(OH)2 or MCO3, has the greater molar solubility in water at 25°C? Justify your answer with a calculation.

(d) MCO3 decomposes at high temperatures, as shown by the reaction represented below.MCO3(s) ⇋ MO(s) + CO2(g)

A sample of MCO3 is placed in a previously evacuated container, heated to 423 K, and allowed to come to equilibrium. Some solid MCO3 remains in the container. The value of Kp for the reaction at 423 K is 0.0012. (i) Write the equilibrium constant expression for Kp of the reaction.

(ii) Determine the pressure, in atm, of CO2(g) in the container at equilibrium at 423 K.

(iii) Indicate whether the value of ∆Gº for the reaction at 423 K is positive, negative, or zero. Justify your answer


ANSWER #2

(a) 2+ We know this because 2+ 2 OH− ions and one CO32− ion carry a 2− charge, thus on M ion must be 2+.

Save part (i) for unit G(b i) pH = 9.15, thus pOH = 4.85 (since pH + pOH = 14), and since pOH = −log [OH−], thus [OH−] = 10−4.85 = 1.4 × 10−5 M

(b ii) Since the chemical equation would be M(OH)2 ⇋ M2+ + 2OH−

The solubility product expression would be Ksp = [M2+][OH−]2

(b iii) [M2+] = ½ [OH−] = ½[1.4 × 10−5 M] = 7.0 × 10−6 M

Ksp = [M2+][OH−]2 substitute: Ksp = [7.0 × 10−6] [1.4 × 10−5]2 = 1.4 × 10−15

(c) From part (b iii) for M(OH)2 the [M2+] = [7.0 × 10−6]

For MCO3, Ksp = (s)(s) solve for s s = 7.4 ×10−14 = 2.7 ×10−7M

Since 7.0 × 10−6 > 2.7 × 10−7, the M(OH)2 has the greater molar solubility.

(d i) Kp = PCO2( )

(d ii) PCO2 = 0.0012atm

(d iii) ∆Gº = −RTlnK K = 0.0012 < 1, thus lnK is negative; therefore ∆Gº is positive.


3. Several reactions are carried out using AgBr, a cream-colored silver salt for which the value of the solubility-product constant, Ksp, is 5.0 × 10−13 at 298 K. (a) Write the expression for the solubility-product constant, Ksp, of AgBr.

(b) Calculate the value of [Ag+] in 50.0 mL of a saturated solution of AgBr at 298 K.

(c) A 50.0 mL sample of distilled water is added to the solution described in part (b), which is in a beaker with some solid AgBr at the bottom. The solution is stirred and equilibrium is reestablished. Some solid AgBr remains in the beaker. Is the value of [Ag+] greater than, less than, or equal to the value you calculated in part (b) ? Justify your answer.

(d) Calculate the minimum volume of distilled water, in liters, necessary to completely dissolve a 5.0 g sample of AgBr(s) at 298 K. (The molar mass of AgBr is 188 g mol−1.)

(e) A student mixes 10.0 mL of 1.5 × 10−4 M AgNO3 with 2.0 mL of 5.0 × 10−4 M NaBr and stirs the resulting mixture. What will the student observe? Justify your answer with calculations.

(f) The color of another salt of silver, AgI(s), is yellow. A student adds a solution of NaI to a test tube containing a small amount of solid, cream-colored AgBr. After stirring the contents of the test tube, the student observes that the solid in the test tube changes color from cream to yellow. (i) Write the chemical equation for the reaction that occurred in the test tube.

(ii) Which salt has the greater value of Ksp: AgBr or AgI? Justify your answer.


ANSWER #3

(a) Ksp = [Ag+][Br−] (AP has made it clear that the ions must have charges, and square brackets, not parentheses are a must.)

(b) In this problem, the 50 ml is a distractor.

Let s = [Ag+] = [Br−] Ksp = (s)(s) solve for s s = 5.0 ×10−13 = 7.1×10−7M

(c) The of [Ag+] after the addition of distilled water is the same as in part (b), 7.1 × 10−7 M. The concentration of ions in solution that is in equilibrium with a solid does not depend on the volume of the solid. In other words, as long as there is solid on the bottom, and equilibrium has been established, the concentration of the solution will be constant.

(d) 5.0gAgBr × 1mol188gAgBr

= 0.0266molAgBr 0.0266molAgBrV

= 7.1×10−7M V = 3.7 ×104 L

(e) In order to determine if a precipitate forms, solve for Q. Remember that solutions are diluted when poured together, thus you need to use the dilution equation (M1V1 = M2V2) to determine the concentration of each ion in solution.

[Ag+ ] =10.0ml( ) 1.5 ×10−4M( )

12.0ml= 1.3×10−4M [Br− ] =

2.0ml( ) 5.0 ×10−4M( )12.0ml

= 8.3×10−5M

Q = [Ag+ ][Br− ] = 1.3×10−4M( ) 8.3×10−5M( ) = 1.1×10−8M

Since Q > Ksp (1.1 × 10−8 > 5.0 × 10−13), a precipitate will form.

(f i) If the precipitate was cream-colored to begin with, AgBr(s), but then changes to cream-colored, AgI(s) must have formed.

AgBr(s) + NaI(aq) → AgI(s) + NaBr(aq) OR AgBr(s) + I−(aq) → AgI(s) + Br−(aq)

(f ii) AgBr(s) has the greater value of Ksp. When two ions are present I− and Br−, the precipitate will consist of the least soluble salt. Because the color of the precipitate in the test tube turns yellow, must be the precipitate, thus it has the smaller of the two values.

OR you could use the following justification:

Since AgBr(s) is dissolving, AgBr ⇋ Ag+ + Br− thus Ksp describes this reaction.

Since AgI(s) is forming Ag+ + I− ⇋ AgI thus the inverse of Ksp describes this reaction, 1Ksp

The overall reaction shown in (f i) is the sum of the two reactions above, the Keq for the over all reaction is the product of

the K’s for the two reactions; KspofAgBr ×1

KspofAgI Since the reaction is spontaneous, the Keq for the over all reaction

must be greater than 1, which means that Ksp of AgBr must be larger than the Ksp of AgI


4. Answer the following questions relating to the solubilities of two silver compounds, Ag2CrO4 and Ag3PO4. Silver chromate dissociates in water according to the equation shown below.

Ag2CrO4(s) ⇄ 2 Ag+(aq) + CrO42−(aq) Ksp = 2.6 × 10−12 at 25°C

(a) Write the equilibrium-constant expression for the dissolving of Ag2CrO4(s).

(b) Calculate the concentration, in mol L−1, of Ag+(aq) in a saturated solution of Ag2CrO4 at 25°C.

(c) Calculate the maximum mass, in grams, of Ag2CrO4 that can dissolve in 100. mL of water at 25°C.

(d) A 0.100 mol sample of solid AgNO3 is added to a 1.00 L saturated solution of Ag2CrO4. Assuming no volume change, does [CrO4 −2] increase, decrease, or remain the same? Justify your answer.

In a saturated solution of Ag3PO4 at 25°C, the concentration of Ag+(aq) is 5.3 × 10–5 M. The equilibrium-constant expression for the dissolving of Ag3PO4(s) in water is shown below.

Ksp = [Ag+]3[PO43−]

(e) Write the balanced equation for the dissolving of Ag3PO4 in water.

(f) Calculate the value of Ksp for Ag3PO4 at 25°C.

(g) A 1.00 L sample of saturated Ag3PO4 solution is allowed to evaporate at 25°C to a final volume of 500. mL. What is [Ag+] in the solution? Justify your answer.


ANSWER #4(a) Remember that in equilibrium expressions, solids and pure liquids are not included. Solubility product expressions are

never a quotient, always a product because the left side of the reaction will always be a solid. Square brackets must be used, and charges on ions must be shown.

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(b) We should think about this problem as a RICE Box, in which we do not need to concern ourselves with the silver chromate. All Ksp expressions in which there are three ions, turn into 4x3 problems.

(c) In the previous problem x is solved as the concentration of chromate ion that forms, but it stoichiometrically it also represents the molarity of the silver chromate that dissolves. Remember that molarity is per liter, and in the problem we are asked to determine the mass that dissolves in 100 ml.

8.7 ×10−5mol Ag2CrO4

1L× 331.7g

1mol Ag2CrO4

× 0.1L = 0.0029g Ag2CrO4

(d) The [CrO42−] will decrease. Adding [Ag+] will make Q (the equilibrium expression with non equilibrium values) greater than K. To reestablish equilibrium, the reaction will proceed from right to left (the reverse reaction), decreasing Q and returning to equilibrium. LeChatelier’s Principle is at work, since “stress” (increased silver concentration) upset the equilibrium, in order to reestablish equilibrium, the reaction must shift left to relieve this stress.

(e)

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Since we know [Ag+] = 5.3 × 10−5, we know that [PO43−] will be ⅓ of that value.

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(g) [Ag+] = 5.3 × 10−5 . The concentration remains the same because even though the volume became smaller, silver phosphate ions will crystallize into solid at the bottom, maintaining the same concentration in the saturated solution.


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1. Solve the following problem related to the solubility equilibria of some metal hydroxides in aqueous solution. (a) The solubility of Cu(OH)2 is 1.72 × 10−6 gram per 100. milliliters of solution at 25°C.

(i) Write the balanced chemical equation for the dissociation of Cu(OH)2(s) in aqueous solution.

(ii) Calculate the solubility (in moles per liter) of Cu(OH)2 at 25 °C.

(iii) Calculate the value of the solubility-product constant, Ksp, for Cu(OH)2 at 25°C.

(b) The value of the solubility-product constant, Ksp, for Zn(OH)2 is 7.7 × 10−17 at 25°C.

Save part (i) for unit G

(i) Calculate the solubility (in moles per liter) of Zn(OH)2 at 25°C in a solution with a pH of 9.35.

(ii) At 25°C, 50.0 milliliters of 0.100-molar Zn(NO3)2 is mixed with 50.0 milliliters of 0.300-molar NaOH.

• Calculate the molar concentration of Zn2+(aq) in the resulting solution once equilibrium has been established. Assume that volumes are additive.

• Calculate the concentration of nitrate ions in the resulting solution once equilibrium has been established. Assume that volumes are additive.


ANSWER #5(a i) Remember that solubility reactions are always written with the solid on the left side.

Cu(OH)2 ⇋ Cu2+ + 2 OH−

(a ii) You are told the mass that dissolves in 100. ml, turn this into moles per liter, for molar solubility.

1.72 ×10−6g0.10L

× 1mol97.57g

= 1.76 ×10−7M

(a iii) The stoichiometry of the balanced equation above, tells us that the concentration of the hydroxide ion is twice the concentration of the copper ion. Since [Cu2+] = 1.76 × 10−7, then [OH−] = 3.52 × 10−7

Ksp = [Cu2+ ][OH − ]2 substitute values from above Ksp = [1.76 ×10

−7M ][3.52 ×10−7M ]2 = 2.18 ×10−20

(b i) Remember that pH + pOH = 14 and −log[OH−] = pOH

pH = 9.35 ➙ pOH = 4.65 ➙ [OH−] = 2.24× 10−5 (carry extra SF into the next part of the problem)

Ksp = [Zn2+ ][OH − ]2 substitute and solve 7.7 ×10−17M = [Zn2+ ][2.24 ×10−5M ]2 ➙ [Zn2+ ] = 1.5 ×10−7M

(b ii) The easiest way to solve this type of problem is to first “pretend” that the reaction precipitates as completely as the quantities of reactants will allow.

First use the M1V1 = M 2V2 equation to solve for the concentration of the zinc and hydroxide ions immediately after they are poured together and would be diluted by each other. Since equal volumes are poured together, the volume doubles, causing the concentration to be halved, thus [Zn2+] = 0.05 M and [OH−] = 0.15 M.

Zn(OH)2 ⇋ Zn2+ + 2 OH− Again, pretending for just a moment that the reaction goes to completion....the stoichiometry of the reaction tells us that the 0.05 M of Zn2+ will react with 0.1 M of OH−, leaving 0.05 M of excess OH−.

However, this would leave the Zn2+ as essentially zero molarity....and that can’t really happen in an equilibrium, thus we must use the solubility expression and the concentration of the OH− (as if we are doing a “common ion” problem) to determine just how small (nearly zero) the Zn2+ ions would be.

Ksp = [Zn2+ ][OH − ]2 substitute in the excess OH− 7.7 ×10−17M = [Zn2+ ][0.05M ]2 ➙ [Zn2+ ] = 3.1×10−14M which

of course is practically zero.

SPECIAL NOTE

Solving the problem in the manner above, is an approximation. If we consider the [Zn2+] = x then the quantity of [O] doesn’t actually react all the way down to 0.05 M but would be larger by 2x, thus the reaction would be more correct if substituted in this way 7.7 ×10−17M = [x][0.05 + 2x]2 , however if you go to all the trouble to solve this equation, you would find that the problem would solve out nearly to the same values. If x = 3.1 × 10−14, then 2x = 6.2 × 10−14 which can not possibly be added to 0.05, significant figures will not allow this.

You might ask yourself, why bother to solve for [Zn2+] at all, if it is so nearly zero? Compared to zero, that quantity is significant, but compared to 0.05, it is not significant.


2. In a saturated solution of MgF2 at 18°C, the concentration of Mg2+ is 1.21 × 10−3 molar. The equilibrium is represented by the equation below.

MgF2(s) ⇄ Mg2+(aq) + 2 F−(aq)

(a) Write the expression for the solubility-product constant, Ksp, and calculate its value at 18° C.

(b) Calculate the equilibrium concentration of Mg2+ in 1.000 liter of saturated MgF2 solution at 18°C to which 0.100 mole of solid KF has been added. The KF dissolves completely. Assume the volume change is negligible.

(c) Predict whether a precipitate of MgF2 will form when 100.0 milliliters of a 3.00 x 10−3 molar Mg(NO3)2 solution is mixed with 200.0 milliliters of a 2.00 × 10−3 molar NaF solution at 18°C. Calculations to support your prediction must be shown.

(d) At 27°C the concentration of Mg2+ in a saturated solution of MgF2 is 1.17 × 10−3 molar. Is the dissolving of MgF2 in water an endothermic or an exothermic process? Give an explanation to support your conclusion.


ANSWER #6(a) The concentration of F is twice the M because of the stoichiometry of the reaction MgF2 ⇋ Mg2+ + 2 F−

Ksp = [Mg2+ ][F− ]2 ➙ Ksp = [1.21×10

−3 ][2.42 ×10−3 ]2 ➙ Ksp = 7.09 ×10−9

(b) When solid KF is added, it immediately dissociates since it is completely soluble. This increases the [F−], making this a “common ion” type problem. Increasing [F−] will decrease the Mg. We now know the Ksp so we can use that to solve this problem.

7.09 ×10−9 = [Mg2+ ][F− ]2 ➙ 7.09 ×10−9 = [Mg2+ ][0.1]2 ➙ [Mg2+ ] = 7.09 ×10−7

(c) Whenever you are asked if a precipitate will form you should be solving for Q and comparing Q to K. Also whenever two solutions are poured together, you must remember that each solution becomes diluted, thus you must use McVc = MdVdtwice (in the equation “c” stands for concentrated, and “d” stands for dilute). Remember to add the two volumes to get total volume.

McVc = MdVd for Mg(NO3)2 0.003M( ) 100ml( ) = Md( ) 300ml( ) ➙ for Mg(NO3)2 [Mg2+] = 0.001 M

McVc = MdVd for NaF 0.002M( ) 200ml( ) = Md( ) 300ml( ) ➙ for NaF [F−] = 0.00133 M

Q = [Mg2+ ][F− ]2 ➙ Q = [0.001][0.00133]2 = 1.77 ×10−9 ➙ Q < Ksp, thus no precipitate forms.

(d) At this higher temperature, the amount of MgF that dissolves is smaller indicating that the equilibrium has shifted to the left when energy is added. This tells us that the reaction, MgF2 ⇋ Mg2+ + 2 F− must be exothermic because when the stress of head is placed on the system, it shifts to relieve that stress by moving away from the energy on the right, shifting left, thus exothermic.


1. The solubility of iron(II) hydroxide, Fe(OH)2, is 1.43 × 10−3 gram per liter at 25 °C.(a) Write a balanced equation for the solubility equilibrium.

(b) Write the expression for the solubility product constant, Ksp, and calculate its value.

Save part (c) for unit G

(c) Calculate the pH of the saturated solution of Fe(OH)2 at 25 °C.

(d) A 50.0-milliliter sample of 3.00 × 10−3 molar FeSO4 solution is added to 50.0 milliliters of 4.00 × 10−6 molar NaOH solution. Does a precipitate of Fe(OH)2 form? Explain and show calculations to support your answer.


ANSWER #7

(a) Fe(OH)2 ⇋ Fe2+ + 2 OH−

(b) Ksp = [Fe2+][OH−]2 1.43×10−3g1L

× 1mol89.87g

= 1.59 ×10−5M Fe(OH )2 = 1.59 ×10−5M Fe2+

1.59 ×10−3MFe2+ × 2OH −

1Fe3+ = 3.18 ×10−5M OH − substitute: Ksp = [1.59 × 10−5] [3.18 × 10−5]2 = 1.61 × 10−14

Save part (c) for unit G

(c) pOH = −log [OH−] pOH = −log[3.18 × 10−5] = 4.498 (since pH + pOH = 14) pH = 9.502

Alternatively Kw = [H+][OH−] thus

H +⎡⎣ ⎤⎦ =Kw

OH −⎡⎣ ⎤⎦ H +⎡⎣ ⎤⎦ =

1×10−14

3.18 ×10−5⎡⎣ ⎤⎦= 3.14 ×10−10 pH = − log[H + ] pH = − log[3.18 ×10−5 ] = 9.502

(d) Whenever you are asked if a precipitate will form you should be solving for Q and comparing Q to K.

Whenever two solutions are poured together, you must remember that each solution becomes diluted, thus you should use McVc = MdVd twice (in the equation “c” stands for concentrated, and “d” stands for dilute).

Remember to add the two volumes to get total volume, and as you can see the volume is simply doubled, thus the concentrations are halved.

Thus upon pouring together, [Fe2+] = ½[3.0 × 10−3] = 1.5 × 10−3 M and [OH−] = ½[4.6 × 10−6] = 2.0 × 10−6 M

Q = Mg2+⎡⎣ ⎤⎦ F−⎡⎣ ⎤⎦2

substitute: Q = 1.5 ×10−3⎡⎣ ⎤⎦ 2.0 ×10−6⎡⎣ ⎤⎦2= 6.0 ×10−15

Q < Ksp, (6.0 × 10−15 < 1.61 × 10−14) thus no precipitate forms.


2. At 25°C the solubility product constant, Ksp, for strontium sulfate, SrSO4, is 7.6 × 10−7. The solubility product constant for strontium fluoride, SrF2, is 7.9 × 10−10.(a) What is the molar solubility of SrSO4 in pure water at 25°C?

(b) What is the molar solubility of SrF2 in pure water at 25°C?

(c) An aqueous solution of Sr(NO3)2 is added slowly to 1.0 liter of a well stirred solution containing 0.020 mole F− and 0.10 mole SO42− at 25°C. (You may assume that the added Sr(NO3)2 solution does not materially affect the total volume of the system.)(i) Which salt precipitates first?

(ii) What is the concentration of strontium ion, Sr2+, in the solution when the first precipitate begins to form?

(d) As more Sr(NO3)2 is added to the mixture in (c) a second precipitate begins to form. At that stage, what percent of the anion of the first precipitate remains in solution?


ANSWER #8(a) SrSO4 ⇋ Sr2+ + SO42− in an saturated solution [Sr2+] = [SO42−], call them “s”

Ksp = [Sr2+ ][SO42− ] 7.6 ×10−7 = [s][s] s = [Sr2+ ] = [SO4

2− ] = SrSO4[ ] = 8.7 ×10−4M

(b) SrF2 ⇋ Sr2+ + 2 F− in an saturated solution [F−] = 2×[Sr2+]

Ksp = [Sr2+ ][F− ]2 7.9 ×10−10 = [s][2s]2 s = [Sr2+ ] = SrF2[ ] = 5.8 ×10−4M

(c i) You can NOT use the “s” values in the previous parts because in this problem you are adding Sr2+ ions into a solution that already has SO42− and F− ions in it. This means that you must run a “common ion” problem to calculate the concentration of Sr ions required to just reach equilibrium, which would be the moment at which precipitate would start to show up.

Ksp = Sr2+⎡⎣ ⎤⎦ SO42−⎡⎣ ⎤⎦ 7.6 ×10−7 = Sr2+⎡⎣ ⎤⎦ 0.1[ ] Sr2+⎡⎣ ⎤⎦ = 7.6 ×10−6M

Ksp = Sr2+⎡⎣ ⎤⎦ F−⎡⎣ ⎤⎦2

7.9 ×10−10 = Sr2+⎡⎣ ⎤⎦ 0.02[ ]2 Sr2+⎡⎣ ⎤⎦ = 2.0 ×10−6M

The calculations above give the concentration that Sr would need to reach (as you were dripping it into the solution) in order to see a precipitate.

The smaller concentration required for SrF2 forms a precipitate first.

(c ii) This question asks for an answer that you have already calculated. From the calculations in part (c i), when the concentration of [Sr2+ ] = 2.0 ×10−6M then a precipitate will begin to form.

(d) This is a tricky question. It tells you that more and more Sr2+ is added. This means that the Sr2+ will continue to precipitate with the F−, but as the F− begins to run out, the [Sr2+] begins to increase. When [Sr2+] reaches the level required for the SrSO4 precipitate to form in a solution with [SO42−] = 0.10 M (7.6 × 10−6 as calculated in part (c i)), we can calculate the [F−] that would be present by using the [SO42−] in the SrF2 Ksp expression.

Ksp = [Sr2+ ][F− ]2 7.9 ×10−10 = [7.6 ×10−6 ][F− ]2 [F− ] = 1.0 ×10−2M this would be the concentration that the F has

dropped to when the [SO42−] rose to the amount required to see the more soluble SrSO4 precipitate show up.

This does not quite answer the question, which asked what percentage of the F is remaining compared to the starting F concentration which was 0.02 M.

%F−still in solution = [F− ]present [F− ]originally

= 1.0 ×10−2M0.020M

×100 = 50%


Unit F FR (part B) Solubility Equilibrium,...

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