**Solubility Practice Questions**

Note:Ksp Values can be obtained from any online resources or your text book.

Trial Ksp is also sometimes referred to as Qsp

Conditions for precipitation are Qsp > Ksp.

If Qsp = Ksp the system is in equilibrium

If Qso < Ksp there will be no precipitation taking place if the two solutions are mixed.

**Practice Questions**:

1. One liter of water is able to dissolve 2.15 x 10^{-3} mol of PbF_{2}. What is the K_{sp} for PbF_{2}?

2. The molar solubility of CoCO_{3} in a 0.10 M Na_{2}CO_{3} solution is 1.0 x 10^{-9} mol/L. What is K_{sp} for CoCO_{3}?

3. The molar solubility of PbF_{2} in a 0.10 M Pb(NO_{3})_{2} solution is 3.1 x 10^{-4} mol/L. Calculate K_{sp} for PbF_{2}.

4. What is the molar solubility of AgBr in water?

5. What is the molar solubility of Ag_{2}CO_{3} in water?

6. What is the molar solubility of AgI in 0.20 M NaI solution?

7. What is the molar solubility of Fe(OH)_{3} in a solution with a hydroxide ion concentration of 0.050 M?

8. Will a precipitate of CaSO_{4} form in a solution if the Ca^{2+} concentration is 0.0025 M and the SO_{4}^{–} concentration is 0.030 M? For CaSO_{4}, K_{sp} = 2.4 x 10^{-5}.

9. Will a precipitate form in a solution containing 3.4 x 10^{-4} M CrO_{4}^{-2} and 4.8 x 10^{-5} M Ag^{+}?

10. Will a precipitate of PbSO_{4} form if 100 mL of 1.0 x 10^{-3} M Pb(NO3)_{2} solution is added to 100 mL of 2.0 x 10^{-3} M MgSO_{4} solution?

11. Will a precipitate of PbCl_{2} form if 50.0 mL of 0.10 M Pb(NO_{3})_{2} solution is added to 20.0 mL of 0.040 M NaCl solution?

12. Barium Sulfate is so insoluble that it can be swallowed without significant danger, even though Ba^{2+ }is toxic. At 25^{o}C, 1.00L of water dissolves only 0.00245 g of BaSO_{4}. What is the K_{sp} of Barium Sulfate?

13. A student prepared a saturated solution of CaCrO_{4} and found that when 100 mL of this solution was evaporated, 0.416 g of CaCrO_{4} was left behind. What is the value of K_{sp} for this type?

14. Copper (I) Chloride has K_{sp} = 1.9 x 10^{-7}. Calculate the molar solubility of CuCl in

a) pure water

b) 0.020 M HCl

c) 0.020 M MgCl_{2}0.020 M Au(NO_{3})_{3}

15. At 25^{o}C the value of K_{sp} for LiF is 1.7 x 10^{-3} and for BaF_{2}, 1.7 x 10^{-6}. In terms of moles per liter, which salt is the more soluble in water? Calculate the solubility of each in these units.

16. Chalk is CaCO_{3} and at 25^{o}C it’s K_{sp} = 4.5 x 10^{-9}. How many grams of CaCO_{3} dissolve in 100 mL of water?

17. What is the molar solubility of Mg(OH)_{2} in 0.10 M NaOH. K_{sp} of MgCl_{2}is 7.1 x 10^{-12}.

18. Does a precipitate of PbCl_{2} form when 0.0100 mol of Pb(NO_{3})_{2} and 0.0100 mol of NaCl are dissolved in 1.00L of solution.

19. Silver acetate, AgCH_{3}COO, had K_{sp} = 4 x 10^{-3}. Does a precipitate form when 0.010 mol of AgNO_{3}and 0.30 mol of Ca(CH_{3}COO)_{2} are dissolved in a total volume of 1.00 L of solution.

20. Does a precipitate of PbBr_{2} form if 50.0 mL of 0.010 M Pb(NO_{3})_{2} is mixed with

a) 50.00 mL of 0.010 M KBr?

b) 50.0 mL of 0.10 M NaBr?

21. Would a precipitate of silver acetate form if 18.0 mL of 0.10 M AgNO_{3} were added to 40.0 mL of 0.024 M NaC_{2}H_{3}O_{2}, K_{sp} = 4 x 10^{-3}.

22. Suppose that 50.0 mL of 0.10 M AgNO_{3} were added to 50.0 mL of 0.050 M NaCl solution

a) What weight of AgCl would be formed?

b) What would be the final concentrations of all the ions contributed by these salts?

23. Drinking water often contains dissolved chloride and/or calcium ions. Devise a procedure to test which of these ions is present in a sample of tap water.

24. A solution contains a least 0.8 mol/L of each of the following ions: Ag^{+}, Ba^{2+}, Mg^{2+}. Use a flow chart to illustrate a procedure that could be used to separate these ions from each other.

25. A solution contains a least 0.8 mol/L of each of the following ions: Pb^{2+}, Ag^{+}, Mg^{2+}, Fe^{2+}. Use a flow chart to illustrate a procedure that could be used to separate these ions from each other.

26. For the following equations, write the final balanced equation with the proper precipitates and ions.

a) MgS + Zn(NO_{3})_{2} =>

b) Pb(ClO_{3})_{2} + KI =>

c) Ca(OH)_{2} + FeSO_{4} =>

d) Li_{2}CO_{3} + AgCH_{3}COO =>

**Solubility Practice Questions: Solution**

**1.**

PbF_{2} <=>Pb^{2+} + 2F^{–} C = n/v

2.15×10^{-3} 2.15×10^{-3} 4.3×10^{-3} C = 2.15×10-3 mol

1L

Ksp = [Pb^{2+}][F^{–}]^{2} C = 2.15×10-3 mol/L

Ksp = (2.15×10^{-3})(4.3×10^{-3})^{2 }

Ksp = 3.98×10^{-8 }

**2.**

CoCo_{3} <=>Co^{2+} + CO_{3}^{2-} Na_{2}CO_{3} =>2Na^{+} + CO_{3}^{–}

1×10^{-9} 1×10^{-9} 1×10^{-9 }0.1 0.2 0.1

Ksp = [Co^{2+}][Co_{3}^{2-}]

Ksp = (1×10^{-9})(1×10^{-9} + 0.1)

Ksp = 1×10^{-10 }

**3.**

PbF_{2 }<=>Pb^{2+} + 2F ^{–} Pb(NO_{3}) =>Pb^{2+} + 2NO_{3 }^{– }_{}

3.1×10^{-4 }3.1×10^{-4 }6.2×10^{-4 }0.1 0.1 0.2

Ksp = [Pb^{2+}][F^{–}]^{2 }

Ksp = (3.1×10^{-4 }+ 0.10)(6.2×10^{-4})^{2 }

Ksp = 3.86×10^{-8 }

**4.**

Ksp = 5.0×10^{-13 }AgBr Ag^{+} + Br ^{– }

x x x

Ksp = [Ag^{+}][Br ^{–}]

5.0×10^{-13} = x^{2}

x = 7.07×10^{-7 }

∴ Molar solubility = 7.07×10^{-7 }

**5.**

Ag_{2}CO_{3 }<=>2Ag^{+} + CO3 ^{– }Ksp = 8.1×10^{-12 }

2x x

Ksp = [Ag^{+}]^{2 }[CO_{3 }^{–}]

8.1×10^{-12 }= 4x^{3 }

x^{3} = 2.015×10^{-12 }

x = 1.3×10^{-4}

∴ Molar solubility is 1.3×10^{-4 }

**6.**

AgI <=>Ag^{+} + I ^{–} NaI=> Na^{+} + I ^{– }

x x x 0.2 0.2

Ksp = [Ag^{+}][ I ^{– }]

8.3×10^{-17 }= (x)(x + 0.2) Negligible because Ksp is very small.

x = 4.15×10^{-16 }

∴ Molar solubility is 4.15×10^{-16 }

**7.**

Fe(OH)_{3 }<=>Fe^{3+ }+ 3OH ^{– } OH ^{– }= 0.05

x x 3x

Ksp = [Fe^{3+}][ OH ^{– }]^{3 }

1.6×10^{-39 }= (x)(3x + 0.05)^{3 }Negligible because Ksp is so small

Since the value of Ksp is very small the value of x will be negligible hence can be ignored

x = 1.28×10^{-35 }

∴ Molar solubility is 1.28×10^{-35 }

**8.**

Trial Ksp = [Ca][SO_{4 }]

Trial Ksp = (0.0025)(0.03)

Trial Ksp = 7.5×10^{-5 }Ksp = 2.4×10^{-5}

∵ Trial Qsp > Ksp

∴ Precipitate will form

**9.**

Ksp = [Ag^{+}]^{2 }[CrO_{4 }^{-2 }] Ag_{2}CrO_{4 }=>2Ag^{+}** + **CrO_{4}^{2- }

Trial Ksp = (4.8×10^{-5 })^{2 }(3.4×10^{-4 })

Trial Ksp = 7.83×10^{-13 }

∵ Trial Qsp < Ksp

∴ No precipitate will form

**10.**

Pb(NO_{3})_{2} Pb^{+2 }+ 2NO_{3}^{ – }MgSO_{4 }=>Mg^{2+ }+ So_{4 }^{2 – }

C_{1}V_{1 }= C_{2}V_{2}^{ }C_{1}V_{1 }= C_{2}V_{2}^{ }

(1.0×10^{-3 })(100) = C_{2 }(200) (2.0×10^{-3 })(100) = C_{2 }(200)

C_{2 }= 5.0×10^{-4 }C_{2 }= 1×10^{-3 }

PbSO_{4 }=> Pb^{+} + SO_{4}^{2- }

Trial Qsp = [Pb^{2+}][SO_{4}^{2 – }] Ksp = 6.3×10^{-7}

Trial Qsp = (5.0×10^{-4})(1×10^{-3})

Trial Qsp = 5×10^{-7 }

∵ Trial Qsp < Ksp

∴ No precipitate will form

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