Tutorial 10-Solutions
5 Pages
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Tutorial 10-Solutions

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5 Pages
English

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Chemistry 12 Tutorial 10—Solutions Chemistry 12 Tutorial 10—Ksp Calculations Solutions 1. Calculate the solubility of SrF in moles/Litre in water. 2 -s +s +2s 2+ -Equilibrium Equation: SrF Sr + 2F (where s = molar solubility) 2(s) (aq) (aq) 2+ - 2Ksp = [Sr ] [F ] 2Ksp = s (2s) 2Ksp = s x 4s 3Ksp = 4s 3s = Ksp 4 -9 -3s = K s p = 4 . 3 x 1 0 = 1.0244 x 10 M 3 3 4 4 -3Because the Ksp was 2 SD’s, the answer will be: Molar Solubility = 1.0 x 10 M ********************************************************** 2. Calculate the solubility of Ag CO in grams/Litre. 2 3 -s +2s +s + 2-Equilibrium Equation: Ag CO 2Ag + CO (where s = molar solubility) 2 3(s) (aq) 3 (aq) + 2 2-Ksp = [Ag ] [CO ] 3 2Ksp = (2s) x s 2Ksp = 4s x s 3Ksp = 4s 3s = Ksp 4 –12 -4Ksp s = = 8 . 5 x 1 0 = 1.286 x 10 M 3 3 4 4 Chemistry 12—Tutorial 10 Solutions Page 1 of 5 pages Chemistry 12 Tutorial 10—Solutions -4Molar Solubility = 1.286 x 10 M To get solubility in g/L: -4 -21.286 x 10 mol x 275.8 g Ag CO = 3.5 x 10 g/L (or 0.035 g/L) 2 3 L ...

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Chemistry 12 Tutorial 10—Solutions
Chemistry 12
Tutorial 10—Ksp Calculations
Solutions


1. Calculate the solubility of SrF in moles/Litre in water. 2

-s +s +2s
2+ -Equilibrium Equation: SrF Sr + 2F (where s = molar solubility) 2(s) (aq) (aq)

2+ - 2Ksp = [Sr ] [F ]

2Ksp = s (2s)

2Ksp = s x 4s

3Ksp = 4s

3s = Ksp
4

-9 -3s = K s p = 4 . 3 x 1 0 = 1.0244 x 10 M 3 3
4 4

-3Because the Ksp was 2 SD’s, the answer will be: Molar Solubility = 1.0 x 10 M

**********************************************************

2. Calculate the solubility of Ag CO in grams/Litre. 2 3
-s +2s +s
+ 2-Equilibrium Equation: Ag CO 2Ag + CO (where s = molar solubility) 2 3(s) (aq) 3 (aq)

+ 2 2-Ksp = [Ag ] [CO ] 3

2Ksp = (2s) x s

2Ksp = 4s x s

3Ksp = 4s

3s = Ksp
4

–12 -4Ksp s = = 8 . 5 x 1 0 = 1.286 x 10 M 3 3
4 4
Chemistry 12—Tutorial 10 Solutions Page 1 of 5 pages Chemistry 12 Tutorial 10—Solutions

-4Molar Solubility = 1.286 x 10 M

To get solubility in g/L:

-4 -21.286 x 10 mol x 275.8 g Ag CO = 3.5 x 10 g/L (or 0.035 g/L) 2 3
L 1 mol Ag CO 2 3

**************************************************
-5 3. At a certain temperature, the solubility of SrCO is 7.5 x 10 M. Calculate the Ksp 3
for SrCO . 3
Molar Solubility

-5 -5 -5 -7.5 x 10 M +7.5 x 10 M +7.5 x 10 M
2+ 2- SrCO Sr + CO 3(s) (aq) 3 (aq)

2+ 2- Ksp = [Sr ] [CO ] 3

-5 2 = (7.5 x 10 )

-9 Ksp = 5.6 x 10

*************************************************
Answers to Self Test on Tutorial 10

1. The Ksp is just a K for an ionic substance dissolving in water. eq

2. Give the Net-Ionic Equation and the Ksp expression for each of the following
dissolving in water. (The first one is done as an example.)

Substance Net-Ionic Equation Ksp Expression

+ 2- + 2 2-Ag SO Ag SO 2Ag + SO Ksp = [Ag ] [SO ] 2 4(s) 2 4(s) (aq) 4 (aq) 4


2+ 2- 2+ 2-CaCO CaCO Ca + CO Ksp = [Ca ] [CO ] 3(s) 3(s) (aq) 3 (aq) 3


2+ 3- 2+ 3 3- 2 Ca (PO ) Ca (PO ) 3Ca + 2PO Ksp = [Ca ] [PO ]3 4 2(s) 3 4 2(s) (aq) 4 (aq) 4


+ - + -AgClO AgClO Ag + ClO Ksp = [Ag ] [ClO ] (aq) 3 (aq) 33(s) 3(s)


Chemistry 12—Tutorial 10 Solutions Page 2 of 5 pages Chemistry 12 Tutorial 10—Solutions
3. a) Calculate the molar solubility (solubility in moles/Litre) of Fe(OH) in water. 2

-s +s +2s (where “s” is the molar solubility)
2+ - Fe(OH) Fe + 2OH 2(s) (aq) (aq)

2+ - 2 Ksp = [Fe ] [OH ]

2 Ksp = s (2s)

3 Ksp = 4s
-17 4.9 x 10Ksp -6 3 3 s = = = 2 . 3 x 1 0 M 4 4


- b) What is the [OH ] in a saturated solution of Fe(OH) ? 2

-s +s +2s (where “s” is the molar solubility)
2+ - Fe(OH) Fe + 2OH 2(s) (aq) (aq)

- - -6 -6 Since [OH ] = 2s ; then [OH ] = 2(2.3052 x 10 ) = 4.6 x 10 M

4. Calculate the solubility of BaCO in grams per Litre. 3

-s +s +s where “s” is the Molar Solubility
2+ 2- BaCO Ba + CO 3(s) (aq) 3 (aq)

2+ 2- Ksp = [Ba ] [CO ] 3

2 Ksp = s

-9 -5Ksp s = = 2.6 x 10 = 5.099 x 10 M

Solubility in g/L:

-5 -2 5.099 x 10 mol/L x 197.3 g/mol = 1.0 x 10 g/L


5. The solubility of PbI at a certain temperature is 0.70 grams per Litre. 2

a) Calculate the solubility in moles/Litre

-3 -3 0.70 g x 1 mol = 1.5 x 10 mol/L (unrounded = 1.518 x 10 M)
L 461.0 g


Chemistry 12—Tutorial 10 Solutions Page 3 of 5 pages Chemistry 12 Tutorial 10—Solutions


b) Calculate the value of Ksp for PbI at this temperature 2

-3 -3 -3 -1.518 x 10 M + 1.518 x 10 M + 3.036 x 10 M
2+ - PbI Pb + 2 I 2(s) (aq) (aq)

2+ - 2 Ksp = [Pb ] x [I ]

-3 -3 2 -8 Ksp = 1.518 x 10 ( 3.036 x 10 ) = 1.4 x 10

6 It is found that 0.043 grams of MgCO is all that can dissolve in 100.0 mL of solution 3
at a certain temperature. From this information, calculate the Ksp for MgCO at this 3
temperature.

Plan: g mol M (molar solubility) Ksp

-4 0.043 g MgCO x 1 mol MgCO = 5.1008 x 10 mol MgCO 3 3 3
84.3 g MgCO3


-4 -3 [MgCO ] = 5.1008 x 10 mol = 5.1008 x 10 M (This is the molar solubility) 3
0.1000 L

-3 -3 -3 -5.1008 x 10 M +5.1008 x 10 M +5.1008 x 10 M
2+ 2- MgCO Mg + CO 3(s) (aq) 3 (aq)

2+ 2- Ksp = [Mg ] [CO ] 3

-3 2 = (5.1008 x 10 )

-5 Ksp = 2.6 x 10


Chemistry 12—Tutorial 10 Solutions Page 4 of 5 pages Chemistry 12 Tutorial 10—Solutions

2+ - 7. Two separate experiments were done with combinations of Cu and IO ions. Use 3
the information given to fill in the missing value.

2+ - The Net-Ionic Equation for equilibrium is: Cu(IO ) Cu + 2IO 3 2(s) (aq) 3 (aq)

2+ -Experiment # [Cu ] [IO ] 3

1 0.00327 M 0.00654 M


2 0.00240 M ?


Use the results from Experiment #1 to calculate the Ksp for Cu(IO ) : 3 2

2+ - 2Ksp = [Cu ] [ IO ] 3

2 = 0.00327 x (0.00654)

-7 = 1.399 x 10

2+ -Use the Ksp and [Cu ] to calculate [IO ] in Experiment #2: 3

2+ - 2Ksp = [Cu ] [ IO ] 3

- 2[IO ] = Ksp 3
2+ [Cu ]
-7 - Ksp 1.399 x 10[IO ] = = 3 2+[Cu ] 0.00240


- -3[IO ] = 7.63 x 10 M 3




Chemistry 12—Tutorial 10 Solutions Page 5 of 5 pages