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Category > Chemistry Posted 25 Aug 2017 My Price 8.00

The solubility Ksp

Table 1. The solubility Ksp determined using the acid-base titration.

Volume (mL)

Trials

Observations:

  • There were some water droplets left in the Erlenmeyer flask after washing and drying.

1

2

3

4

Volume Ca(OH)2solution

10

10

10

NA

Initial burette reading, Vi (mL)

0.00

0.10

0.00

NA

Final burette reading, Vf (mL)

29.1

29.1

28.7

NA

Volume of HCl solution delivered

29.1

29.0

28.7

NA

HCl Concentration: 0.0176 M

Temperature: 23C

Table 2. The solubility Ksp determined using the EDTA titration.

Volume (mL)

Trials

Observations:

  • There were some water droplets left in the Erlenmeyer flask after washing and drying
  • The solution turned purple at 24.5mL (1st trial),
  • The solution turned purple at 21.75mL (2nd trial).
  • The solution turned purple at 26.1 mL (3rd trial).
  • The solution turned purple at 26.5mL (4th trial).

1

2

3

4

Volume Ca(OH)2solution

10

10

10

10

Initial burette reading, Vi (mL)

0.00

0.00

0.00

0.00

Final burette reading, Vf (mL)

27.75

24.45

29.60

30.10

Volume of HCl solution delivered

27.75

24.45

29.60

30.10

EDTA Concentration: 0.007 M

Temperature: 23C

Determination of two values for the solubility (in mol/L) of calcium hydroxide:

Acid base titration:

Ca(OH)2 (s) ↔ Ca2+ (aq) + 2OH- (aq)

Trial 1:

(0.010L) (C1)  = (0.0176) (0.0291 L)

C1                   = (0.0512 mol/L) / 2 (stoichiometric ratio)

                        = 0.0256 mol/L

Trial 2:

(0.010L) (C1)  = (0.0176) (0.029 L)

C1                   = (0.05104 mol/L) / 2 (stoichiometric ratio)

                        = 0.0255 mol/L

Trial 3:

(0.010L) (C1)  = (0.0176) (0.0287 L)

C1                   = (0.0505 mol/L) / 2 (stoichiometric ratio)

                        = 0.0253 mol/L

Average C1     = (Trial 1 + Trial 2 + Trial 3) / 3

= [(0.0256 + 0.0255 + 0.0253) mol/L]/3

= 0.02546 mol/L

EDTA titration:

Trial 1:

(0.010L) (C1)  = (0.007) (0.02775 L)

C1                   = (0.0194 mol/L)

Trial 2:

(0.010L) (C1)  = (0.007) (0.02445 mL)

C1                   = (0.0171 mol/L)

Trial 3:

(0.010L) (C1)  = (0.007) (0.02960 mL)

C1                   = (0.0207 mol/L)

Trial 4:

(10.00mL) (C1)           = (0.007) (30.10 mL)

C1                               = (0.0211 mol/L)

Average C1     = (Trial 1 + Trial 2 + Trial 3 +Trial 4) / 4

= [(0.0194 + 0.0171 + 0.0207 + 0.0211) mol/L] / 4

= 0.0196 mol/L

Determination of two values of Ksp for calcium hydroxide:

Acid base titration:

Ksp =[Ca2+] [OH-]2

Ksp = [0.02546] [0.05092]2

Ksp = 6.60 x 10-5

EDTA titration:

Ksp =[Ca2+] [OH-]2

Ksp =[0.0196][0.0392]2

Ksp = 3.01 x 10-5

Are my answers right based on the information provided? also lab manual says we should perform at least 3 trials with 2 trials that falls within 3%. does that mean that for calculation, do i take only the two trials that fall within 3% or take all trials?

 

Answers

AccountingQueen
(3)
Status NEW Posted 25 Aug 2017 02:08 PM My Price 8.00

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