301 Name. Date Lab Section, Postlaboratory Problems-Experiment 24- Solubility Pr

Question

301 Name. Date Lab Section, Postlaboratory Problems-Experiment 24- Solubility Product 1. What are some of the errors involved in the total volume measurements in Part A? 2. In our laboratories at Modesto Junior College, we generally obtain sizeable percent differences (about 100%) between the He values obtained using the method of Part A and the titration method of Part B. We have not been able to determine the source of this discrepancy although we have considered complications from hydrolysis and/or dissociation of the hydrogen tartrate ion. If you also obtained significant differences, suggest possible explanations for them. Concerning the applicability of the titration method, comment on if and how you could use it to determine K, for each of the following 3. a. barium hydroxide b. calcium sulfate c. copper(Il) iodate

Explanation / Answer

1. Errors possible here are the flask is not calibrated before and therefore we would get incorrect volume from volume mesurement experiment.

2. Error arising in the experiment could be due to adding excess water to the solution, spilling of the solution accidentally.

3. Applicability of titration methods to determine Ksp values

a. barium hydroxide (Ba(OH)2

Ba(OH)2(s) <==> Ba2+(aq) + 2OH-(aq)

Ksp = [Ba2+][OH-]^2

1 mole Ba2+ and 2 mole OH- in solution

Titration of OH- with HCl gives OH- concentration and thus Ba2+ can be calculated and followed by Ksp.

b. calcium sulfate (CaSO4)

CaSO4(s) <==> Ca2+(aq) + SO4^2-(aq)

Ksp = [Ca2+][SO4^2-]

1 mole Ca2+ and 1 mole SO4^2- in solution

titrate Ca2+ with EDTA. 1 mole Ca2+ consumes 1 moles EDTA

So, we can easily calculate concentration of Ca2+ from EDTA consumed.

c. copper(II) iodate (Cu(IO3)2)

Cu(IO3)2(s) <==> Cu2+(aq) + 2IO3-(aq)

1 Cu2+ and 2 IO3- in solution.

Ksp = [Cu2+][IO3-]^2

IO3- concentration is determined by reaction with KI (excess) first to form I2. Released I2 is titrated with Na2S2O3.

1 mole of IO3- gives 3 moles of I2.

1 mole of I2 reacts with 2 moles of Na2S2O3

So concentration of IO3- in solution can be determined and thus Cu2+

d. lead(II) chloride (PbCl2)

PbCl2(s) <==> Pb2+(aq) + 2Cl-(aq)

Ksp = [Pb2+][Cl-]^2

tiration with NaOH of Cl- in solution.

Solution has, 1 Pb2+ and 2 Cl- in it, so knowing Cl-, Pb2+ and Ksp can be calculated

e. lithium carbonate (Li2CO3)

Li2CO3(s) <==> 2Li+(aq) + CO3^2-(aq)

2HCl + CO3^2- --> H2CO3 + 2Cl-

Titration with HCl of CO3^2- in solution. 1 mole CO3^2- consumes 2 moles HCl

[Li+] = 2[CO3^2-]

So Ksp can be calculated

f. silver acetate (AgCH3COO)

AgCH3COO(s) <==> Ag+(aq) + CH3COO-(aq)

Ksp = [Ag+][CH3COO-]

titration with KSCN, to react with Ag+ in solution,forms AgSCN

[Ag+] = [SCN-]

So Ksp can be calculated

g. potassium nitrate is completely soluble in water. No Ksp for it.

h. silver chloride (AgCl)

AgCl(s) <==> Ag+(aq) + Cl-(aq)

Ksp = [Ag+][Cl-]

Titration with NaOH, neutralize Cl- is solution.

[Ag+] = [Cl-]

therefore,

Ksp can be calculated

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