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23 Nov 2019

The equilibrium constant for the reaction of dissolved molecularchlorine with water to give hydrogen ions, chlorine ions, and HOClis 4.5 x 10^-4, where as usual the concentration of water isincluded in the K value. If the pH of the solution is determined byother processes so that the amount of hydrogen ion contributed bythe chlorine reaction is negligible, calculate the fraction of theoriginal 50 ppm chlorine which remains as Cl₂ at pH of2.

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This is for the lab practical I take on Tuesday night. If you could answer each one and show your work/ explain that'd be amazing so I can do well on the actual practical on Tuesday! Thank you!

Chemistry 211 Lab Final Study Guide Spring 2015

Useful Information:
1 ampere =coulomb/sec
Faradayâs constant = 96,500 coulombs/mole of electrons
Molar mass of Au = 197.0 g/mole, Molar mass of Cu = 63.55 g/mole
m = moles , ' â

c kg

ÎTf

= Tf

â Tf

where Tf is for the pure solvent and Tf

is for the mixture.

ÎT = (K

)(m ) ,

m = (

(mass of

solute, g )
)( ).

f f c

c molar

mass, g

mol â1

mass of

solvent, kg

7. Next question

(a) If you weigh 6.355 grams of copper, add concentrated nitric acid to dissolve it, and then dilute the solution to a total volume of 50 mL, how many moles of copper are in this solution?

(b) For the equilibrium reaction 3H2(g) + N2(g) â 2NH3(g), what is the equilibrium constant expression, Kc, for this reaction? (5pts)

8. Suppose at some temperature, the equilibrium concentration of H2 is 0.45M, the equilibrium concentration of N2 is 0.38M, and the equilibrium concentration of NH3 is 0.42M. What is the value of the equilibrium constant for the reaction 3H2(g) + N2(g) â 2NH3(g) at this temperature? (Show your work. No credit will be given if your work is not shown.)

9. Next question
(a) If you need 0.0534 moles of NaOH to reach equivalent point in the titration with acetic acid, how many mL of 0.09876M NaOH will you need?

(b) The Ksp of Cu(OH)2 is 2.2x10-20, what is the solubility of Cu(OH)2?

10. Next question
(a) In non-sulfide qualitative analysis of representative cations, the Group A cations comprising of Pb2+,

Hg2+, and Ag+ give a precipitate with solution.
(b) From a solution containing mixture of Pb2+, Fe3+, and Al3+, Pb2+ is removed by adding HCl solution. To the remaining cation mixture, on making the solution basic with NH3 (pH=9 â 10), both the cations form precipitate. Further analysis of the cation mixture containing Fe3+ and Al3+ is carried out by adding NaOH solution.

Aluminum hydroxide (dissolves/remains undissolved) in NaOH solution while Ferric hydroxide ( dissolves/remains undissolved).
(c) Fe(OH)3 precipitate is dissolved in minimal amount of aqueous HCl which gives a

yellow solution. To this solution addition of solution (at pH = 4 â 5) yields a dark blue solid known as Prussian Blue.

(d) Describe the method for checking for completeness of precipitate in qualitative analysis of ions.

Silver bromide, AgBr (s), is an essential reagent in black and white film developing. It is,

however, only sparingly soluble in water. AgBr (s) has K = 5.0 Ã 10-13, making it difficult to

rinse AgBr from the film negative with water.

Instead, excess AgBr is removed by an aqueous solution of sodium thiosulfate (Na2S2O3), which forms the complex ion Ag(S2O3)23-:

Ag+ (aq) + 2 S2O32- (aq) Ag(S2O3)23- (aq) Kf = 4.7 Ã 10+13 a) To see how this helps, first determine the molar solubility of AgBr (s) in water.

b) The large formation constant of Ag(S2O3)23- (aq) means that almost all of the silver is complexed with the thiosulfate. To determine how much, calculate the [Ag+] at equilibrium in a 1.0 L solution that initially contains 0.001 M silver ions and 0.200 M sodium thiosulfate. Do so by setting up an ICE table and assuming that x is small compared to 0.200 (but not 0.001). Show that this results in a value of x = [Ag(S2O3)23-] = 0.001. In other words, all of the Ag+ complexes.

c) Your result from part b suggests that no Ag+ remains in solution. This obviously canât be correct, since it would result in an infinitely large reaction quotient that wouldnât be equal to an admittedly large equilibrium constant. To determine the correct [Ag+], use the equilibrium concentrations determined in part b as your initial concentrations in a new ICE table. Then use Kf to determine [Ag+]. [This approach is called a stoichiometric shift and is useful when a reaction starts with only reactants but goes almost to completion. Is essence, we are approaching the equilibrium from the direction of all Ag(S2O3)23- and no Ag+ as opposed to the direction of all Ag+ and no Ag(S2O3)23-.]

d) Before we determine the solubility of AgBr (s) in a thiosulfate solution, we need to know the appropriate equilibrium constant. To find it, determine the value of the equilibrium constant for the reaction:

AgBr (s) + 2 S2O32- (aq) <-->Ag(S2O3)23- (aq) + Br - (aq) Kc = ??

e) Finally, calculate the molar solubility of AgBr (s) in 1.0 M sodium thiosulfate. In other words, what is [Ag(S2O3)23-] for the reaction given in part d when the initial concentration of thiosulfate is 1.0 M?

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