Ans to (Oct 31) Qns

Posted by Kyo on October 30, 2007 at 17:42:54:

In Reply to: (Oct 31) Qns posted by Kyo on October 30, 2007 at 17:40:35:

Ans to Q11
46%
(hints : the tricky part abt this qn is the "50cm3 of resulting soln". The first steps are of course to write out the 2 balanced redox reactions. Next, using the molarity of H2O2 found from titration with KMnO4, you can find the no. of mol of H2O2 present in 75cm3 of this same "resulting solution". Then, subtract this calculated no. of mol of H2O2 from total no. of mol of H2O2 present initially in the 75cm3 soln, to obtain no. of mol of H2O2 that reacted with the NO2. From here, simply apply stoichiometry to obtain no. of mol, volume at s.t.p, and hence % by volume, of NO2 gas.)


Ans to Q12
660 kPa2 (3 sig fig)
(hint : look carefully at the state symbols)


Ans to Q13
a)


b) Rate determining step is the slow step NO2F --> NO2 + F (ie. formation of the F radical), since the 2nd step involves the highly reactive F radical and hence is fast.
rate equation = k [NO2F]
Because the 1st elementary equation is clearly a unimolecular reaction (ie. involving only 1 molecule), the reaction is first order in the reactant NO2F.

c)


d) 120 min
(hint : 1st order reactions have constant half life)


Ans to Q14
8.356 dm3
(hint : electrolysis of water only; the Cr2O3(aq) is but a red herring)


Ans to Q15
Draw a labelled diagram of an electrolytic cell (not a voltaic/galvanic cell) powered by a battery. The anode and cathode (of known mass) are both made of copper metal, the electrolyte is CuSO4 (aq). A current I, (read from the ammeter) is passed through for t seconds. The cathode is then dried and reweighed.

No. of mol of Cu deposited by (I x t) coulombs of electricity = (gain in mass of Cu cathode / molar mass of Cu)
==> Coulombs of electricity required to deposit 1 mol of Cu = (I x t) / (gain in mass/63.5) coulombs.
Since 2 mol of e- are required to deposit 1 mol of Cu,
==> charge on 2 mol of e- = (I x t) / (gain in mass/63.5) coulombs
==> 2 x Avogadro's constant x (1.6 x 10-19) coulombs = (I x t) / (gain in mass/63.5) coulombs
==> Avogadro's constant = (1.98 x 10^20 x (I x t)) / (gain in mass)


Ans to Q16
a) From the graph's titration end point, 50cm3 of MgCl2(aq) reacts completely with 20cm3 of NaOH(aq). By stoichiometry, mole ratio of MgCl2 to NaOH is 1:2.
==> [MgCl2] = (0.5 x 1.0 x 20/1000) / (50/1000) = 0.2 mol/dm3.
From the graph, after adding 10cm3 of NaOH(aq), pH = 9 ==> pOH = 5 ==> [OH-] = 1 x 10^-5 mol/dm3.
At 10cm3 of NaOH(aq), molarity of Mg2+ should be at half the original concentration (since it *corresponds* to the point of half-neutralization for an acid-base titration).
==> [Mg2+] = 0.5 x [MgCl2] = 0.1 mol/dm3.
Ksp = [Mg2+][OH-]2 = 0.1 x 10^-10 = 1 x 10^-11 mol3/dm9

b) Vol of excess NaOH(aq) = 30-20 = 10cm3
No. of mol of OH- ions = 1.0 x (10/1000) = 0.01
[OH-] = 0.01 / (30cm3 of NaOH + 50cm3 of MgCl2) = 0.125 mol/dm3
==> pOH = -log10(0.125) = 0.903
==> pH = 13.1

c) Initially, pH rises steeply as NaOH(aq) is added because ionic product of Mg(OH)2 < Ksp of Mg(OH)2, and hence molarity of free OH- ions increases, causing pH to rise.
At pH 9, ionic product > Ksp and any additional OH- added is precipitated out as Mg(OH)2 solid.
At 20cm3 of NaOH(aq), there is no more Mg2+ ions available to precipitate out the OH- ions, and hence as more NaOH(aq) is added, molarity of free OH- ions increases, causing pH to rise.
The pH finally levels off at approximately 13.1 because this approaches the upper pH limit of the alkaline NaOH(aq) solution. Bear in mind that continuing to add more of the NaOH(aq) solution not only adds OH- ions, but also adds more water solvent at the same time.

Follow Ups:

• (01 Nov) Qns - Kyo 10/31/07 (12)
  • Ans to (01 Nov) Qns - Kyo 11/01/07 (11)
    • (02 Nov) Qns - Kyo 11/01/07 (10)
      • Ans to (02 Nov) Qns - Kyo 11/02/07 (9)
        • (03 Nov) Qns - Kyo 11/02/07 (8)
          • Ans to (03 Nov) Qns - Kyo 11/03/07 (7)
            • (04 Nov) Qns - Kyo 11/03/07 (6)
              • Ans to (04 Nov) Qns - Kyo 11/03/07 (5)
                • Final Day (05 Nov) Qns - Kyo 11/03/07 (4)
                  • Spot the Chemistry Errors in this YouTube video! - Kyo 05/22/08 (3)
                    • Balance the equation : NO2 + H2O ---> N2O + HNO3 - Kyo 06/25/08 (2)
                      • NH3 vs PH3 ; H2O vs H2S - Kyo 06/26/08 (1)
                        • Br2 (aq) vs Br2 (CCl4) ; and Methylbenzene vs Phenol - Kyo 07/06/08 (0)

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