1. Xtra Edge February 2012 - Career Point

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2. 61V Hence E1º = = 0.52 V 5 Now the reaction corresponding to the potential E2º is BrO3 – + 2H2O + 6e – = Br – + 6OH – ...(4) This reaction can be obtained by adding the following three reactions. BrO3 – + 2H2O + 4e – = BrO – + 4OH – (Eq.2) BrO – + H2O + e – 1 = Br2 + 2OH 2 – (Eq.3) 1 Br2 + e 2 – = Br – Hence ...(5) º reaction( 4) º reaction( 2) ∆ G = ∆ G + ∆ G º reaction( 3) + ∆ G º reaction( 5) or – 6F(E2º) = – 4F(0.54 V) – 1F(0.45 V) – 1F (1.07 V) = (– 3.68 V) F 3. 68 or E2º = = 0.61 V. 6 3. What is the solubility of AgCl in 0.20 M NH3 ? Given : Ksp(AgCl) = 1.7 × 10 –10 M 2 K1 = [Ag(NH3) + ] / [Ag + ] [NH3] = 2.33 × 10 3 M –1 and K2 = [Ag(NH3)2 + ]/[Ag(NH3) + ][NH3] = 7.14 × 10 3 M –1 Sol. If x be the concentration of AgCl in the solution, then [Cl – ] = x From the Ksp for AgCl, we derive K sp −10 [Ag + ] = − [ Cl ] = 1. 7× 10 M x If we assume that the majority of the dissolved Ag + goes into solution as Ag(NH3)2 + then [Ag(NH3)2 + ] = x Since two molecules of NH3 are required for every Ag(NH3)2 + ion formed, we have [NH3] = 0.20 M – 2x Therefore, ⎛ ⎜ . 7× 10 + 2 [ Ag ][ NH ] ⎜ 3 = ⎝ x + [ Ag( NH3 ) 2 ] − = 6.0 × 10 –8 M 2 From which we derive Kinst = 2 2 2 M ⎞ ⎟ ( 0. 20M − 2x) ⎠ x 10 1 2 ( 0. 20M − 2x) 6. 0× 10 M = = 3.5 × 10 2 −10 2 x 1. 7× 10 M 2 which gives x = [Ag(NH3)2 + ] = 9.6 × 10 –3 M, which is the solubility of AgCl in 0.20 M NH3 −8 2 4. Potassium alum is KA1(SO4)2.12H2O. As a strong electrolyte, it is considered to be 100% dissociated into K + , Al 3+ , and SO4 2– . The solution is acidic because of the hydrolysis of Al 3+ , but not so acidic as might be expected, because the SO4 2– can sponge up some of the H3O + by forming HSO4 – . Given a solution made by dissolving 11.4 g of KA1(SO4)2.12H2O in enough water to make 0.10 dm 3 of solution, calculate its [H3O + ] : (a) Considering the hydrolysis Al 3+ + 2H2O Al(OH) 2+ + H3O + with Kh = 1.4 × 10 –5 M (b) Allowing also for the equilibrium HSO4 – + H2O H3O + + SO4 2– with K2 = 1.26 × 10 –2 M 11. 4 g Sol. (a) Amount of alum = = 0.024 mol − 1 474. 38 g mol 0. 024 mol Molarity of the prepared solution = 3 0. 1dm = 0.24 M Hydrolysis of Al 3+ is Al 3+ + 2H2O Al(OH) 2+ + H3O + 2+ + [ Al( OH) ][ H3O ] Kh = 3+ [ Al ] If x is the concentration of Al 3+ that has hydrolyzed, we have ( x)( x) Kh = 0. 24M − x = 1.4 × 10–5 M Solving for x, we get [H3O + ] = x = 1.82 × 10 –3 M (b) We will have to consider the following equilibria. Al 3+ + 2H2O Al(OH) 2+ + H3O + H3O + + SO4 2– HSO4 – + H2O Let z be the concentration of SO4 2– that combines with H3O + and y be the net concentration of H3O + that is present in the solution. Since the concentration z of SO4 2– combines with the concentration z of H3O + , it is obvious that the net concentration of H3O + produced in the hydrolysis reaction of Al 3+ is (y + z). Thus, the concentration (y + z) of Al 3+ out of 0.24 M hydrolyzes in the solution. With these, the concentrations of various species in the solution are 3+ Al 0. 24 M− y−z 3 y O H + + Thus, Kh = + 2H2O 2− SO4 0. 48 M−z 2+ + Al( OH) + H XtraEdge for IIT-JEE 34 FEBRUARY 2012 y+ z z 4 HSO − + H2O 3 y O ( y + z)( y) = 1.4 × 10 ( 0. 24M − y − z) –5 M ...(i)
z 1 K2 = = y( 0. 48M − z) 1. 26× 10 From Eq. (ii), we get ( 0. 48M) y z = 2 ( 1. 26× 10 M) + y − Substituting this in Eq. (i), we get −2 M ...(ii) ⎛ ⎞ ⎜ ( 0. 48M) y ⎟ ⎜ y + ⎟ y −2 ⎝ ( 1. 26× 10 M) + y ⎠ = 1.4 × 10 ⎛ ⎞ ⎜ ( 0. 48M) y ⎟ ⎜ 0. 24 − y − −2 ⎟ ⎝ ( 1. 26× 10 M) + y ⎠ –5 Making an assumption that y
Page 3 and 4: Volume - 7 Issue - 8 February, 2012
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(0, 4) ( 2 3, 2) (4, 0) ⎡ 2 4 2
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28. (a) Acetylene is first oxidized
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XtraEdge for IIT-JEE 105 FEBRUARY 2
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